7 Adams W.W., Loustaunau P.-an Introduction to Groebner Bases (Graduate Studies in Mathematics, Vol 3)-American Mathematical Society(1994)

An Introduction to Grôbner BasesWilliam W. Adams Philippe Loustaunau Graduate Studies in Mathematics Volume 3 Editorial Board James E. Humphreys Robion C. Kirby Lance W. Small 2000 Mathematics Subject Classification. Primary 13PIO. ABSTRACT. Grobner bases are the priroary tocl for doing explicit computations in polynomial rings in many variables. In this book we give a leisurely introduction te the subject and its applications suitable for students with a little knowledge of abstract and linear algebra. The book contains not only the theory over fields, but also, the theory in modules and over rings. Library of Congress Cataloging-in-PubIication Data Adams, William W., 1937": An introduction to Grôbner bases/William W. Adams, Philippe Loustaunau. p. cm. ~(Graduate studies in mathematics, ISSN 1065-'7339; 3) Includes bibliographical references and index. ISBN (l.8218-3804-0 IL Title. III. Series. 1. Grobner bases. I. Loustaunau, Philippe, 1958QA25l.3.A32 1994 512' .4-dc20 94-19081 CIP Copying and reprinting. Individual readers of this publication, and non profit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any materiai in this publication (including abstracts) is permitted only under license from the American Mathematical Society. Requests for sueh permission should be addressed to the Assistant to the Publîsher, American Mathematical Society, P.O. Box 6248, Providence, Rhode Island 02940-6248. Requests can aIse be made by e-mail ta reprint-permission«lams. org. © Copyright 1994 by the American Mathematical Society. AlI rights reserved. Reprinted with corrections in 1996. The American Mathematical Society retains ail rights except those granted to the United States Government. Printed in the United States of America. The paper used in tbis book is acid-free and falls within the guidelines established to ensure permanence and durability. This publication was typeset by the authors, with assistance from the American Mathematical Society, using A,A.1S- TEX, the American Mathematical Society's TEX macro system. 109876543 04 03 02 01 00 § To my wife Elizabeth and our daughters Ruth and Sarah WWA To my wife Yvonne and our children Eileen and Gareth PL Contents Preface Chapter 1. Basic Theory of Grübner Bases 1.1. Introduction 1. 2. The Linear Case 1.3. The One Variable Case 1.4. Term Orders 1.5. Division Algorithm 1.6. Grübner Bases 1.7. S-Polynomials and Buchberger's Algorithm 1.8. Reduced Grübner Bases 1. 9. Summary Chapter 2. Applications of Grübner Bases 2.1. Elementary Applications of Grübner Bases 2.2. Hilbert Nullstellensatz 2.3. Elimination 2.4. Polynomial Maps 2.5. Sorne Applications to Algebraic Geometry 2.6. Minimal Polynomials of Elements in Field Extensions 2.7. The 3-Color Problem 2.8. Integer Programming Chapter 3. Modules and Grübner Bases 3.1. Modules 3.2. Grübner Bases and Syzygies 3.3. Improvements on Buchberger's Algorithm 3.4. Computation of the Syzygy Module 3.5. Grübner Bases for Modules 3.6. Elementary Applications of Grübner Bases for Modules 3.7. Syzygies for Modules 3.8. Applications of Syzygies 3.9. Computation of Hom vii ix 1 1 7 JO 18 25 32 39 46 50 53 53 61 69 79 90 97 102 105 ll3 ll3 ll8 124 134 140 152 161 171 183 viii CONTENTS 3.10. Free Resolutions Chapter 4. Griibner Bases over Rings 4.1. Basic Definitions 4.2. Computing Griibner Bases over Rings 4.3. Applications of Griibner Bases over Rings 4.4. A Primality Test 4.5. Griibner Bases over Principal Ideal Domains 4.6. Primary Decomposition in Rix] for R a PID Appendix A. Computations and Algorithms Appendix B. References Well~ordering 194 201 202 212 225 237 246 259 275 277 279 and Induction List of Symbols Index 283 285 applied mathematics. Grôbner hasis theory is generating increasing interest because of its usefulness in proix . The book can be read without ever "computing" anything. even these ideas are reviewed in the text. Moreover. in particular in Chapter 4) together with the ideals in this ring and the concepts of a quotient ring and of a vector space introduced at the level of an undergraduate abstract and linear algebra course. The book does not assume an extensive knowledge of algebra. and engineering who have sorne acquaintance with modern algebra. Indeed. one of the attributes of this subject is that it is very accessible. This book is also suitable for students of computer science. aU that is required is the notion of the ring of polynomials in several variables (and rings in general in a few places.Preface We wrote this book with two goals in mind: (i) To give a leisurely and fairly comprehensive introduction to the definition and construction of Grobner bases. This is why Griibner basis theory has become a major research area in computational algebra and computer science. Indeed. The theory stands by itself and has important theoretical applications in its own right. most of this material i8 reviewed and basic theorems are stated without proofs. Sorne topies in the later sections of Chapters 2. or get insight into. 3. This is always clearly stated at the beginning of the section and references are given. But the real point here is that cnmputing is the very essence of the subject. the methods introduced in the book without actually doing sorne of the computations in the examples and the exercises by hand or. Except for linear algebra. Computing is useful in producing and analyzing exarnples which illustrate a concept already understood. This book is designed to be a fust course in the theory of Griibner bases suitable for an advanced undergraduate or a beginning graduate student. and 4 require more advanced material. or which one hopes will give insight into a less weIl understood idea or technique. In fact. more often. However. (ii) To discuss applications of Griibner bases by presenting computational methods to solve problems which involve rings of polynomials. using a Computer Algebra System (there are over 120 worked-out examples and over 200 exercises). the reader will not fully appreciate the power of. engineering. Weispfenning [BeWe] and by B. . In the polynomial ring k[x]. the ideas behind the abstract characterization of Gr6bner bases had been around before Buchberger's work. . where k is a field. primarily in Section 3.. . one can compute (using the Euclidean Aigorithm) a single polynomial d = gCd(f"". The basic idea behind the theory can be described as a generalization of the theory of polynomials in one variable.. implementation of algorithms and their complexity are discussed only briefly in the book. Becker and V. This abstract characterization of Grübner bases is only one side of the theory. Then a polynomial JE k[x] is in J if and only if the rernainder of the division of J by d is zero. namely the greatest comrnon divisor of the elements of J. is his algorithm for computing these bases.3. As is the case for rnost topies in commutative algebra. But the true signmcance of Grübner bases is the fact that they can be cornputed. and what gave Grobner basis theory the status as a subject in its own right. For those interested in a geometric treatment of sorne of the theory we recommend the excellent book by D. We have kept our presentation algebraic except in Sections 1. Macaulay [Mac] used sorne of these ideas at the beginning of the century to determine certain invariants of ideals in polynomial rings and Hironaka [Hi]. Bruno Buchberger's great contribution.) = (d). Our chaice of tapies is designed ta give a braad introduction ta the elementary aspects and applications of the subject. In fact ·jt falls far short of the true significance of Grübner bases and of the real contribution of Bruno Buchberger. The interested reader should consult the references. We mention in particular the books by T. Grübner bases are the analog of greatest cornmon divisars in the multivariate case in the following sense.. Indeed. used similar ideas to study power series rings. Although this book is about computations in algebra. Mishra [Mi] which contain a lot of material not in this book and have extensive lists of references on the subject.5. O'Shea [CLOS]. Given any set of generators {f" .. sorne of the issues which might be of interest to computer scientists are outside the scope of this book. . The reader who is interested in going beyond the contents of this book should use our list of references as a way to access other sources.xn ] is in J if and only if the remainder of the division of J by the polynomials in the Grübner basis is zero (the appropriate concept of division is a central aspect of the theory).J. Little and D.xn ] generates J and a polynomial J E k[XI'" .} <:: k[x] for J.1 and 2. and computer science.J.. For example. .. any ideal J can be generated by a single element. . l Wolfgang Grôbner was Bruno Buchberger's thesis advisor. in 1964. Cox. Grübner bases were introduced in 1965 by Bruno Buchberger ' [Bu65].) such that J = (f" .J. J. A Grübner basis for an ideal J <:: k[XI' . Gr6bner basis theory can be presented from a geometric point of view. For example. science.x PREFACE viding computational tools which are applicable to a wide range of problems in mathematics. Elizabeth Rutman. Perpetua Kessy. In Chapter 4 we develop the theory of Grübner bases for polynomial rings when the. Lyn Miller. Garry Helzer. We are careful to give motivations for the definition and algorithm by giving the familiar examples of Gaussian elimination for !inear polynomials and the Euclidean Aigorithm for polynomials in one variable. ltaly. In particular we would like ta thank Beth Arnold. Karen Horn. This book grew out of a series of lectures presented by the fust author at the National Security Agency during the summer of 1991 and by the second author at the University of Calabria. We then develop the theory of Grübner bases for finitely generated modules over polynomial rings. In Chapter 3 we begin by using the concept of syzygy modules to give an improve~ ment of Buchberger's Aigorithm. We go on to show how to use Grübner bases to compute the syzygy module of a set of polynomials (this is solving diophantine equations over polynomial rings). With these. during the summer of 1993. and Eric York. graph theory. We give an outline of the section dependencies at the end of the Preface. coefficients are now allowed to be in a general Noetherian ring and we show how to compute these bases (given certain computability conditions on the coefficient ring). We show how the theory simplifies when the coefficient ring is a principal ideal domain. There are exercises at the end of each section. Other exercises extend the theory presented in the book. sorne doable by hand while others require the use of a Computer Algebra System. Ann Boyle. We would like to thank many of our colleagues and students for their helpful comments and suggestions. A few harder exercises are marked with (*). We close the chapter with three specialized applications ta algebra. . Many of these exercises are computational in nature. In Chapter 2 we present the basic applications to algebra and elementary algebraic geometry. Alyson Reeves. and conclude by showing how to compute the Hom functor and free resolutions. Brian Williams. We also give applications ta determining whether an ideal is prime and to COIDputing the primary decomposition of ideals in polynomial rings in one variable over principal ideal domains. Julie Hawks and the AMS staff for their help in the prepa~ ration of the manuscript. give more efficient methods for com~ puting sorne of the objects from the previous chapter. After Chapter 1 the reader has many options in continuing with the rest of the book. and integer programming. We also want to thank Sam Rankin. we extend the applications frOID the previous chapter.PREFACE xi In Chapter 1 we give the basic introduction to the concept of a Grübner basis and show how to compute it using Buchberger's Aigorithm. xii PREFACE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 3 . PREFACE xiii Chapter 4 . . .. .8 and summarize what we have done in Section 1..xn ] denote the set of ail polyuomials in n variables 2 with coefficients in the field k.. 1I'n. and l f3i E M. We refine the definition of a Grübner basis in an important way in Section 1. zJ. x~n.3y2 + !xz is a polynomial ln Q[x. .. IR..xn ] we have the usual operations of addition and multiplication of polynomials. In Section 1. ({. For example. Such polynomials are finite surns of terms of the farm ax~l .. We consider polynomials f(X1' . For example.. that is. 7 we present the algorithm due to Bruno Buchberger which transformed the abstract notion of a Grobner basis into a fundamental tool in computational algebra. 1 . . In the next two sections we motivate the method of solution of these problems by presenting the method of solution in two familiar special cases. k[Xl. and with respect ta these operations k[xI..xn ] is a k-vector space with basis the set. The basic method in both cases is to use the leading term of one polynomial to subtract off a term in another polynomial... the real numbers. y.. We let klx1. .2 and 1. where a E k. Let k be any field (e. lWe denote by N the set of non-negative integers. _ .)J. two... 'from now on... ..i = 1. of all power products. Also.n}.f = x 2 + y2 . namely the notion of a Grübner basis. C). and the division algorithm for polynomials in one variable. 1. y. f = XI+x~ -1 and 9 = Xl -3X~+~XIX3 are polynomials in three variables.xn ) in n variables with coefficients in k.9. . Basic Theory of Gri:ibner Bases In this chapter we give a leisurely introduction to the theory of Grübner bases.'.3. Introduction.1. Note that in klx1'.n. x~n a power product. N = {a.Chapter 1. 1rn = {xf' ···x~n 1 fli E N. or the complexnumbers.g.5 we go on to generalize the ideas in Sections 1. In Section 1. In the fust section we introduce the reader to the kinds of problems we will be interested in throughout this book. . namely the TQW reduction of matrices of systems of linear polynomials.2. 2Most of the time.xn ] is a commutative ring. 1. .6 to defining the central notion in this book. }. In Section 1. . . but instead will use the variables x. . we will not use variables with subscripts. . the rational numbers.. We caIl X~l .. or three variables. i = 1. whenever we work with just one. yJ and 9 = x .4 we introduce what we mean by the leading term of a polynomial in n variables. or z as needed.1 is a polynomial in Q[x.3. This leads us in Section 1. .3y2) ç IR2 is the intersection of the circle x2 + y2 = 1 and the parabola x = 3y2 in the xy-plane. .Is = o. . . . s}.2 CHAPTER 1. We thus have two ways of viewing a polynomial 1 E k[Xl. l. .. For 1 E k[xj. and the computation of the solutions can drastically improve if the given system of equations is transformed inta a different system that has the same solutions but is "easier" ta solve. . These algorithms solve for one solution at a time.1).1) That is. if k = IR...an). if S ç k[Xl' . Indeed... VU) = {(al.. Xn]. and find an "approximation" ta the solution. . ..an) Ek n 1 I(a" .an) E k n 1 I(a" . l. fI = 0.. . however this neecl not concern us here).) = n:~l VUi). n}.0) and radius 1... . Ta illustrate this.an) ---. This function is called evaluation. an) = 0. an) 1 ai E k.I(a" . Is) = {(a" .an) Ek n . . There are many numeric algorithms for solving non-linear systems such as (1. More formally. but is easier ta solve.1). . . for ail (al. .1.. . recall that the Gauss-Jordan elimination method transforms a system of linear equations inta the so-called row echelon form (see Section 1. x .2). then k = IRn is the usual Euclidean n-space.2.. . The system thus obtained has exactly the same solutions as the original system.. .. .(a" ... x n ] we define VU) ta be the set of solutions of the equation 1 = O. .. More generally. More generally still. we define V(S) = {(a" .. h = 0.) is defined to be the set of ail solutions of the system (1.. For example.. .) A polynomial 1 E k[Xl.. . One is as a formaI polynomial in k[Xl.. xn]. a variety can be the solution set of a number of systems sueh as (1.. V(X2 + y2 .. .. . . .... an) E k n 1 j. this example will be discussed . the variety VU" . x n ] determines a function k n ---. . as we will see below.. .. VU) is ealled the variety defined by f.. VU" . For example. BASIC THEORY OF GROBNER BASES For a positive integer n we define the affine n-space kn = {(a" ...1. and do not take into consideration possible alternate descriptions of the variety (using a different system). ...Is E k[Xl. i = 1. given fI. . (For example. .· . Xn].. They ignore the geometric properties of the solution space (the variety). . the variety V(x2 + y2 1.1.. . Note that VU" . . This "double identity" of polynomials is the bridge between algebra and geometry..1) ç IR2 is the eircle in the xy-plane with center (0.· .an) = O} ç kn. i n = 1. . k defined by (al. Xn ] and the other is as a function k n ---.. . k (it should be noted that if k happens to be a finite field then two different polynomials ean give rise ta the same function. .an) =Ofor alliE S}.. ).J.J. The remainder of this chapter is devoted to finding this "better" generating set for I (which will be called a Grübner basis for I).. . This means that the system ft = 0.. 118)' And by "better" we mean a set of generators that allows us to understand the algebraic structure of I = (f" .1. ° .m... . that is. .. Thus we have that V(I) = V(J" ... .. . and contrast it with the solutions of the finite system (1.y2. h = 0. then hj E I...j. .) is an ideal in k[X1"" . i = l. the solutions of the infinite system of polynomial equations (1..1..s. since fi E l for i = 1. .'" ...1.js) and the geometric structure of V (f" .. we consider the variety V(I). Conversely.. then V(f" . . We will now look at the problem from a different perspective..2) j = O..··· . . .y) = (x + xy. ..3). and hence a variety is determined by an ideal./s): (J" . Sa... .3) ft = 0.1. The set {J" ..j. . . l A solution of System (1. . and if j is any element of I.) = V(I) = V(f{.s} .. . ) ff = D.. if we have I = (J" .j.ls).js = O. .1.j.1..js) will be a better generating set for the ideal I = (J" . We will develop an analogous procedure for System (1..y]. if (ab'" an) E kn is a solution of System (1.j E I.. then j(a" . The desired "better" representation for the variety V(J" .3). This will be done by considering the ideal generated by polynomials ft. For exampIe..j" denoted (J" . .. then so is j + 9 and if j E I and h is any polynomial in k[X1'" . we will have a "better" representation for the variety V(h.js} is called a generating set of the ideal I. .Xn].Xn]..1.. Hence (a" .. .1) which will give us algebraic and geometric information about the entire solution space of System (1.. in k[x. for sorne Ui E k[X1"" . In the case of linear polynomials this "better" generating set is the one obtained from the row echelon form of the matrix of the system. if j. . ~hat is. if we obtain a "better" generating set for the ideal l = (!I. The method for obtaining this information is to find a better representation for the corresponding variety. (x + y.an) is a solution of System (1.y + xy).).1.. Now. ..j. We note that an ideal may have many different generating sets with different numbers of elements. V. ..an) = 0. . not by a particular set of equations. It is easy to check that I = (J" ..X n].) = (f{.. INTRODUCTION 3 more thoroughly in the next section.. Consider a collection.1.X n ]. since f = E:~l Udi. .2). To see how this might help.1)... .js = has the same solutions as the system f{ = 0.. 9 E I.1.m· .x 2. We define the set I(V) of polynomials .) = {t ~=1 Udi 1 Ui E k[X1.x) = (x. . of points of the affine space k n .) better..2) will clearly be a solution of System (1. .. .lB) for sorne h.xn]. . it is desirable to solve the following problems: given J = (f"". lB E k[Xl. given h.. then the solution set of the infinite system j" = 0. an) E V We will examine the exact relationship between these two descriptions later. . Understanding the relationship between these two maps allows us to go back and forth between algebraic and geometric questions.1) states that any ideal J in k[Xl. in addition to the map { Subsets of k[Xl"" . E k[Xl"" . . The latter ideal is defined algebraically as the set of alllinear combinations of h. More specifically..an) = 0 for all (a" . For more on the relationship between J and J(V(I)).lB) = (fi.. .. ID· For this reason and many others.. .. with polynomial coefficients. .. an) EV}. ----> >--> { Ideals of k[Xl.. It is the bridge from geometry back to algebra sinee.. .. For now. . but they are not in J. l.. of a finite generating set for the ideal (j" 1 .. . It is easy to see that J ç J (V(J)) . we will be interested in the exact relationship between the ideal J and the ideal J (V(J)).). . . -. In order to find the "better" generating set discussed abov€. Indeed. Xn].\ E A.'" ." _. Xn]} V J{V). we will need to determine whether two finite sets of polynomials in k[Xl.xn]} ----> >--> { Varieties of k n } S we now have a map { Subsets of k n } V(S).. and so x and y are in the ideal J (V(I)). and if. y2) ç k[x. We will prove the Hilbert Basis Theorem at the end of the section.4 CHAPTER 1.Xn] by J{V) = {f E k[Xl.\ E A we have a polynomial lÀ E k[Xl.l. Another consequence of this result is that if A is an infinite set and for all . In particular.O)}. but equality does not always hold.. . . .\ E A is. Xn]....Xn].. . xn] (in particular the ideal J(V)) has a finite generating set. . if J = (X2. . in fact. It is easy to verify that the set J(V) is an ideal in k[Xl"" . .. we will need to determine whether (f" . with polynomial coefficients). the solution set of a finite system. BASIC THEORY OF GROBNER BASES in k[Xl"" . see Section 2. the Hilbert Basis Theorem (Theorem 1. Xn]. . It would seern that this ideal is very different from the ideal (f" ... while the former ideal is defined by the geometric condition that 1 is in J{V) if and only if I(a" . namely.··· . .'" . we note that the ideal J(V) can be put in the form (h.i.xn] give rise to the same ideal.\ E A) (this ideal is defined to be the set of ail finite linear combinations of the j". For example..1. xn]1 I(a" .x n ]. y]. .2. then V(I) = {(O. an) = 0 for ail (a" . lis E k[Xl1'" .'" . The construction of the ideal J(V) above is a very important one. . lB) and 1 E k[Xl"" .. . . We now turn our attention to the Hilbert Basis Theorem.xnJ. when are the corresponding evaluation functions V(I) --. if J . . . We would like to answer the following question: for J. as pointed out above.. We saw earlier that J determines an evaluation function k n . .1. . . Observe that ". It guarantees the termination of our algorithms and also. k equal? We note that this is related to the ideal l (V(I)) introduced earlier.. . .. .1 (HILBERT BASIS THEOREM).g E J. . c:.. .gis identically zero. + u2h + . ." PROBLEM 2. In connection with this construction. PROBLEM 4.an) E V(I).. ..xnJ. and let JE k[Xl' . . and hence the evaluation functions V (I) ----> k determined by J and gare equa!. h c:. we would like to solve the following problems: PROBLEM 3..j. c:. it guarantees that every variety is the solution set of a finite set of polynomials. . The discussion above is related to another problem that deals with a certain algebraic construction.. We denote the set of equivalence classes by k[Xl' . THEOREM l. . Determine a set of coset representatives of k[Xl' . ... . This result is crucial in everything we will be doing throughout this book.. then the evaluation function V(I) ----> k defined by f ... INTRODUCTION 5 PROBLEM 1. Determine a basis for k[Xl' .. + u.x n ].xnJ/ J as a vector space over k (which may or may not be finite).. . k[Xl' .. J..xnJ. Before we go on to the proof we would like to make a couple of definitions.).xnJ such that l = (f" . gin k[Xl. . Let l be an ideal of k[Xl. k defined by (aIl'" .. .. 12 c:. . In c:. the word "determine" is informally understood ta mean that one can give an algorithm that can be programmed on a computer.... In the ring k[Xl' .. . Determine whether J is in 1.··· .. (ii) If I. E k[Xl' ..1... we say that fis congruent to g modulo J. and an ideal J of k[Xl. then there exist polynomials Ir.xnl/ J are ofthe form f + J and are called cosets of J.xnJ/ J.::::" is 311 equivalence relation on k[Xb ... E k[Xl' . .. Indeed. Elements of k[Xl.. An ideal l in a general ring R which satisfies Condition (i) is said to be finitely .f. determine Ul. . .xnJ and is called the quotient ring of k[Xl' ..l.U. . . If J E l.. .xnJ by J.· REMARK: In this book. It is also a vector space over k.xnJ.. f(a" .. This is the so-called "ideal membership problem...... if f .an)' We now consider the restriction of this function to V(I). xnJ. . denoted f == 9 (mod J).an) --.an) 1---7 f(al... is an ascending chain of idealB of k[xj.xnl/ J.. that is. we consider the evaluation fonction V(I) --... ...X n ]. Also.. ... k defined by (a" .xnJ such that f = ud... then there exists N such that IN = IN+l = I N +2 = .xnJ/ J is a commutative ring with the usual operations of addition and multiplication inherited from k[Xl..gis in the ideal I(V(I)).Xn ]. .an) for all (a" . .. .xnJ we have the Jollowing: (i) If lis any ideal of k[Xl' .· ... Recall that given J and 9 in k[Xl' . .. ..1. Then there exists h E l with 12 r. ç: In ç: . for sorne fI. 1= (fI. assume ta the contrary that there exists an ideal l of R that is not generated by a finite set of elements of R. .J.6 CHAPTER 1.··· . then the other also holds. and sa l ç: IN. Also. say Ii = (ril.... For each n 2: 0. be an ascending chain of ideals of R. Since for i = 1. it is enough ta show that J is finitely generated. THEOREM 1. Condition (ii) is sometimes referred ta as the Ascending Chain Condition.· ... r is the coefficient of xn). there exists Ni such that fi E IN. . and any commutative ring R satisfying that condition is called a Noetherian ring.rit.f. BASIC THEORY OF GROBNER BASES generated. The reader may skip the proof and proceed directly ta the next section. then sa is Rlxl. PROOF.Since the ideals In are increasing. It turns out that if either of the two conditions in Theorem 1. ç: In ç: .'~I In.. By Condition (i).).. . . this is the content of the next theorem. For the reverse implication. it is easy ta see that 1 is an ideal of R. is an ascending chain of ideals of R. Let N = maxlSiS' Ni.1. . Let us first assume Condition (i). then fi E IN for ail i = 1. THEOREM 1.. If R is a Noetherian ring.t i1 let /ij be a . and Condition (ii) follows. The remainder of this section is devoted ta the proof of Theorem 1. .1. .2.. ..3. 12)· We continue in this fashion. The following conditions are equivalent for a commutative ring R: (i) If l is any ideal of R.t (fI)· Thus (h) ç (fI. ç: h ç: .1 holds. In the next two sections we will illustrate the discussion of this section using two examples: !inear systems and polynomials in one variable.).. Let fI E 1. E R such that 1= (fI. Let R be a Noetherian ring.. each Ii is finitely generated. ..). fi is in l.1. . . then there €Xists N such that IN = IN+I = IN+2 = . by Theorem 1. then there €Xist elements fI. for ail n 2: O.N and j = 1. These will be fundamental motivations for the general constructions we will develop in the remainder of this chapter. . (ii) If h ç: h ç: h ç: ..2. 0 We now state and prave a more general version of the Hilbert Basis Theorem. That is.. . .. S. Thus 1= IN..1.. PROOF. or ta have a finite generating set. By Theorem 1.. It is easy ta see that In is an ideal of Rand that In ç: In+ 1. the ring R is Noetherian if and only if every ideal in R has a finite generating set.1. and we get a strietly ascending chain of ideals of R which contradicts Condition (ii).. there exists N such that In = IN for ail n 2: N. and let J be an ideal of Rlxl. f.8. Now for i = 1. Consider the set l = U:. Since R is Noetherian. fs E R.1.2. define In = {r E R 1 r is the leading coefficient of a polynomial in J of degree n } U {O} (that is. and let h ç: 1.. ... we have r = L:~::1 SjTnjl for sorne Sj E R.3) X { + y y - z o O.2!J = Y + 4z. the solutions to the system (1.2) + + y z o O. If n > N.1) fI = 0. D 1 Using a simple induction on n and the above result. h = 12 .xnJ is Noetherian (first noting that the field k is trivially Noetherian). .9 has degree at most n . Then the polynomial 9 = ~~:j sjlnj is of degree n. Theorem 1. Let!J = x+y-z and 12 = 2x+3y+2z be linear polynomials in !R[x.1. we can easily show that k[xj. That is. let 1 E J.fs = 0.1 and. that is. 1. zJ. We prove by induction on n that 1 E J'. 12). in fact. and hence f is also in J*. THE LINEAR CASE 7 polynomial in J of degree i with leading coefficient Tij' To complete the proof of the theorem it suffices to show that J = (fij Il <:: i <:: N. 1 1 3y + 2z We now perform row reduction on the matrix associated with this system: -1 ] 4 .1 are in J'.9 has degree at most n . a method ta change a generating set for the ideal l = (!J.2. Thus 1 . We consider the ideal l = (!J.2. Consider the following examples. then 1 E ID. by induction. 1 . This amounts to creating a new polynomial..2. + 4z and are easily obtained parametrically as: x = 5z and y = -4z. EXAMPLE 1.2.1 is the well-known row reduction which changes System (l. In tllls section we consider the system (1.2. y. 12) and the variety V(fj. Clearly J* c:: J. By induction..2.1 is true. f . and is in J*. where each fi is linear. Conversely. In this case. and replacing 12 . If 1 = 0 or n = 0. The Linear Case. We subtracted twice the first row from the second row and replaced the second row by this new row. then. So let J* = (fij 1 <:: i <:: N.9 E J*. and assume that all the elements of J of degree at most n . Now let n > 0. Therefore 1 is in J*. the algorithmic method to answer ail the questions raised in Section 1. The last matrix is in row echelon form. and r = L~~l SjTNj.1) to row echelon form.9 is in J*.2. . If n S N.N iNj has degree n. The solutions of System (l. The row reduction process is. then rEIn = IN..1. and hence 1 E J*.1. for sorne Sj E R.2. Thus 1 . and let the degree of 1 be n. Let T be the leading coefficient of f.l <:: j <:: t i ).. The polynomial 9 = L~~l SjX n . . since rEIn.l <:: j <:: ti). 12) into another generating set.1 and is in J.2) are the same as those of the following system (1. and is in J*. has leading coefficient T. leading coefficient T. . This ean be written in long division form 2 x +y - z 1 2x + 3y + 2z 2x + 2y .2.2. y. and sinee 12 = 2ft + 13 we see that 12 E (ft. 12 = x + 2y + 3z. 13) and the variety V (f. 17z. The row reduction is as follows: [~ (1.7z -------Jo - j.z. as in the above. 12). sinee 13 = h-2ft we see that 13 E l = (ft. namely 2x. to eliminate a term from 12. that is. . and we write h~h The new polynomial !3 that was created can be vi€wed as a remainder of a certain division: we used the first term of h. the division stops and the remainder is exactly h. Repeated use of the reduetion steps. The praeess by whieh the polyuomial 12 was replaced by 13 using ft is ealled reduction of 12 by ft. similar to that in Example 1. that is. ] [0 1 -1 1 2 3 .. will be denoted by j"f. it makes it easy to determine V (I). 13). the solutions to the system (1. 13) is {ft. -lOy . .2). f.4) { 3~ ----.8 CHAPTER .5) 1 2 -4 -~ ] [r -~ -10 -7 ] ----. o 0 -17 This says that a new generating set for l = (ft.2. 7 3 -------Jo+ -1 z.h) and so l = (f" 12) = (f" 13)· This praeess simplifies the generating set of the ideal l and allows for an easy resolution of System (1. BASIC THEORY OF GROBNER BASES by h The original ideal l is equal to the ideal (f" 13)· lndeed. h. Since this first term of h cannot eliminate any other terrns.2.2.-17z}.1. + - Y 2y 4y + + 1 2 z 3z 2z = 0 = 0 = O. of h by 12 and ft in succession. z]. namely X. f This amounts to a division. Let ft = y .2z y+4z whlch gives us 12 = 2ft + h When the system has more than tWQ equations. 12.4y + 2z be linear polynomials in lQI[x.2.h. and 13 = 3x . Note that the polyuomial -17z is obtained by the following reduetions: f3 -----j. EXAMPLE 1. the division (or reduction) of a polynomial may require more than one polynomial. We eonsider the ideal l = (f" h. Also. First. used in the· second reduction in (1. but would have eliminated z first. We note that in OUT example the order is such that x is first followed by y and then z and so the leading term of f. y.2. We will concentrate on Example 1.2. -4. the reduetions in (1.2. if SO. Second. = -z + y. + 3fz -17z. when we row-reduce a matrix there iB an order on how to proceed ta introduce zeros: first we introduce zeros into the first column (that is.1. We would like ta "extract" from these exarnples sorne general ingredients that will be used in the general situation of non-linear polynomials.2. any polynomial Jean be reduced to a polynomial in z alone by the division process using both fI and 13 in a way similar to that used in (1. 13) = P.5). We would have used the same row reduction method.2. is y. and then we introduce zeros in the second column (we eliminate y) etc.7 z leaving the remainder of -17 z and giving us Equation (1. The process of row reduction vi€wed in this light gives us a way ta solve the problems posed in Section 1. This had the effect of using the leading terms of fI and h to eliminate terms in 13 and in -lOy .4) and then we chose to eliminate y from the new third equation. Let us concentrate on Example 1.5) and the coefficient "-10" of f. the leading term of fz is x. 12) = V(fI. but they would have been just as good for our purpose of solving System (1.6) 13 = -lOf.2. We would have wound up with a different set of equations in fQW echelon form.4y + 3x. because the leading term of fI is x and the leading term of 13 is y.6) (the remainder is in z alone).2. but there must be an order. THE LINEAR CASE 9 Note that we have (1. and the leading term of 13 is 3x. 1) 1 À E IR}. we imposed an order on the variables: we chose ta eliminate x first from the third equation of (1.1. we have a very clear description of the solution space: V(I) = V (f. That is. We next turn to the question of determining whether a polynomial f E k[x.5) were obtained by subtracting multiples of fI and fz. So the order does not matter.2.2. any polynomial in z alone cannot be reduced using division by fI and h. say f.(5. we eliminate x). This issue becomes essential in our generalization of these ideas.6).1. z] is in l and.5). First. . fz = 3z + 2y + x and J" = 2z . Note that -17z cannot be reduced further using the leading terms of fI and 12. In Qur case. The division process allows us to write J as a linear combination of fI and 13 plus a remainder in a similar fashion to Equation (1.2.2.2.2. The coefficient "3" of h is the multiple of fz used in the first reduction in (1. We could have written the variables in the polynomials in a different order. express it as a linear combination of the elements in the generating set. it is a line in ]R3. It is not too hard to see that J E l = (f" 12) = (f" 13) if .4). then y. is the multiple of f. 1. Prove the last statement made about Example 1./s) = (g1"" .. For 0 of j E k[x]. The theory of polynomials in one variable is a good illustration of the more general theory that will be presented in the remainder of this chapter.3.zl/J Îs the set of all cosets of powers of z.2. if f = anx n + an_lX n .an E k and an of 0. then y. is the largest exponent of x that appears in f.gt ta obtain a basis of the k-vector space k[X1. denoted !c(f)... The leading term of j. Finally one could also check that the basis of the vector space k[x. Following what was done for Example 1. as in Example 1. . solve the problems posed in Section 1.3.. BASIC THEORY OF GROBNER BASES and only if this remainder is zero.2. lt(f) = anxn and le(f) = an. Let j = x 3 ..1 in the last paragraph of the section. Let B be a row echelon form for the matrix A and assume that B has t non-zero IOWS.l + . EXAMPLE 1. .2. 1. Let 91.y . h = x + y + 3z + t. tl. ) x n ] corresponding ta the IOWS of A.2.2.1.4. In this context we will use the wellknown Euclidean Algorithm to solve the problems mentioned in Section 1..gt be the polynomials corresponding ta the non-zero rows of B. y. + alX + ao.1 for this set of polynomials.js be the linear polynomials in k[Xl" . fs = 2x . In this section we consider polynomials in k[x]. with aD. . 3i . Sa. then deg(f) = n.2.1. The statements made in this paragraph may be a little diflicult to verif'y or appreciate at this point but will become clear later. denoted deg(f).xnl/(f" . 1.js). The main tool in the Euclidean Algorithm is the Division Algorithm (also known as long division of polynomials) which we illustrate in the next example.'" ..1. Let f" .2. .2X2 + 2x + 8..10 CHAPTER 1. Let A be an s x n matrix with entries in a field k. The leading coefficient of j... denoted lt(f). f.2.t. We divide j by 9 to get the quotient ~x and the remainder x+ as follows: i 2. zl/ J is the set of all cosets of powers of z. = x . y. Assume that we now eliminate z first. Use the polynomials gl. .2.2. polynomials in one variable. ExercÏses 1.'" . then x. that is. What is the row echelon form of the matrix? Use this ta give another basis for the vector space lQI[x. The One Variable Case. zlj J. is the term of j with highest degree.2. x second and z last. the problems posed in Section 1.2y + z + t. Prove that (f1.y. we recall that the degree of j. and j4 = 2x + 2y + z + t. Repeat this eliminating y first. Consider the following polynomials in lQI[x. is the coefficient in the leading term of f.z . y.. 1. prove that a basis of the vector space lQI[x.. In doing this we will present sorne of the standard material concerning k[xl but will present this material using notation that will be more immediately generalizable to the study of polynomials in many variables.1. . z. Solve .gt). . . and 9 = 2x 2 + 3x + 1 be in lQI[xl. Namely.3..1 for Example 1. !!..1. In general if we have Iwo polynomials f = anXn+an_lXn-l + . with r = 0 or deg(r) < deg(g). -+b1x+bo.. Using the notation introduced above. Repeated use of reduction steps.3. we note Ihal the factor of 9 in this product is and so we get h = f . in the reduction f -. Moreover rand q are unique (q is caUed the quotient and r the remainder).. The idea was to multiply 9 by an appropriate term) namely ~x... Going back to Example 1.Ux .+ r. Then for any f E k[x]. THEOREM 1.i)g + (2fx + 3i)· Let us analyze the steps in the above division.. We first multiplied 9 by ~x and subtracted the resulting product from f. as in the above. f~h~r.!!.. the polynomial h has degree strictly less than the degree of f. When we continue this process the degree keeps going down until the degree is less than the degree of g.2. h. 80 that the leading term of 9 times this term canceled the leading term of f. Let 9 be a non-zero polynomial in k[x].1. will be denoted f -. We note that.~xg = _~X2 + ~x + 8. We call h a reduction of f by 9 and the process of computing h is denoled 4 + . after this first cancellation we repealed the rocess on h = _lx2 + ~x +8 by subtracting lt(h)g = _lx2 . Thus we have the first half of the following standard Iheorem.lli 4 ::f.. +alx+aO and 9 = bmxm+bm_lXm-l + . After this first cancellation we obtained the fusl remainder h = f .l from h to P 2 2 It(g) 2 4 4 oblain the second (and in tbis example the final) remainder r = 2fx + This can be WTitten using our reduction notation "t.. there exist q and r in k[x] such that f = qg + r. . THE ONE VARIABLE CASE 11 2x 2 +3x+l x 3 -2x 2 +2x+8 _IX2 2 _~X2 + ~x+ 8 2 - ~X 27 X - ± and so we have f = (~x .3.. h.3.::~~l 9 as the first remainder. with n = deg(f) ?: m = deg(g). then the first step in the division of f by gis to subtract frOID f the product ~xn-mg.!!.l f -. Using this idea repeatedly (that is. We will repeat Example 1. INITIALIZATION: q := 0.4" (_~x2 + ~x + 8) . r := J = x 3 .3).3. Now let l = (f.1. sinee h = J . 0 Observe that the outline of the proof of Theorem 1. so we can replace J by h in the generating set of J.3.. As mentioned above this is denoted J -"..(2x'+3x+l) = -~x'+~x+8 Second pass through the WHILE loop: q := 2: X ' r := x+3 d The WHILE loop stops since deg(r) = 1 < 2 = deg(g). BASIC THEORY OF GROBNER BASES PROOF.3.1.g E k[x] with 9 OUTPUT: q.+ 2X2 .3. Ilsing Theorem 1.1..3. The proof of the uniqueness of q and r is an easy exercise (Exercise 1.Xx'. + --z?. which we present as Algorithm 1.1.3..2.3.t~' (2x' + 3x + 1) = 1 _l-x 2 1 7 '1 .2 repcatedly) we cau prove the followillg result._ 0 x3 _ 1 q. EXAMPLE 1.3. O"e Variable Division Algorithm The steps in the WHILE loop in the algorithm correspond to the reduction proeess mentioned above. It is repeated until the polynomial r in the algorithm satisfies r = 0 or has degree strietly less than the degree of g. We obtain the quotient q and the remainder r as in Example 1.= 2: X . h. r := WHILE r J op 0 AND lt(r) deg(g) :S deg(r) DO q:= q+ lt(g) lt(r) r:= r . g) and suppose that J -".2"x r:= (x 3 -2x'+2x+8).g).2 gives an algorithm for eomputing q and r. This algorithm is the well-known Division Algorithm.lt(g)g ALGORITHM 1.+ r..:~i~i g.1 following Algorithm 1.12 CHAPTER 1. it is easy to see that l = (h.3. INPUT: J. This idea is similar to the one pres€llted for lillear polynomials studied in Section 1.. r sueh that op 0 qg + rand J= r = 0 or deg(r) < deg(g) INITIALIZATION: q := 0.-.2x' + 2x + 8 First pass through the WHILE loop: . The proof of the existence of q and r was outlined above.3. Then. . that J = qg +r for sorne q. .qh.3. the system of equations (1. Every ideal oJ k[x] is generated by one clement"' 13 PROOF. with one oJ JI. with one of JI.S.h) = (g). then h divides g. say J = (JI.) = (g). 9 divides both Ir and h NOW.4. with r = 0 or deg(r) < deg(g) = n. 12 E k[x]. = 0 with fi E k[x]. We now investigate how to compute the polynomial 9 of Theorem 1.'" . D Observe that the polynomial 9 in the proof of Theorem 1. 12) Jor aU q E k[x].4 says that k[x] is a PlV. For example.4 is the "best" generating set for the ideal J = (JI. Thenged(fl' 12) = gcd(fl . • !erg) = 1 (that is.!. J = qg. This follows from the fact that if J = (gl) = (g.!2 not zero. U.4 is unique up to a constant multiple. Sinee Ir.. Therefore r = 0.). If r oJ 0. 12 E (g).3. Let Ir. where (JI. there exists 9 E k[x] such that (h. i = 1.3.2. has precisely the same set of solutions as the single equation g = 0.. ged(fl.. . Equality follows from the fact that 9 is in J.J. • if h E k[x] divides JI and h. 9 is monie).3.4. E k[x] such that 9 = Uril + u2h. 12 not zero. .). .3.4.6. Let JI. denoted gcd(fl. For any J E J we have. is the polynomial 9 such that: • 9 divides bath h and 12. . 12)· The algorithm for eomputing ged's is called the Euclidean Algorithm.3.h notzero. and this contradiets the choice of g. and we are done.3. THE ONE VARlABLE CASE THEOREM 1. and l ç: (g).1.3. We reeall that the greatest eommon divis or of h and 12. 12).J.3. 12). we may assume that !erg) = 1. We will show that 9 = ged(h. Il depends on the Division Algorithm discussed above and the following facto LEMMA 1.3. We will first foeus on ideals J ç: k[x] generated by two polynomials. By Theorem 1. there exist UI.. Therefore Theorem 1. 12))· Then PROOF..3.5. by Theorem 1. hl = (gcd(lr. Sinee 9 is unique up ta a constant multiple. 3Recall that an ideal generated by one element is called a principal ideal. D As a consequence. . if we have an algorithm for finding gcd '8. then we can actually find a single generator of the ideal (JI.. let h be sueh that h divides bath h and h Since 9 is in the ideal (JI. Let 9 E J be sueh that 9 oJ 0 and n = deg(g) is least. with one oJh. 12 E k[x]. then gl divides g2 and g2 divides gl' We see that the polynomial 9 in the proof of Theorem 1. 12). or PID. 12). Let J be a non-zero ideal of k[x].1) Ir = 0. 12) exists and (JI. We further recall PROPOSITION 1. r E k[x].qg E J. then r = J . and an integral domain for which every ideal is principal is called a principal ideal domain. Thus h divides g. we have gCd(f" hl = gcd(h .7. g). consider the following EXAMPLE 1. Let h = x 3 .14 CHAPTER L BASIC THEORY OF GROBNER BASES PROOF. Therefore. (gcd(h. It is easy to see that (h.6. INPUT: h. 12 not zero OUTPUT: f = gCd(f" 12) INITIALIZATION: f:= h.1 g:= -2x+2 Second pass through the WHILE loop: x 2 _ 1 -~2 X _ 1 -~2 0 X 3 x 2 _1 f:= -2x+2 . 12). The reader should note that the algorithm terminates because the degree of r in the WHILE loop is strictly less than the degree of g. then gcd(h. INITIALIZATION: f := x 3 .3x + 2. by Proposition 1. and hence the degree of r is strictly decreasing as the algorithm progresses. by Lemma 1. The last step in the algorithm ensures that the final result has leading coefficient 1 (that is.3.2.g:= 12 WHILEgofODO f ~ + r. 12)· 0 We give the Euclidean Algorithm as Algorithm 1.O) = leC!) f. Euclidean Algari/hm To illustrate this algorithm.qh. and sinee the gcd of two polynomials is de/ined to have leading coefficient l. sinee at each pass through the WIDLE loop. as long as 9 of O.-qh.h)) = (J"h) = (J. Also.5.gh.3.- 1 !c(!) f ALGORITHM 1. is monic). the algorithm does give gCd(f" 12) as an output.3. 9 := x 2 .h))· .h) = (JI .Thus sinee the generator of a principal ideal is unique up to constant multiples.2. where r is the remainder of the division of f by 9 f :=g g:= r f ·.3. we have gCd(f" 12) = gcd(f.3x + 2 and 12 = x 2 -1 be polynomials in <Qi[x].3. 12 E k[x].1 First pass through the WHILE loop: -3x+2 --+ -2x+2 f:= x 2 . which is the previous r.h) = (gcd(h -gh. with one of h.h) = gcd(f.g) = gcd(r. When 9 = 0. 3. . Let J" .. J. . .. where r is the remainder of the division of J by 9 (that is.x. J. J ~+ r. then gCd(f" .). . J.. J..1) was the key to solving the problems mentioned in Section 1.1) we first compute 9 = gCd(J" . The remainder of that division is zero if and only if J is in the ideal I = (f" .. J8))' PROOF.1 for the special case of polynomials in one variable.3.. . The proof of statement (i) is similar to the proof !'lf Proposition 1.i = 1. (12. by (i). as desired. Recal! that the greatest common divisor of s polynomials J" .s. . . (ii) iJ s 2" 3. J.. . .... I = (J" . We then use the Division Algorithm to divide J by g. . be polynomials in k[x].).. To prove statement (ii).. .5. The computation of gCd(J"".. the cosets of 1. the coset representative of the element J + I in the quotient ring k[x]/ I is r + I. we first compute 9 = gCd(J"".). Also. a polynomial at a time.) = (g). J.). .) is done by induction. J. .xd-" where d = deg(g).) = (f" hl· Again. gCd(J"".x 2. . To decide whether a polynomial J is in the ideal I = (f" . J.8. J" denoted gcd(J" ..1. .) = (gCd(f"". Then. J.3. i = 1. In the current section we saw that the concept of reduction leading ta the Division Algorithm (Algorithm 1. 9 is monic). J.)). to solve System (1. • lc(g) = 1 (that is..3.3. ..3. with r = 0 or deg(r) < deg(g)). . . THE ONE VAmABLE CASE 15 g:= 0 The WHILE loop stops J:= k(f)J= ~2J=x-1 Therefore gcd(J" h) = x .))... PROPOSITION 1. We now turn our attention to the case of ideals generated by more than two polynomials. . with not all of the Ns zero. 0 With the ideas developed in this section we can now solve al! the problems raised in Section 1. J. . .). • if h E k[x] divides J. .3. as is easily seen from part (ii) of Proposition 1.....) = gCd(J" gcd(h. As noted before. even though we have already used the notion of ordering in the concepts of degree and leading . . Using the notation introduced earlier: JE I = (g) if and only if J ~+ O. J.. form a basis for the k-vector space k[xllI (Exercise 1. Then (i) (f"". s...) = (h). is the polynomial 9 such that: • 9 divides fi. J.1. .8. . . It then suffices to solve the single equation 9 = O. J... and hence (f" . then h divides g.. Finally.1. by (i).).) = gCd(f" h) = gCd(J" gcd(h.6). ../. J.. let h = gcd(h.. . . In the last section (the linear case) we saw that there were two ingredients for our solution method: a reduction algorithm (in that case it was row reduction) and an order among the terms. We have not yet stressed the importance of the ordering of the terms in the one variable case. . xn ) = al a. the ordering is forced upon us.. Let j" 12. i. a.xn].. the terms that we introduce (coming from . . = x 6 . .3).3.] 1.:i~ig.j.3.' + (g)} is a k-vector space basis for k[x]/(g).2.3.4x 3 + x 2 .1 (as in Example 1.8 with k(x" . ..3. for n 2:: 2. Show that in k[x.8... Prove that a system of equations J = 0. x + (g).1. in general. 9 = 0 with two relatively prime polynomials J. Let J.9..6. where we recall that k(x) denotes the field of fractions of k[x].3) to divide J = 2x 5 . Apply yOliI algorithm to the polynomials j.3.8.2x .3.1.3.3.4 is false. 1. = x 5 -2x 4_X 2+ 2x. when we compute r .b1"O}. show that the ideal (x. to be polynomials in k[x]. x 4 + 2x 3 + 2x2 . 1.. .3.3.X + 2 by 9 = x 2 + X + 1.bEk[x]. Prove that q and r obtained in Theorem 1.e. and h = x6-2x5+X4_2x3+X2_2x. + Ud2 = h. Find a single generator for the ideal 1 = (x" -l. This can only OCCliI if the powers of x are ordered so that xn < xm if and only if n < m (Exercise 1.g E IC[x.7.2). + u2h. where j.2. In effect. Let 9 E k[x] be of degree d..3. Let 9 E k[y] be irreducible.b1"O}. u" U2 E k[x] such that J = gcd(f" 12) and J = u. [Hint: View J and 9 in k(x) [y] and use the Gauss Lemma: J and 9 are relatively prime when viewed in k[x.xn ] # {O}.3.11.16 CHAPTER 1.bE k[x" .. . x d .4.e. 1. .3. y]. . y] and h is not in <C. i. k(x" .3.. 1.. Modify Algorithm 1.4. with unknowns U2. 1. y] cannot be generated by a single element.5.2x . y] be such that J rt (g). Exercises 1. y] if and only if they are relatively prime in k(x )[y]. We consider the equation u.10... k[x" . y] has at most finitely many solutions. Is x 5 +X3+ X 2 -7 E I? Show that x 4 +2x 2 -3 Eland write x 4 +2x 2 -3 as a linear combination of x 6 .3 E lQI[x] of Exercise 1.j.2x .1 and x 4 + 2x 3 + 2X2 . y]. FoUow Algorithm 1.3. Show that the above equation is solvable if and only if 9 = gcd(f" 12) u" . Prove that (j.. We note that the condition n < m is equivalent to the statement that xn divides xm.y].xn)[y]. k(x)={~la. Compute gcd(f" 12. BASIC THEORY OF GROBNER BASES term. show that if hE IC[x. in the Division Algorithm. . 1. y) c:: k[x.xn ].Xn ]. g) is infinite. Show that. 1.3.g) n k[x" . in order for the algorithm to terminate...] 1. h) using Proposition 1. . Indeed. h E k[x]. [Hint: Use the hint of Exercise 1. That is. Prove that if J and 9 have a non-constant common factor in IC[x. Generalize this exercise to the case where h E C[Xl 1 ' " . In particular.2 to output J. and let J E k[x" . where we recaU that k(x" .:i~ig) must be smaller than the leading term of r which has been canceled. then V(f. 1.xn ] is not a PID.3. Prove that {1 + (9).12 = x 4 + 2X3 + 2x2 .xn ) denotes the field of fractions of k[x" .h = x 7 +x6 -2x 4 -2x3+x+1.3.2 are nnique.3.. then the equation h = 0 has infinitely many solutions.. Theorem 1.3. 9 E k[x.3. .xn .. 3.1. i. and let !c(g)'J = gq + r be as in a. where r = 0 or deg(r) < deg(g). 9 = lOx 3 ~ 23x 2 . (This is the partial fraction decomposition of rational functions. 9 = (y2 + 2y + 1)x 2 + (y3 + 3y2 . T'). g. b. where J. where i 2 = -1.r E R[x] such that !c(g)'J = gq + r.20x . Generalize a and b to the case of s polynomials fI. Let J.1) has to be modified. Use e ta compute gcd(f. Cive an algorithm for computing such UI and U2.g be primitive polynomials in R[x]. then there exist unique UI.3.. THE ONE VARIABLE CASE 17 divides h.) e. g). . Ct E R and Tf primitive. Moreover.) ~ deg(g). .. c.y3 + y2 _ 2)x + (_y2 + 2y). (ii) J = (-2 + 4i)x3 + (5 + 3i)x 2 . U2 E k[x] that satisfY the equation above and such that deg( UI) < deg(f2)deg(g).2y) E Z[x. then deg( U2) < deg(f. f.Xn]. where R is a unique factorization domain (UFD). 1. b. Use d ta compute the partial fraction decomposition of x-3 x 3 + 3x 2 + 3x + 2· f. In this exercise we present a "pseudo" division algorithm for polynomials in R[x] .) and deg(u2) < deg(h). A polynomial JE R[x] is called primitive ifits coefficients are relatively prime. The polynomials q and rare called the pseudo-quotient and the pseudo'Temainder respectively.J.g = 2X2 + (1 + i)x + (1 + i) E (Z[i])[x].g).2)x + (y3 ~ y2 . and e = deg(f) . the Division Algorithm (Algorithm 1. R a UFD. d. Use the above to give an algorithm for computing gcd(f. a. Use this algorithm to find q and r in the following cases: (i) J = 6x 4 . Use b ta show that if fI and 12 are relatively prime. Use d ta give an algorithm for computing gcd(f. Find U" U2 which satisfy b. h = x 2 . Let fI = x 3 .g E R[XI. Let J. where r' is the primitive part of r.3 E Z[x].g). Prove that there exist polynomials q.deg(g) + 1. h. d.g E R[x] be such that 9 # 0 and deg(f) 2: deg(g). U2 E k[x] such that h Ul U2 -~-+­ fIh ~ fI 12' with deg(uI) < deg(f.4x + 3 E Ql[x]. Prove that if 9 = gCd(f" 12) divides h. then for every hE k[x] such that deg(h) < deg(fI) +deg(h). y]. When the coefficients of polynomials in one variable are not in a field k. Give an algorithm for computing q and r. Use g to compute gcd(f. g) for the examples in c.2ix + (-1 + i). .g) = gcd(g. e.. r = ar'. where J = (y2 + y)x 3 + (_y3 ~ y+ 1)x 2 + (C.12. c. .12 = x 2 + X .3..1. there exist UI.2. . if deg(h) < deg(f.e.)+deg(h}-deg(g). E k[x J. Prove that gcd(f.llx 3 ~ 3x 2 + 2x. if ai ~ f3i for aU i = 1. . "power product" will always refer ta a product of the Xi variables. . where j3 = ((3" . Therefore. Sometimes we will denote xf' . That is. the orders in the linear and one variable cases were used ta define a division (or reduction) algorithm. x~n 1 (3i E N. Also. Term Orders. x~n by x(3. we arranged the terms of the polynomials in increasing or decreasing order. First recall that the set of power products is denoted by ']['n = { xf' . BASIC THEORY OF GROBNER BASES 1... n} ..4. Thus the order must be a total order.x/3 E 1I'n. if xa: divides x/3.3..1. In the linear case. DEFINITION 1. then y.2 and 1. and hence we must be able to compare any two power products. in 1I'n.3. . that meant that the reduced polynomial T was obtained by eliminating a leading variable using g.4. There are many ways ta order 'JI'n. in the divisions described in Sections 1. exactly one of the following three relations must hold: xa < x/3. + T. then we should have xa: ~ x f3 . We would like ta emphasize that. (3n) E l'ln. ..6). . xo.4.. thus the order had to extend divisibility relations (see the discussion at the end of the previous section). By a term order on 1I'n we mean a total order satisfying the following two conditions: < on '['n .5 and Theorem 1. throughout this book. that is.. that meant that the remainder polynomial T had degree less than that of g. For example.2 and 1. given any xo.2.4. the reduction ------7+ described in Sections 1.18 CHAPTER 1. and it turns out that these conditions are captured in the following definition (thls will be justified in Proposition 1. We will also always assume that the different terms in a polynomial have different power products (so we never write 3x 2y as 2x2y + x 2y). we computed with x first. In more than one variable we need an order analogous to the ones used in these two special cases and this will be the focus of this section. In the one variable case we used the highest degree term first and thls was required by the procedures used.3 must stop after a finite number of steps. 80 every power product is a term (with coefficient 1) but a term is not necessarily a power product. Recall that whenever we had f --'!. The particular order used was unimportant but did have to be specified. It was important in the last two sections to specif'y an order on the power products.. in Section 1. . . . etc.. there is no infinite descending chain x a1 > X U2 > X U3 > ... and "term" will always refer ta a coefficient times a power product. i = 1. = x/3 or X U > x/3. or equivalently. However. that is. the polynomial r was such that its leading power product was less than the leading power product of g: in Section 1.n. An order that satisfies aU these conditions is called a term order. for the reduction to be finite.. we need that the order be a well-ordering. we already know sorne properties that a desirable order must satisf'y. Moreover. then xa: ~ x f3 . DEFINITION 1.2. The easy verification that they are term orders will be left to the exercises (Exercise 1..3). vEN.. then we have 1 < x < x 2 < x3 < . xr DEFINITION Xl 1. It is important to note that we need ta specify the order on the variables. < y2 < ... x(3 of 1. (ii) If X U < x(3... we give three examples of term orders.f3= ({J" . for aU x"Y E 1fn.) Note that. if we use the lexicographie order with x < y. < y < xy < x 2y < . > Xn..3. or z instead of the subscripted variables above... .4. for this order. satisfy ai Sa.. we have 1< X2 < x~ < x~ < . (We deliberately altered the order of X and y from what was probably expected to emphasize the point that an order on the variables must be specified.{Jn) ENn we define L~=l or Qi < L~=l f3i L~=l Qi = L~=l f3i and xO: < x/3 with respect ta lex with Xl > X2 > . in the case of two variables Xl < {Ji· and X2. then XUx"y < x(3x"Y. As noted before. . > X n as foUows: For we define the first coordinates ai and {Ji in Q and f3 fram the left..1.. We define the lexicographical order on 1fn with Xl > X2 > .. which are difJerent.1.4. ... We define the degree lexicographical order on 1fn with > X2 > ..4. < xi < .4. For example...an )... Before we prove that the basic properties we discussed above follow from the conditions in Definition 1. TERM ORDERS 19 (i) 1 < x(3 for aU x(3 E 1fn. y... We will always denote this order by "lex".4. < Xl < X2Xl < X~Xl < . This will be of importance later on (see Section 2.3). We emphasize again that we always need ta specify the arder on the variables. we will usually use X. is always greater than X21 for an non-zero /-1. > X n as follaws: For Q= (a" . when we do examples in a smaU number of variables... We will always denote this order by "deglex". with this order.. That is.4.~<~<~<~~<~~<~<···.This must be especially kept in mind when we consider examples involving Xl Yl z (see Exercise 1.4. we first order by total degree and we break ties by the lex order. we need to prove that any term order extends the clivisibility relation and is a well-ordering. . That is. We define the degree reverse lexicographical order on 1I'n with x. Or. we always need ta specify the order of the variables. In the case of two variables Xl and X2l we have 1<~<x. satisfy ai > (3.4. It is easy to see that in the case of two variables deglex and degrevlex are the same orders (Exercise 1. y] with x < y. as the following example shows: XiX2X3 > XIX~ with respect ta deglex with Xl > with respect ta degrevlex with Xl X2 > X3 but XîX2X3 < XIX~ > X2 > X3· This order tUTns out ta be extremely good for certain types of computations.5. if X U divides x i3 then XU <:: x i3 .4). DEFINITION 1.1.1). Again.4.4 refers to the smallest variable. However 1 if there are at least 3 variables. The three examples given above are the ones we will use the most.20 CHAPTER 1. we have 1 < x < y < x 2 < xy < y2 < x 3 < x 2y < xy2 < y3 < .4.4.4. this is not the case anyrnore.. using the degree lexicographie ordering in k[x. > X2 > .9. has all the properties discussed before that definition. we have Xl > X2 > . > Xn. For xc" x i3 E 1I'n. PROPOSITION 1. as defined in Definition 1.<~<x. BASIC THEORY OF GROBNER BASES 80.4.. Note that the term "right" in Definition 1. There are many other orders on lI'n which we will see later in both the exercises and the text... We will always denote this order by "degrevlex".... We want to observe that a term order. > X n as follows: For we define L~=l Qi < L~=l (3i or 1:7=1 Œi = cr L~=l f3i and the first coordinates Œi and (Ji in and fj from the right. We will see that each order has different properties and which order we use will depend on the problem we want to solve. We now return to the general definition of a term order. which are different. The important property that this order possesses is given in Exercisè 1. . To finish this section.1. since if we had equality. THEOREM PROOF.3) is divisible by sorne ::. 1XO'i+l).6. Then there exist . that is.1.4. We define: • Ip(J) = x U ' .i. . > write our polynornials in this way. We will always try to . So if we go back to the chain of ideals (1.cCXj for sorne j ..xnl when we have a terru order on 'Ir". If we expand each Uj as a linear combination of power products. j = 1.4.(::cal. by Proposition 1.4. .1). ZUi E 1['n.: i. • It(J) = a.4. . By 2: 1 and so by Condition (ii) we Now from the Hilbert Basis Theorem (Theorem 1. for every subset A of1I'n. Every term order on 1I'n is a weU-ordering.. Then for all f E k[Xl...4.c ai ) =F. . X U :':: x i3 . i. with f # D.e a j .4.X n ]. 4We will say that we have a terru order on k[Xl.5 and this contradicts (1. .3) xO'.1) we can prove 1.6 will be used throughout this book for many proofs in a manner described in Appendix B.eUj.4.:ca i E ']l'n. But x a i+ 1 must appear as the power product of a term on the right-hand side of Equation (1. we now see that this chain is a strictly ascending chain of ideals in k[Xl"" . x U ' .4. TERM ORDERS 21 By assumption there is an x" Condition (i) in Definition 1. we fix sorne notation.::.4. i. l ::. there e:rists X U E A such that for aU x i3 E A. j :::. First we choose a terrn order4 on k[Xl"" . D PROOF.1.l ::::. Bueh that (1.ai E k. as desired..xn ].2) We first note that (x al .4. Suppose to the contrary that the given term order is not a wellordering...xn ]. the leading term of f. E x" 1I'n such that x i3 = xUx". i = 1. .we may write where 0 #. This is a contradiction to the Hilbert Basis Theorem (Theorem 1.1)..'" . j :::. j ::. ' .xn ].I:=:. 0 Theorem 1.4. and x cq > X U2 > . then i (1. • le(J) = a" the leading coefficient of f.1 we have have x(3 = xO:a{Y 2: ::cO:..2. . the leading power product of f..3).1) This defines a chain of ideals in k[x" . we see that each term in UjX CXj is divisible by X Uj • Thus every term of the right-hand side of Equation (1. Therefore x U '+' is divisible by sorne ::.4.'" X Ur .2). where Uj is a polynomial in k[Xl l • • • . and hence xO:i+ 1 ~ ::.i+ 1 - -~ ~U j=l J 'X°l'j .xn ] (1. lex.Us in k[Xl l . and It(f) may change..le(f) = -2. ) X n be variables.. Note that Ip.Xm together. Let Xl. . • if the arder is degrevlex with x > y > z.. while x 3 y2 . then Ip(f) = x 2 yz..4.. we have 1 < x a . 1 Xn. 1.4..1.6. Prove that for ail x a oJ 1 in 'lI'n. x n ] if and only if It(f) E k[Xi' . and using the given term order on them. Prove that f E k[Xi. 1. and leading power product of f with respect to deglex.8xy3z3 E Ql[x..) Ip(ui)). Consider the polynomial f = 3x 4z.4. . > XnLet i E {l. . x n ]. .2x3y4 + 7x'y'z3 . k[x]. called the homogeneous components of x"y2 -x 2yz' +y2z).1. . + f. Determine the leading term. In the polynomial ring in one variable. the one such that 1<x<x2 <x3 < . Show that it must be the usual one. and lt(f) = -2x 3 . .(lp(f. . . Show that lex. and assume that < is also a well-ordering. • if the arder is deglex with x > y > z. . leading coefficient. Ip(f g) = Ip(f) Ip(g). BASIC THEORY OF GROBNER BASES We also define Ip(O) = le(O) = It(O) = O. 1. and It(fg) = It(f) It(g). show that Ip(!1UI + . .. le(f) = 3.] 1.3. .) 1.. Show that in k[x.. y]. ..6). Repeat the exercise with x < y < z.4. and lt(f) = 2x 2 yz. Also. Given polynomials f I l ' " . 1. and It are multiplicative..u. and let m < n. .xn]. .4.4..z5 is homogeneous sinee every term has total degree 5.4. le(f g) = le(f) le(g). le. let < be a term order. that is. .1s and u}... le(f) = 2. For example. deglex and degrevlex are the same orders.4..g. Prove that any term order on power products in the variables Xi l • • • l X m is the restriction of a term order on power products in the variables Xl. let f = 2x'yz + 3xy3 . Let < be a total arder on 'lI'n satisfying condition (ii) in Definition 1. i.2. We cali a polynomial f E k[XI' ..9. deglex and degrevlex are term orderings. x'y' z + xy4 . n}. grouping the variables Xl.4. . . (This is a partial converse of Theorem 1. if we change the term arder.) ~ max'<i<. y.4. x. Does equality necessarily hold? (Prove or disprove. then Ip(f) = x 3 . and lt(f) = 3xy3. le(f). . and degrevlex with x > y > z.x 2 yz2 and y2 z. z]. .7.2X 3 : • if the order is lex with x> y > z.. > .e. 1. Let f be a homogeneous polynomial and let the term ..[Hint: Use the idea of lex..4.4. Let f E k[XI' .] and consider the lex arder with Xl > x. then Ip(f). Exercises 1...x 2 yz2 + y2 Z is not hornogeneous. 1. then lp(f) = xy3. x n ] homogeneous provided that the total degree of every term is the same (e. the latter polynomial is the SUffi of the two homogeneous polynomials x 3 y2 .8.5..22 CHAPTER 1. .4. b. 1. is a total order.. the first Uj sueh that Uji cl 0 satisfies Uji > O. Show that the partial <u can be extended to a term order..] 1.'" .14.12. (Mora-Robbiano [MoRo]) Let We define an order in <!J!m as follows: ("". prove there is a subset G ç J such that J = (G) and for ail subsets F ç J with J = (F) we have G ç F... .) Now we define an order <u in k[Xl" . [Hint: Prove first that if J = (xn" . .xn =) then for fJ E !\In we have x!3 E J if and only if there is an i such that xn.a' u m ) < (fJ· U" . Let J ç k[Xl"" ..11) contains a unique minimal generating set. . the first Uj such that UJi cl 0 satisfies Uji > O. Show that revlex is not a term order on k[Xl l ' " . then <u is a term order if and only if for all i...Jn) E !\In we define xn < x!3 if and only if the first coordinates "i and ...4.. . Prove that every monomial ideal J (see Exercise 1. .. Ui is the lisual dot product in Qn. (3 E Nn .Ji in a and fJ from the right which are different satisfy "i > ./ E N r•. . (Dickson's Lemma) Prove that the result of Exercise 1. <u. The revlex ordering is detined as follows: For 0: = (""'" '''n)..+ !\In). Prove that if the vectors Ul. TERM ORDERS 23 ordering be degrevlex with Xl > X2 > .. .4. .Ji from the left which are different satisfy Qi < . d.13. e. [Hint: First show that a polynomial f E J if and only if each term of f is in J.. > Xn· Prove that In divides f if and only if X n divides lt(f)..Ji. .4. . Prove that <u is a transitive relation..) 1. i=l (By a + !\In we mean {a + 'Y l 'Y E !\In}. /3.4. Show more generally that f E (Xi.xn ] be an ideal generated by (possibly infinitely many) power products (such an ideal is called a monomial ideaQ. where 0: . Prove that xo.Jm) if and only if the first ai. 1. .10.J" .fJ· u m ).] 1. .xn] as follows: for a. That is. .. <lt x/3 implies xct: x1 <u x f3 x' for aU 0:..X O ".11 is equivalent ta the following statement: Given any A ç Nn there exist 0:1.Um be vectors satisfying: for aU i. ." '''m) < (.. Prove that there exist a b .am E A such that j m A ç U(a. . . divides x!3. ..Ji' (Note that this is just lex on <!J!m. c. xO: <u x/3 implies xCl' <u' x/3.4.Ct'/1l.. a. .11.xn ). .•. that is. <u'... Prove that if the veetors Ul.U m span Qn. Let Ul.1.4.).J" .Xn].U m span Qn 1 then the order.xn ) if and only if It(f) E (Xi..4.fJ = (. E Nn Bueh that l = (XO'l. .. xn <u x!3 <==? (a· U" ... where Ci E k .17.. We will use the following result from linear algebra (see. 1.. ..4. .. . 1 Œin) E Mn 1 i = 1.r. .. . .4. ta the deglex.. u for aU = 2.e. c.. First we define the convex hull of the vectors al.ae.16. the hyperplane. < xn? Saffie question with Xl > X2 > . let {er.. ..n and 01 . 1. b.. . Let A be any r x n matrix with rational entries.24 CHAPTER 1. Show that the first alternative in the lincar algebra theorem above is equivalent to the condition that the zero vector is in the convex hull of the rows of A together with the vectors ei.] There is a geometric way to view the linear algebra theorem used above. ei is the vector in Qn with all coordinates equal to 0 except the ith coordinate which is equal to l..r. What vectors Ur. Show that there is no term ordering on Q[x. i. r (compare with Exercise 1. orthogonal to u has ail the rows of A on one side. .. Let f = 2X4 y 5 + 3x 5 y' + x 3 y 9 E Q[x. be power products in k[Xl1 .. Conclude that Xl is the leading term of f with respect to sorne term order if and only if the zero vector is Ilot in the conv€x hull of the ... Assume that there is a term order < such that Ip(J) = X....:. . .<>r) = {teiŒi Ci . en} be the standard basis for Qn. . 1 O.. Consider the vectors Qi = (ail)' . . .T. ..{O}. U = 2:7=1 aljUj > L:7=1 CYpjUj = Qe . • There exists a column vectar U E IQn with non-negative coordinates such that the coordinates of the vector Au are all positive. yJ.. [Hint: Consider the matrix A whose rows are the vectors al . then exactly one of the following two alternatives holds: • there exists a row vector v E with non-negative coardinates such that the coordinates of the vector vA are ail negative or zero. . . i = 1..un) E!Qn such that Ui 2: 0 for i = l" .4. for example... > Xn.. BASIC THEORY OF GROBNER BASES 1.14 give rise ta the lex. . and to the degrevlex term orderings with Xl < X2 < . n.. that is.15...Un E !Qn in Exercise 1. a r as follows: conv(ŒI"" .T.4.14). and tei = 1}. and so L has the convex hull of the rows of A and the ei'8 on one side... . e ror a.r.xn ] and let f = 2:~~1 c. yJ such that Ip(J) = x4 y S. . Use the above result to show that there is a vector u E Qn with nonnegative coordinates such that D!l·U > ae·u for f = 2.4. In this exercise we show that there exists a vector u = (Ul) ..Xi . . x~in 1 i = 1.i = l. L. for i = l.. . [Ga]): THEOREM. (*) Let Xi = x~i1 .. 1=1 1=1 AIso. . Note that the second alternative irnplies that there is a vector u which makes an acute angle with every row of A. for n = 3.. . The basic idea behind the algorithm is the same as for linear and one variable polynomials: when dividing 1 by Ir. Let Show that lp(g) = XO c. Let us first look at the special case of the division of 1 by g.. The theorem states that every symmetric polynomial is a polynomial in the elementary symmetric polynomials.2 and 1. where 1.. Use the well-ordering property of term orders to complete the proof of the existence of h and so ta prove the Fundamental Theorem of Symmetric Polynomials.xn ] such that 1= h(O"l.3 we had a division algorithm. Fix the lex terrn ordering on k[Xl"" .. XIX2 + XIX3 + X2X31 and XIX2X3 are symmetric..(ln = X... Use it in the case = 2 to write + as a polynomial in CTI = Xl + X2 and CT2 = XIX2. x n ] is called symmetrie provided that when the variables of 1 are rearranged in any way. . d. 2: Œn- b. [Hint: See Exercise 1. let CTl = Xl + X2 + .] d. + In-lX nl ··· .cg is a symmetric polynomial. Division Algorithm.. . 150 we want to cancel terms of 1 using the leading terms of the fi's (so the new terms which are introduced are smaller than the canceled terrns) and continue this process until it cannat be done anymore..r and ej. alsa referred ta as a reduction process. part e.1: a division algorithm in klxl. .4. We fix a term order on klxl.xn] be a symmetric polynomial. .X2'" Xn. . Let lt(f) = cx a where a = ((lI.. DIVISION ALGORlTHM 25 vectors al ~ ai) i = 2. Let 1 E klxl.'" .. Recall that a polynomial 1 E klxl...'" .xn ] with Xl> X2 > .xn]. Use the above to determine all the possible leading terms of 1 = 2x4 y 5 + 3x 5 y 2 +x 3 y 9 _ x7 y .cg) < lp(f) and that 1 . For example. xn]. Now observe that lp(f . ... .. + X n1 (T2 = XIX2 +XIX3 + . the resulting polynomial is still equal to f.5. In Sections 1. a. We will define a division algorithm in klxl. Show that (Xl 2: Q2 2: . We need to show the existence of a polynomial hE klxl....(ln) E lin and C E k.'" . . ln.. In this section we study the second ingredient in our solution method for the problems mentioned in Section 1.'" .. j = 1..O"n). . . Note that the above proof yields an algorithm for computing h given the symmetric polynomial f.. In this exercise we prove the Fundamental Theorem of Symmetric Polynomials. x n ] that extends both of the algorithms seen in the previous sections. Xl + X2 + X3. These polynomials are called the elementary symmetric polynomials.18. 1. . g E klxl. n xf xi 1. . > Xn.xn]. . ..1. For general n.4.14.5. . Another way of saying this is that we are not allowed to write f = 4x 2 y + 2x2y2 X + 4 y 3 .8xy .ix.1. (We are not allowed to cancel.. Y. namely -:1-x3 .2 and 1. We can continue this process and subtract off ail terms in f that are divisible by lt(g). We will also say X > Y. yi.. let the order be deglex with y > x.26 CHAPTER 1. sinee.x 2--> -x . we say that f f -"-+ h.g.~X2 . to denote power products or terms instead of the more cumbersome .3x . (We observe that in the special cases presented in Sections 1.x . usually X. let f = 6x 2 y .h in k[Xl' .1 and just cancel 4x y). We can think of h in the definition as the remainder of a one step division of f by 9 similar to the one seen in Section 1.-x.3x --> -x + -yx . written oF 0. Y or Z. if and only iflp(g) divides a non-zero terot5 X that appears in f and It must be strongly emphasized that in tills definition we have subtracted from f the entire term X and we have replaced X by terms strictly smaller than X.VCX or aœO< unless we need to make an explicit reference to the exponeot o. where h = -3xy3 2 X + 4y 3 -1. Given f. 22 424424 Note that in the last polynomial. We now consider the term order deglex with x > y so that now lt(g) = y3 and so f -"-+ h. with 9 reduces ta h modulo 9 in one step.1. Also. in fact X = ltU).-x 2 . We could write this reduction process in long division format as 5From now on we will use capitalletters. 2 2 EXAMPLE 1. We note that in this latter case we canceled the term X = 4y 3 from f which is not the leading term of f.X + 4y3 . then f -"-+ h. Then f 9 --> --yx 2 1 7 2g137 11 9 7 g 1 3 + -yx .2. BASIC THEORY OF GROBNER BASES DEFINITION 1.5. for term X. Let f = y x + 4yx . .3 we considered only the case where X = ltU).) For example. 9 = 2y + x + 1 E Q[x.5.xnl. If the term order is lex with x > y. yi.. . where now h = 6x 2 y . say. only 4x 2 y. in this case X = 6x y is the term of f we have canceled using lt(g) = 2xy. no term is divisible by lp(g) = y and so this procedure cannat continue.3.1 and 9 = 2xy + y3 be polynomials in Q[x. provided that Ip(X) > Ip(Y). x n ] with . Let fI = yx . . . h 2 f ---4 -----j. . ... . fs} ifr = 0 or no power product that appears in r iB divisible by any one of the lp(fi)...4.. y]. . x n ] such that f" h 1 ----? J.5.X E lQl[x.. Let f. .··· . of 0 (1 SiS s).y.. The reduction pro cess allows us to de:fine a division algorithm that mimics the Division Algorithm in one variable.5... Let F = {J" h}. E k[XI' . r cannat be reduced modulo F... f. A polynomial r is caUed reduced with respect ta a set of non·zero polynomials F = {J" .5.xn ]. and let F = {J" . and 80 we extend the process of reduction defined above to include this more general setting. Let the arder be deglex with y > x. DIVISION ALGORlTHM 27 2y + X +1 1 y2 x + 4yx _~yx2 2 3x 2 y2X+ hX2 + hX + ~yx - 3x2 _lyx2 _ lX 3 _ lX 2 4 4 4 2 4 lx 3 _2X 2 _ 'lX In the multivariable case we may have ta divide by more than one polynomial at a time. fI... and fI. .5. -----+ h t -1 -----+ EXAMPLE 1. In other words.5. . . DEFINITION 1. .h. fs}.... f = y 2 x.1. F f" • •• J. with J... then we caU r a remainder for f with respect ta F. s. Given f. h = y2 .. f" h . s} and a sequence of polynomials h" . Then sinee h Y2 x il y 2 -------t ---)0 X. denoted f ---->+ h. DEFINITION 1.6. We say that f reduces ta h modulo F. .l E k[XI. DEFINITION 1. i = l.it E {l.. .. . h t . . if and only if there exist a sequence of indices ib i2.. ._.3.5. If f ~+ rand r is reduced with respect ta F.fs be polynomials in k[XI. .. . + .1. + . lp(r)) = lp(f)... lt(fi) . . J.1 correspond to the single quotient q in Algorithm 1.xnl with Ji of 0 (1 <:: i <:: s) OUTPUT: Ul.lt(h) f . . have the same definition: no term of r is divisible by the leading term of any divisor. U2 := 0.3. . .} and ma. we have s different quotients in Algorithm 1. Js as opposed to dividing J by a single polynomial 9 in Algorithm 1. lp( us) lp(f.) divides lp(h).3... E klxl. .5.. the multivari- able Division Algorithm. . BASIC THEORY OF GROBNER BASES fi of 0 (1 <:: i <:: s).. J.. .. r := 0. ) x n ]) such that J = Ud.5. and Algorithm 1. Us := 0. assumed an ordering among the polynomials in the set {fI. In Algorithm 1. . .5.. Us in Algorithm 1. .1. fI. in effect.28 CHAPTER 1.··· .1.x(lp(ul) lp(f.= J WHILEhofODO IF there exists i such that lp(fi) divides lp(h) THEN choose i least such that lp(f. U" r such that J = Ud. .5. This algorithm is given as Algorithm 1. The remainders.10. Multivariable Division Algorithm Note that in Algorithm 1.5. + usfs + r. and a remainder r E k[Xl' . denoted by r in both algorithms.. The quotients Ul. once the leading term of ris not divisible by lt(g).3.xnJ.1. .'" .3.1.. the one variable Division Algorithm in Section 1. ELSE r:=r+lt(h) h:=h-lt(h) ALGORITHM 1.5..). INPUT: J. INITIALIZATION: Ul := 0.3.1 because we are dividing J by s different polynomials fI. It is informative to consider the similarities between Algorithm 1.1. h .. . we also know that no other term of r l .) divides lp( h) lt(h) Ui := Ui + lt(fi) h '= h . ...Us E klxl.5.1 we have.1. Js} when we chose i to be least such that lp(f.)... . + UsJs + rand r is reduced with respect to {fI. This is an important point and will be illustrated in Example 1. this algorithm returns quotients U"". ... ) divides Ip( h) = yx Ul "= Ul + ~yx = lyx . We recompute Example 1.lx 2 + lx . The order is deglex with y > x. EXAMPLE 1.1.~t~x' (2y + x + 1) 3 + lyx _ llx 2 lx 4 2 4 Third pass through the WHILE loop: y = [p(ft) does not divide Ip(h) = x 3 :=r+lt(h) = h:= h -It(h) = ~yx . r := 0.\ h y.ix h := h .2 but now we follow Algorithm 1.1 we start with r = f and subtract off multiples of 9 until this occurs. yJ. DIVISION ALGORITHM 29 is divisible by It(g).It( h) = 0 iX3 - ix3 - .3x2 First pass thmugh the WHILE loop: y = Ip(ft) divides Ip(h) = y 2 x ·-u 1 + ~ lyx U 1·2y -"2 h := h ft 2 = (y x + 4yx . So in Algorithm 1. h := y 2 x + 4yx . (2y + x + 1) 1 2 Ul := Ul .5.1.3x 2 ) _ x . Let F = {ft}..) does not divide Ip( h) = x 2 r := r + It(h) = ~x2 h := h -It(h) = -ix Sixth pass through the WHILE loop: y = Ip(f.. where ft = 2y + x + 1 E Q[x.7.5.~(2y + x + 1) - ±x Fifth pass through the WHILE loop: y = Ip(f. 80 we start with h = f and r = and subtract off the leading term of h when we can or add the leading term of h into r when we cannot.'4'X2 Fourth pass through the WHILE loop: y = [p(f.fx 2) .5. and we have obtained the remainder. This simple property is not true in the multivariable case. Let f = y 2x + 4yx . and 80 build up the remainder. necessitating the introduction of the extra polynomial h in Algorithm 1.lW.h h . INITIALIZATION: UI := 0..4"X 1 = - - (_~yx2 + ~yx .1.3x 2) .))ft = Gyx = _~X2 -!. - - +~ !t(h) f !t(h) l _lyx2 = 2YX .'/. 2y2 44 r ix3 h:= h .3x2 = -hx2 + Second pass through the WHILE loop: y = Ip(ft) divides Ip(h) = yx 2 ° 1.) does not divide Ip( h) = x r := r + It(h) = ~x2 .3.5.5.3x 2. !!.} and J in k[XI"" .)). .Ul + lt(hl . Let F = {J" 12}.y ) h . with r reduced with respect to F and Ip(J) = max( max (lp(Ui) Ip(J. . r := 0.8.5.J.y2 Second pass through the WHILE loop: yx = Ip(j. THEOREM 1. INITIALIZATION: Ul := 0. + r. 12 = y2 ._ lt(h) _ Ul .2. Given a set oJ non-zero polynomials F = {J" .lp(r)).. .5.Xn] such that J = Ulj.t:Ss .h . the Division Algorithm (Algorithm 1. Let J = y 2 x. + .J.'" ... U2 := 0.r E k[Xl1'" .5.. + U.+12+x. EXAMPLE 1. (yx . and we get ° and J=yj.) 1 .) divides Ip(h) = y 2 x . y].!!.0lf lt(h) 2 .) does not divide Ip(h) = y2 y2 = Ip(12) divides Ip(h) = y2 lt(h) 1 U2 11 7 1 9 7 := U2 + It(J..us.x) . The order is deglex with y > x.x Third pass through the WHILE loop: yx = Ip(j. and we have and J=(-yx--x 2 +-x)(2y+x+1)+(-x3 --x 2 --x).xn].y.Y 2 . = yx .0lJ ItU..y' (2 h .1) produces polynomialE Ul.Y 2 x .) = .. l:=.Y :E.x E <Ql[x.) does not divide Ip(h) = x y2 = Ip(12) does not divide Ip(h) = x r :=r+lt(h) =x h:= h-It(h) = The WHILE loop stops. 2 4 4 4 2 4 Note that these are the same steps we used in Example 1.30 CHAPTER 1. where j.JJ' yx .9..h . BASIC THEORY OF GROBNER BASES The WHILE loop stops.5.y2 Y . h := y 2 x First pass through the WHILE loop: yx = Ip(j. This difficulty already occurred in the one variable case. we see that if r J J.2y . Using lex with x> y > z > w. = x-y2w.!:. . x 2 . + 12.Xn ].. Consider the polynomial J = y 2x . At each stage of the algorithm. J may be in the ideal (fI.h = Z-W 3 . if J = x... J . where . However if we reverse the order of J. andJ. = x 2 and h = x2_x.. We first observe that the algorithm terminates.. Let J = x 3 y3 + 2y2.6). and since h = J at the beginning of the algorithm. Ip(h i+1) < lp(h i ). To do this in the multivariable CaBe is the subject of the next section. whereas J = J. of the h's in the algorithm. but J is in the ideal (fd2).e. the list of the hi's must stop.. = yx . .)) or by subtracting off lt(h i ) (in case no lp(fj) divides lp(h i )).J.yJ. then f is in (h. y2 - JE X J2.. DIVISION ALGORlTHM 31 PROOF. since the term order is a well-ordering (Theorem 1. that is.9. may not be zero as the following example shows..h = y2 .3.2. To pIove the second part. Let J = x 2y2 -w 2. divide J by J"h.x.5.hJ4 to obtain a remainder r and an expression as in Theorem 1.10. = 2xy2 + 3x + 4y2.5..y. where J. we note that from what we did abov€. 1 /s). . EXAMPLE 1.x is reduced with respect to F.4. Therefore. we have J .9.9. as weil. we obtain Ui by adding terms l~'c~)).h = y-zw. but the remainder of the division of J by J" .5.r E (J" . This is because we compute hi+l from hi by subtracting off i:~~. Repeat this exercise reversing the Iole of J. J. . we use 12 first in the Division Algorithm) then J J2. the leading term of h is subtracted off until this can no longer be done. That is.. .2 E <Ql[x. f2l h.. So the remainder of the division of J by F is non-zero. 1. the converse iS'not necessarily trU€j that is.x E <Ql[x.5. for each i.J4 = w 3 -w E <Ql[x. .5.wJ.5.. For example. J = yJ.J. .+ 0 and indeed.z.1.5.:(~)) li cancels the leading term of h. then J is reduced with respect to {J" h}. using f4l h. . = 0. and x 2 .1..(lt(h i ) + lower terms).h = y2 .J.y. Exercises 1. Prove that given a set of non-zero polynomials F ç k[XI' . fI· 1. h2 .J.). namely x = gcd(X 2 . Using lex with x > y. Repeat this exercise reversing the role of fI. .. yJ. Set F = {J"h}. and the ideal J = (J"h) ç <Ql[x. However.yJ. h.x).) divides lp(h. divide J by J" 12 to obtain a remainder r and an expression as in Theorem 1.J. we get a sequence hl.5. ris obtained by adding in terms lt(h) and so lp(r) <:: lp(f) . 0. there . Using the deglex term order with y > x and the Division Algorithm... and 12 (that is..12 E (J" 12)· The difliculty was resolved by finding a better generating set for (J" 12). Moreover. .-'-. f4l i.x. 80 we have that for all i. D With J written as in Theorem 1. and hence J. Now.).i!' = lt(h i ) + lower terms (in case sorne lp(!.. where hi+ 1 is obtained from hi by subtracting off lt(h i ) and possibly sorne smaller terms: hi+l = hi . Thus. It is then irnmediate that lp( Ui) lp(fi) <:: lp(f). we have at any stage in the algorithm lp(h) <:: lp(f).X 2 . (i) G is a Grobner basis for J..!. s are reduced with respect ta F. b. g3 -. then r + s = h.... Grobner Bases. for any finite set of non-zero polynomials F ç: k[Xl. .1. we define the leading term ideal of S ta be the ideal Lt(S) = (lt(s) 18 ES). (iii) f E J if and only if f = I:. Let F = {fI.] 1. gi+l. then fg --++ rB." ... Let f. 1.xn ]. Let J be a non-zero ideal of k[Xl" ..+ O.6. fs} ç: k[Xl' . In thls section we finally define the fundamental abject of this book..6. then fg --++ O. .6..5.32 CHAPTER 1..2. . t} such that Ip(gi) divides Ip(f)..6. where h.:~l higi with Ip(f) = max. .5. g.r.<:i<:t(lp(hi) Ip(gi)). namely. BASIC THEORY OF GROBNER BASES can be no infinite chain g. x n ] and let F be a collection of non-zero polynomials in k[Xl. [Compare with Theorem 1. F F b..6.!.. 6 Another term which is commonly used in the literature is standard basis..!. .} contained in an ideal J. .4.:~l UiJi with Ip(f) = max'<:i<:slp(uiJi).6. DEFINITION 1. h. In order to do thls we need to make the following definition.xn]. (ii) f E J if and only if f ~+ o. we have a. Show that for any polynomials f. there exists i E {1. ...2 part (iii). A set of non-zero polynomials G = {g" . . Give an example that shows this does not imply that f -. and 9 ~+ s. is caUed a Grübner basis6 for J if and only if for aU f E J such that f '" 0.!. g2 -.. . We note that it is not clear from this definition that Grübner bases exist.. 1. with fi'" 0 (1 ~ i ~ s).5.] 1.5. .. The foUowing statements are equivalentfor a set of non-zero polynomials G = {g". We first present three other characterizations of a Grôbner basis. . We will prove this in Corollary 1... x n ].. . Disprove the following: a. S E k[Xl. . . x n ]...xn ] such that f = I:... a Grübner basis... .g. then F F F F F F F f + g--++ r + s... . .. and for any power product X E 'II'n..!. then there are no non-zero polynomials in J reduced with respect ta G. and let f E k[Xl'... c. If f E F.. -. .g E k[Xl.} ç: J. If f --++ g. If f +g ~+ h.. For a subset S of k[Xl.....9 is that we may not be subtracting off leading terITIS in gi -.5. f ~+ T. (iv) Lt(G) = Lt(J).xn ]. then Xf --++ Xg. r. THEOREM 1... . In other words. Xn ]. g. If f --++ rand 9 --++ s. . . if G is a Grobner basis for 1.5. [Hint: The new point here that did not occur in Theorem 1. If f --++ rand 9 --++ s.. Writing f as in the hypothesis. as desired. For the reverse inclusion it suffices to show that for all f E I. and hence f E (9'. . By Theorem 1. ..9')· 0 PROOF. Hwe expand the right~ hand side of Equation (1.9.9'} is a Griibner basisforthe idealI..xn ].1). where the sum is over all i such that lp(f) = lp(h i ) Ip(9i).. This is a contradiction to the fact ° ° that r is reduced with respect to G. and let f E klx" .9. (ii) =? (iii). Then.xnl and Xi E 8..6.+ o. (1.9'). 0 COROLLARY 1.t} such that Ip(9i) divides lp(r). lt(f) E Lt(G). we see that (iii) follows from Theorem 1.2) .4. IfG = {9'. (i) =? (ii). there exists r E klx" ... reduced with respect to G. if f E I and r oF then r E I and by (i). f -'2.6.+ 0. f= LhiXi' i=l where hi E klx" .. Clearly. .+ 0). and hence .6. Let I be an ideal generated by a set 8 of non-zero terms. such that f -'2. then I = (9'. For the reverse inclusion. Then lt(f) is in Lt(G).. .. Let f E klx" . LEMMA 1. The result follows immediately.5. Then f is in I if and only if for every term X appearing in f there exists Y E 8 such that Y divides X. there exists a finite subset 80 of 8 such that I = (80 ). . then f E I. . Thus r = and f -'2..) It(9i).. for sorne hi E klx" . is also divisible by sorne Ip(9i). there exists i E {l. (iii) =? (iv). For f E I... it immediately follows that lt(f) = L ° lt(h. . . Moreover.xn ]. and sinee the process of reduction is exactly the same as the Division Algorithm...3.+ 0..Clearly.. .1. GROBNER BASES 33 PROOF. we know by hypothesis that f -'2.r E I and so f E I if and only if r E I. ç I. (iv) =? (i). PROOF. if r = (that is..R. . . by Theorem 1. we see that each term is divisible by sorne Ip(9i).6. the only term in the left~hand side. f -'2.. Thus f . since each 9i is in I. If f E I.. Conversely. Thus lt(f). we see that every term is divisible by sorne terrn . since the lt(f)'s generate Lt(I)..) For the next coroUary we first need sorne information about the special nature of ideals generated by terms. Let f E I. let f E I..5.1) lt(f) = L i=l hi It(9i).. . .2)..2..6.6. Lt(G) ç Lt(I). . then (1..6. Clearly (9'...xnl..··· . .+ r. Hwe expand the right~hand side of Equation (1. for i = l.9.6.xn ].. DEFINITION 1.xn ]..). (Note that we have not assumed that cX lt(g. BASIC THEORY OF GROBNER BASES Xi in S. We will prove condition (ii) in Theorem 1. then each such X is in J = (S). Since f .. 2::=1 f - c1X1gii -S+ T. theu..34 CHAPTER 1. . .6. 9 - cXgi ~+ r. and hence f is in J.. of such divisors is dearly a generating set for J.).x n ] is such that 9 ~+ G T. (Of course. that r is unique. so is Tl . .g.2. and hence every term of the left-hand side must also be divisible by sorne term Xi E S.. .6. 0 COROLLARY 1. by the Hilbert Basis Theorem (Theorem 1.g. by hypothesis.lt(g. Then.1.) actually cancels a term in g. We must show that r = O. . Let f -S + Tl and f -S+ r2. E J. We will expand our terrninology a little.. . the remainder of the division of f by G is unique.6.r2 are bath in (G) = (g" ..6.) CLAIM: If c E k is non-zero. for each i E {I"" 1 t}. Conversely. . We first aBsume that G is a Grübner basis.xnl has a Grobner ba- PROOF. where r is reduced.}.2.7. .6. Suppose that f -S+ r such that r is reduced. we know.4 the leading term ideal Lt(I) has a finite generating set which can be assumed ta be of the form {lt(g.. X E ']l'n is a power product. But then Tl .g. . we see that 0 . sis. 0 We now give a fifth characterization of a Gr6bner basis.6. since f E I.. To see this. and f . Moreover r. .. we can write f = Cv Xl/9i where Cv is in k and is non-zero and Xv E 1['n and each iv E {l. So.6.6. If we let G = {g" .2 (ii). 80 now we can apply the daim to 9 = C2X2gi2 f - C 1 X 19i 1 to = obtain f - C1X1gil ~ ~+ T. and 9 E k[Xl" .) We note that if the claim is true we are done. applying the daim ta 9 = f.T2 = 0.r2 is reduced with respect ta G...xn ] is a Grübner basis if and onZy if il is a Grobner basis for the ideal (G) it generates. So let f E (G)...t} (this can be done by writing f = L:~l higi and writing each hi as a SUll of terms).r.T2. We say that a subset G = {g" . and r2 reduced with respect to G. In order to prove the last statement we note that.1).g.. Conversely. if for every term X appearing in f there exists a term Y E S such that Y divides X. J has a finite generating set. we see that l' . . with r.5. By Lemma 1.} be a set of non-zero poZynomiaLs in k[X1"" . then we have Lt(G) = Lt(I) and hence G is a Grübner basis for J by Theorem 1. The finite set. by Theorem 1. By the first part of the lemma each term in each member of this generating set is divisible by an element of S. THEOREM 1.)} with g" . ThenG is a Grobner basisifand onlyifforallf E k[X1"" . Thus. .. using induction. .. assume that remainders upon division by G are unique.g. Let G = {g" .xn ]. Every non-zero ideaZ J of k[X1.} of k[X1"" . PROOF. d = c. the latter being reduced with respect to F.10 that J ~+ 0 and J ~+ x 2 . And 80 9 . Consider the polynomials 91 = Z + x. Set h = 9 .5.2 that the polynomials in Exercise 1. d # 0 and d # c. We will consider three cases.eX 9i ~ + Tl. z does not divide lt(f). But x 2 = Ip(x 2 . z and y do not appear in any term of J.eX 9i ~ 9 ~ + r which is the desired result. Let rI be reduced and assume that 9 .2 do form a Grübner basis).1). z].xE (h. h 2 E lQ![x. GROBNER BASES 35 J . we see r = Tl.CX9i is -clc(9i) which is non-zero and 80 9 .5. F is not a Grübner basis. The theorem is now proved.x we have x 2 .92)' We use the lex term order on lQ![x.h}.L~~I cvXvgi " ~+ r. Let F = {!J. We will give a more systematic way of proving that a set of polynomials is a Grobner basis in the next section.} for an ideal I. We can see this in another way. EXAMPLE 1. Since d # 0 we have 9 ~ h. y.xE lQ![x. as desired. Thus by the definition of a Grübner basis (Definition l.6.h) and J ~+ x 2 .5. and hence z+x divides J. That is. and so J E lQ![x]. and y does not divide lt(f). z] with x < y < z. we get 9 ~ h ~ + T2 and 80 T2 = T.!J = yx-x and 12 = y2 .x.) be the coefficient of X lp(gi) in g.1O. 12). Fis not a Grobner basis. is unique. z]. We showed in Example l.6. Let G = {91. 0 ~+ r which immediately implies that r = 0.+ r also. We use deglex with y> x. y).6..1. So let J = y 2x-x. 0 Although we have in Theorem 1. we saw in Exercise 1.7 that remainders are unique for division by a Grobner basis. We observe that if we have a Grübner basis G = {91"" .cXgi ---->+ rI. d = O. Also. Since y does not appear in J. J = (91. • Case 1.cXgi ~ h. y. EXAMPLE l. We continue Example l. because of the lex term order. since J = y!J + 12 E (!J.92}. since d = c oF 0 we see that 9 ~ 9 .cXgi ~ h --S+ T.6. Namely.dX 9i' Then the coefficient of X Ip(9i) in h is O.6. since we know 9 2.9. as desired. a contradiction to the fact that the only variable occurring in J is x.X)h2' where hl.Z). Thus. Thus by Theorem l.. .5. we may set y = x. PROOF OF THE CLAIM: Define d by letting dlc(g. since the remainder.lt(92)) = (z. and we have J = (Z+X)hl(X. G • Case 2. Then the coefficient of X lp(gi) in 9 .8. • Case 3.x) is not divisible by either Ip(!J) = xy orlp(h) = y2. Thus.6. Then. Suppose to the contrary that there exists J E J such that lt(f) ~ (lt(91). since d # c we have 9 .9.7.x. y. Bueh that T2 is reduced. Let J = (z + x)h l + (y . So if h ~+ T2.2 that the quotients are not necessarily unique (we will see in Exercise 1.X. We will prove that G is a Grübner basis for J. Then. 92 = Y . as desired (by the assumption that the remainder is unique).6. . We note that a Grobner basis with respect ta one term order may not be a Grôbner basis with respect ta a different term order. Show that the polynomials fI. by Theorem 1.4. with lex with x > y do not form a Gr6bner basis for the ideal theygenerate.5). Jz.36 CHAPTER 1. we have already computed (without knowing it!) Gr6bner bases for two special cases. whereas fI. G = {gcd(fI.w E Ql[Xi y. .xn}/1 is r+1. 1. Let fI. the variables being ordered according to the position of their colunm in the matrix of the system of equations (Exercise 1.2. In both cases we do have an algorithm for computing the Gr6bner basis. [Hint: Follow Example 1. z} with x > Y > z.2. In the linear case.6.6. Show that the polynomials 91.6. y. Exercises 1.6. a basis for the k-vector space k[Xl' .. Jz. if we use the lex term order with x > y > z in Example 1. · ...92} is not a Gr6bner basis for 1 (Exercise 1.2 form a Gr6bner basis for the ideal they generate.10).5.5. -17z do. . Also...9.6. . Clearly.).) 1.1. where r is the remainder of the division of J by G. 1.6.6.6. the question now is how do we compute a Gr6bner basis for an ideal I? The results in this section only prove existence. Jz = y . Also.5.fm be non-zero !inear polynomials in k[Xl.xn}/1 is the set of ail cosets of power products that are not divisible by some It(9i) (Exercise 1.. For example.xn} which are in row echelon form. . the polynomials obtained from row reducing the matrix of the original linear polynomials ta row eehelon farm constitute a Gr6bner basis for the ideal generated by these original polynomials. Jz = y2 -2y-2 E Ql[x.6).6. .2 do not form a Gr6bner basis for 1. Ali of these applications will be discussed fully in Chapter 2. J.9.6.y 2 w..)} is a Gr6bner basis for the ideal 1 = (fI. the representative of the element J+1 in the quotient ring k[Xl' . . then {91.7. Let < be any term order in k[x. y}. Show that the polynomials fI = 2xy2+3x+4y2. h in Example 1.zw. BASIC THEORY OF GROBNER BASES then we can solve sorne of the problems posed in Section 1. To decide whether a polynomial J is in 1.1 in a fashion similar to what we did in the one variable case. Show that they form a Gr6bner basis for the ideal .3).3. w} in Exercise 1. with respect to lex with x > y > z > w. The remainder of the division is zero if and only if J is in 1. (See Exercise 1.6. . However. . by Theorem 1. . and the proofs of these results do not indicate any method for finding Gr6bner bases..6.} Show that they do not form a Gr6bner basis with respect to lex with w > x > y > z. J4 = w 3 ..w 3 . Show that the polynomials fI = x .2(ii) (Exercise 1. we use the Division Algorithm and divide J by G.6. h = z .6. 1. z. We will give Buchberger's Algorithm for their computation in the next section..1.9 do not form a Gr6bner basis with respect to lex with x > y > z. In the one variable case. J.92 in Example 1. X n ].6..) Set 1* = J . .6. Prove that Lt(I) is the k-vector space spanned by {lp(f) 1 j E I}. GROBNER BASES 37 1.. Prave that G i8 a Grobner basis for J. 1. 1.} ç: k[x]. then j -S+ T. h E k[Xl" .6.x n ] such that j -S+ 9 and j -S+ h. col O. show that for all X E Lp(I') and Y E ']['n we have XY E Lp(1*). Generalize Exercise 1. Prove that G is also a Grübner basis for J.14. .6.6.xn ] consist of non-zero polynomials. 1.13. For a subset S ç: k[Xl' .16.} be a Grübner basis for J. That is. . a.r E J.. show that if! = (d) is a principal ideal in k[Xl' ..xn ].7. Let G ç: k[Xl' . Show that 1['n i8 a monoid. Let l be an ideal in k[Xl"" . .. . "--S+" confluent provided that for aU J.xn ] and let K be an extension field of k. Prove that F is a Grübner basis if and only if cd E F. . then F ç: lis a Grübner basis for l if and only if cd E F.xn]/J is {X + JI X E ']['n and lp(gi) does not divide X for al! i = 1.g. (Note that we havejust taken the setof alllp(f) not the ideal generated by the lp(f)'s.6.12.xn ] with .t}. for sorne c E k...} is a Grübner " basis. 1['n i8 closed under multiplication. We cal! the reduc- 1.xn ] be an ideal..g.. that is. Prove that tian relation {g" ..} ç: k[Xl.15. Let G be a Grübner basis for an ideal J and let T. (Note: this is just Exercise 1. 1. that i8.. In the polynomial ring in one variable.1.. for sorne cE k.6.hg. 1.} of non-zero terms. .X.8.11. jE k[Xl"" . Let l be an ideal of k[Xl"" . b. . Let Gand G' be two Grübner bases for an ideal J ç: k[Xl. 1. . Let d = gCd(f" . Prove that a basis for the k-vector space k[Xl"" . 1.xn ] set Lp(S) = {lp(f) 1 j ES}.xn ]. g.). consider a set of non-zero polynomials F = {f" . . k[x].{O}.7.. [Hint: Use Theorem 1... Let J ç: k[Xl"" . they generate with respect ta any arder for which the variables are ordered according ta the corresponding colurrms in the matrix.'" .) c.6.6..6. In this exercise we give another equivalent definition of a Grübner basis. (We say that Lp(F) generates Lp(1*) as a monoideal if and only if for al! X E Lp(1*) there exists Y E ']['n and Z E Lp(F) such that X = YZ.xn ]. Let l ç: k[Xl"" .. Show that Lp(1*) is a monoideal of ']['n.6.. . . 1. . Prove that if j .. .j.6 ta principal ideals in k[Xl"" .] Let {g" .. where ris reduced with respect to G.6.xn ] generated by J. col O.xn ] and let G = {g" .xn ] be an ideal generated by a set G = {X" ...) In this exercise we give another equivalent definition of a Grübner basis.6.X n ] sueh that h -S + T and 9 -S + T.j.6. Praye that F ç: J* is a Grübner basis for J if and only if Lp(F) generates Lp(I') as a monoideal.10..6.gt} is a Grübner basis if and only if {hg . ... . Prove that G is a Grübner basis if and only if "-S+" is confluent.. there exists an TE k[Xl"" .X n ]..6. 1..9.xn ] and let 0 # h E k[Xl" . Let J be the ideal of K[Xl"" . .7. Let G be a Grübner basis for an ideal J of k[Xl. . xn ] and let G = {g". gl. 1. Assume that lt<.. Let <h be the order delined by extending < to k[Xl"" . . Then successively set x = 0 then y = 1. Cive an example that shows that there is an ideal J = (h. then we rnay assume that f is homogeneous. and let w be a new varid able. g.'" . Xn' W] delined by Jh = (fh 1 f E J).} and G' = {g~ 1 • ' . ... . b.38 CHAPTER 1. . Xn. Prove that {g" ...'" . .y2 ~x.xn ]. Assume that we are given two term orderings..xn. t. (gi). . for sorne v 2' 0 and that h E J. Let Jbe an idealof k[Xl"" . Let f E k[Xl.. for i = 1. Let J be an ideal of k[Xl. Let {g" . WJ. .xn.(lp(aiJ) lp(gj)).')).. with respect to the same term order. x n ] such that Jh is strictly larger than (fl'. .9).4.} is also a Grübner basis for J with respect to <2 . We deline the homogenization of f to be = w f(".. For an ideal J of k[Xl"" .··· . .6. . where J = (yx ~ X. b E k. Assume that for all gi E G we have gi = L}'~l aijgj where lp(gi) = max. .:) E k[Xl. Prove that G' is also a Grobner basis for 1. Xn].'" .l) E k[Xl. . [Hint: Show that if f is reduced with respect to G and in J.] ..] d.} be a Grübner basis for J with respect to deglex or degrevlex.." . . g.xn .··· . 1 Xn.w] as follows: r Prove that <h is aterrn order in k[Xl"" . Use c to compute a set of generators for Jh.} be a Grübner basis for J with respect to <.xn ] and let G = {g" . Prove that f = w"(ih)h. Also. [Hint: It suffices to show that Lt<h (Jh) = (lt<h (gn. y2 ~ x) by lirst showing that G = {yx~x.xn]. . '. ..xn ]. x n ] have total degree d. Let < be the deglex or degrevlex order in k[Xl. and f = h. (gi) = lt<.J~) ç k[Xl l ' • .18. we deline Jh to be the ideal of k[Xl"" .. Xn' W] we deline gh =g(Xl.. f. . 1.6.g. for 9 E k[Xl"" .19..:Si:s.'" . .g~/} be subsets of l of non-zero polynomials where we assume that G is a Grübner basis for J.6. Prove that r = rf.. Assume that f f -"-++ r and f ~ + r' where r is reduced with respect to Gand r is reduced with respect to Cf. and <2 .x2 ~x} is a Grübner basis for J with respect to deglex with x > y.. a. Note that fh is homogeneous (see Exercise 1.17. . . .w] and that lt<h(fh) = ItU)· c. 1. then f = ax + by for sorne a. xn].g.'} is a Grübner basis for Jh with respect to <h .'" . Let f E k[Xl"" . Let J bean ideal in k[Xl. (yx ~ x) + h2(y2 ~ x) + h3(X 2 ~ x).]. w]. Prove that G h = {g?. If f E Jh... say <. BASIC THEORY OF GROBNER BASES respect to a single term order.) of k[Xl.. . lt<h (gl. For F = {J" .. We will define a reduction procedure that answers the "algebra membership problem" for k-subalgebras of k[Xl' . whether J E kiF]... .. J = L:~l hdi. let L.J. For 9. Let J = (J" . Show that if F consists entirely of terms then F is a SAGBI basis for kiF].xn ] and v = (VI.xn ]. .. . for JE k[Xl. . (xy')V' E IQl[F].4.. xy2} for n <:: m.'" .) ENs and only finitely many CV's are non-zero}. We then present his algorithm. S-POLYNOMIALS AND BUCHBERGER'S ALGORlTHM 39 1. But if J is in J. xv.h E kiF].7.. [Hint: For J = LCv(X')V. Jl:'.6.'" .'" ...xn ] generated by F.xn ] we mean a subring which is also a k-vector space. For a subset F = {J" . that is kiF] = {Lvcvg' . The simplest way for this ta occur is in the F .xy + y.v. .. This result is given in Theorem 1. expand out the xy + y term and show that it suffices ta prove that xnym is in the monoid generated by {x2.) For a k-subalgebra A ç k[Xl. for sorne hi E k[Xl. (Note that we have only taken the set of all Ip(J). In Robbiano and Sweedler [RoSw].. y2. ==? h we write 9 b+ h. If we have 9 ==? 91 ==? 9' ==? . Sa a difficulty arises with elements of J whose leading power products are not divisible by any Ip(J.p(A) as a monoid.xn ] we write 9 b h ta mean there is acE k (c of 0) and v E N' such that It(cFV ) is a term m 9 and h = 9 . and let F = {J" . In this section we first lay the theoretical foundation for the algorithm for computing Grübner bases by presenting Buchberger's Theorem [Bu65. c.X n ]. Bu85].s} such that Ip(Ji) divides Ip(J). 7.. S-Polynomials and Buchberger's Algorithm.. This material is taken from Robbiano and Sweedler [RoSw].cFv..p(A) is a multiplicative submonoid of lI'n.. . an algorithm for computing SAGBI bases is given.20.xn ] is a SAGBI basis for kiF] if and only if for all J E kiF] we have J ==?+ O. .'" .p(F) = {lp(J) 1 J E F}. Prove that F = {x'. (SAGBI stands for Subalgebra Analog to Grübner Bases for Ideals. 7. For F ç k[Xl. d.xn ]. Show that F ç k[Xl' .y] for deglex with y > x. Fix a term order on k[XI J ••• .xn ] show that L. F F F F ..'" .xn ] we denote by kiF] the k-subalgebra of k[Xl"" . where we assume that fi of 0 (1 <:: i <:: s).p(F) generates L.1. In the previous section we defined F ta be a Grübner basis if and only if for all J E J.) a.xn ]. By a ksubalgebrû A ç k[Xl"" .) be an ideal of k[Xl"" . .xn ].Js}. there exists i E {l.Vs) E li' 1 denote by F V = g' .] 1. Hence the difficulty occurs when the largest of the lp(hdi) = lp(hi ) Ip(Ji)'s cancel. .. . f::' Cv E k and v = (VI.xy'} is a SAGBI basis for <QI[F] ç lQl[x. We cali Fa SAGBI brulis for A = kiF] provided L.Js} ç klxl. . .Js} ç k[Xl'''' .4.. hE k[Xl"" .xn]..) b.. This is the problem of determining.. e.... (This problem will be solved using Grübner bases in Section 2.).y'.(y')"'(xy + y)V. Show that 9 b+ h implies that 9 . y].5. The polynomial x 2 .x E Q[x. s.7.xJz.). . EXAMPLE 1.j reduce f using J" we get h 2 = f - It&j) fj· The ambiguity that is introdueed is h2 . note that S(ft. S(gi.g) = ~f . and that it ean be reduced: S(fl.x. . L L S(f.lp(g)). .f" it may happen that sorne term X appearing in f is divisihle hy hoth lp(fi) and Ip(J.g. . . Let f = 2yx . then f is a linear combination. y] with the deglex term ordering with y > x. i = l. We consider the term X = y 2 x in f. Note that X = L = lcm(lp(ft).. 7Recall that the least cornmon multiple of two power products X.x is now redueed -"-+ with respect to {ft. Namely.g) = It(f/ . lp(h)) = y 2 x. Let f = y 2x + l.... ..S(j"J.xh = S(ft. Let ft. 1 ::. with coefficients in k..g E k[Xl. We denote L by lcm(X. Mareover Ip(hf) = y 2x = Ip(~xg) have canceled in 8(f. Let 0 The polynomial oJ f.7. Before we can prove this result.40 CHAPTER 1. Iflp(f) < X... 9 = 3y2 .y.) ---++ G o.4 (BUCHBERGER).yft. we can go ahead with a strategy fot computing Grübner bases.~xg = _~y2 + ~x2.. ft = yx .) for i oJj (henee X is divisible by L = lcm(lp(fi).. Y). . Y divides L and if Z is another power product such that X divides Z and Y divides Z then L divides Z. gt) if and only if for ail i oJ j.hl = It&. Lee L = lcm(lp(f). It turns out that the Spolynomials account for aU ambiguities we need to be concerned about as the next result shows. We have that f y2 + 1 = f . DEFINITION 1.2. and if we It&. . . LEMMA 1. h = y2 . we need one preliminary lemma. .s. with the deglex term ordering with y> x.1. BASIC THEORY OF GROBNER BASES following. . . and that the ambiguity introduced is _y2 + x 2 = yft . h) E (ft. h) ~ x 2 . ... EXAMPLE 1. f..xn ]. . . and f ~ x 2 + 1 = f . i < j ::.7. Then G ..g). .)fi . Now that we have introduced S-polynomials as a way to "cancel" leading terms and to account for the ambiguity in the Division Algorithm. h}.3. and 8(f.7.It(g)g is called the S-polynomial of f and g. of S(j" fj). E k[Xl. Let G = {gl. .xE Q[x.} be a set of non-zero polynomials in k[Xl. Also. h)..y.7.lp(fj)).xn ] be such that lp(f. THEOREM 1. h)..) = X oJ 0 for ail i = l.It&j)fj = . s.!{. If we reduce f using j" we get the polynomial hl = f fi. hut is not zero. . There is another way of viewing S-polynomials.xn ]. . in the division of f hy ft.. Y i8 the power product L such that X divides L.g.$g = h f . . Then L = y 2x. Let f = L:~l cdi with Ci Ek.. a Grobner basis for the ideal l = (gl. Xjgj ) = L hijvgv .7.) :Js c... Set 9 = LiES eiXigi. we are done. Then. gj) E J. + c. gj) ~+ 0 for aU i of j by Theorem 1. there exist dij E k sueh that 9= L i. then S(gi.)(a.1. + csa s = O. ai E k. Now. + C2a2)(.(:.5.. Let S = {i IIp(h.) lp(gi) = X}. We will find a representation of f with a smaller X. since qal + . Now.a. If X = lp(f). + . let us assume that S(gi.5.. al a2 a2 a3 +(c.~JS-l . sinee S(gi.. c.lp(Xj gj )).a..gt} is a Grübner basis for J = (g". + cs-1a'-. but lp(g) < X.j d.if. v=l . + . +(c..4.7.gt)..j J cd.4).Xjg j ) gives a representation ~+ 0 (See Exercise 1.a. 13) + . + . Conversely.6.S(/" 12) + (c. J. PROOF OF THEOREM 1.a. If G = {g".a. We choose ta write f = 2:!=1 hi 9i 1 with x = max (Ip(h i ) Ip(gi)) 1 s: i:st least (here we use the well-ordering property of the term order). + C2 a2)S(h.gj) ~+ 0. + csa. For i E S. We will use Theorem 1.7. since the ci's are in k." .. J.. f. . gj) ~+ 0 for all i of j.lp(gj))' By hypothesis. Thus 2:::=1 Ciai = 0.) + . and so we see from this last equation that S(Xigi..gJ)' X X where Xij = Icm(lp(gi).l h ~.jES. write hi = eiXi + lower terms.) + (CIal + .l 12) + (c. S(gi. Write Ji = aiX +lower terms. Ip(Xigi ) = X.l 13) + .Xjgj). and this will be a contradiction.(:.. for all i E S. S-POLYNOMIALS AND BUCHBERGER'S ALGORITHM 41 PROOF.) clal(.jS(Xigi.. lp(f) < X.. By Lemma 1.. Then the hypothesis says that fi fj) since lp(fi) = lp(fj) = X. Then f can be written in many ways as a linear combination of the 9i '8. + CsJ. X = Icm(lp(Xigi ). Otherwise. so It(Xigi)Xigi ~ It(XjgJ)Xjg j X X X It(gi)gi .l /. Let J E J.2(iii) to prove that G is a Grübner basis for J. ~. D We are now ready to prove Buchberger's Theorem.a. by definition.6.a.lt(gj)gj = Xij S(gi.. .).2. + cs-las-l)S(f. S(fil iJ) = t -. J. This t S(Xig i .-l..:. 7. and h = y2 .7. with respect to Gi-l. . h}· Then 8(h. This is a contradiction.gj)) = v=l max (lp(hijv)lp(gv)). fz.1.1 and obtain a strictly increasing infinite sequence GJ ç G2 ç G3 ç .. Each G i is obtained from G i _ J by adding some h E 1 ta G i . We first need to show that this algorithm terminates.} with gi 7' 0 (1 <.gj) = "Lhijvgv. ..x 2 . 12) = xh .7. r COROLLARY 1. where h is the non-zero reduction. fz.J . Substituting these expressions into 9 abov€. h) = y 2h .7.5. as the algorithm progresses. y] with the deglex term order with x < y.. t). F' THEOREM 1. Then. by Theorem 1. and = {h. Let F = {h.7. do this until there are "enough" polynomials to make all S-polynomials reduce to zero.42 CHAPTER 1. i <.. Sa we add h to F. Ip(8(Xi gi .g. s).9.··· . PROOF. we get = 'E~=1 h~gil with ma. and F' F' 8(fz.. Thus {h. EXAMPLE 1.4. . the following additional equivalent condition for a subset G of k[xJ.t We note that Buchberger's Theorem (Theorem 1.Y is reduced with respect to F.. l::::::v$. BASIC THEORY OF GROBNER BASES where.. Let us fust look at an example..8.lp(Xjgj )) = X.!. 12) -----+ O. y2 _ y.xh = 0. i <.x(lp(Xigi). Buchberger's Algorithm (Algorithm 1.j <. Let G = {gJ. we construct a set Gi strictly larger than Gi .xJ<.} with Ji 7' 0 (1 <.7.7. h) = Yh .t(1p(hD lp(gi)) < X. 0 We have as an immediate Corollary of the proof of Theorem 1. where Ip(8(gi..xn ] to be a Grübner basis. Now 8(h.J. Xjg j )) < ma.4) gives a strategy for computing Grobner bases: reduce the S-polynomials and if a remainder is nOrrzero. . We give Buchberger's Aigorithm to compute Grübner bases as Aigorithm 1.x. Given F = {h.).6. i. h}. and 9 iuto J.yh = y2 .7. .J.. we have t 8(gi. Let h = xy .y2 -----+ O.. Suppose to the contrary that the algorithm does not terminate. t). add this remainder to the list of polynomials in the generating set. .1) will produce a Griibner basis Jor the ideal 1 = (h. 12 = x 2 . . Then 8(h.. of an S-polynomial of two elements of . h} is a let F Grobner basis.x 2h = _y3 + x 2y -----+ x 2y .i<.y E lQl[x.. Then Gis a Griibner basis iJ and only iJ Jor ail i 7' j (1 <. h} G:=GU{h} 1 for ail u E G} ALGORlTHM 1. and hence J = (fI.g:= {{J"h}} First pass through the WHILE loop g :=0 S(J" 12) -S+ x 3 . then S(gi." .x = h (reduced with respect to G) Since h f 0.7. {h h}} G:={J"hh} Second pass througb the WHILE loap g:= {{hh}} S(J"h) -S+ 0 = h Third pass through the WHILE loop .xnJ with J. yJ ordered by the lex term ordering with x < y. Since h is reduced with respect to Gi .. Moreover.. . J. E G} g:= g .1. .. we have that It(h) Thus we get Lt( G .. h = -y + x 2 E lQl[x.gj) -S+ 0 by construction. Let fI = xy .x. This is a strictly ascending chain of ideals which contradicts the Hilbert Basis Theorem (Thearem 1. . Naw we have F ç G ç J.4.7. let h := x 3 . J. .1.. Buchberger's Algorithm Jar Computing Grobner Bases G i .}.) INITIALIZATION: G:= F. Therefore G is a Grübner basis far 1 by Thearem 1.g. g:= {{hJ.9.7.} ç k[Xl"" .l ..g}} S(J.) ç (g". if 9i.. . .{{J. 9j are polynomials in G.1.l ..x g := {{Jd3}.. f 0 (1 Sc i Sc s) OUTPUT: G = {g" . g) -S+ h. S-POLYNOMIALS AND BUCHBERGER'S ALGORlTHM 43 INPUT: F = {fI.gt) ç J.} 1 Ji WHILEgf0DO Choose any {J. where h is reduced with respect to G IFhfOTHEN g := g U {{ u. Thus G is a generating set for the ideal J.1). a Grübner basis for (J"". J. INITIALIZATION: G:= {J"h}. ct Lt(Gi _ l ). g} E g f J. D EXAMPLE 1. ) ç Lt( G 2 ) ç Lt( G 3 ) ç .7. {f" S(ft. {h. in our next exarnple we compute a Grübner basis in the case when k is a finite field. M.h. We use the lex term ordering with x > y to compute a Grübner basis for l = (J" 12) C. we have used the field Ql. {fds}. {ho Js}. Js}. h} g:= {{f"h}.!3}. M.44 CHAPTER 1. To illustrate this point. {f"M. hl· INITIALIZATION: G:= {f" h. h} is a Grübner basis for the ideal (J" 12)· We will conclude this section by giving two more simple examples which i1lustrate Buchberger's Algorithm. hl· So far.7. = x 2 + y2 + 1 and 12 = x'y + 2xy + x be in Zdx. 14}.h}} First pass through the WHILE loop g:= {{f" h}. Let l. {h M} G:= {f" h. M. In this example we consider the field k = Z5 = Z/5Z. 12 = y + x. yi· .!s} {h Is}. {ho M. {f4. The sixth through tenth executions of the WHILE loop will also give Spolynomials which reduce to zero (Exercise 1.h}) let 14 := x' 9 := {{ft. {h.7. yi.4) and thus {ft.ls} := Fourth pass through the WHILE loop 9 := SU" {{h. Let ft = y2 + yx + x'. {h M. Is}} SU" J4) -S+ O.hJ. h 14} Second pass through the WHILE loop 9 {{h. since 9 = 0. h) = x (reduced with respect to {f" h. yi. {h M} Third pass through the WHILE loop 9 {{f" 14}. Thus {f" 12. {fds}} G:= {fl. Id. {J.h) -S+ 0 := := M. {f2. {f2.. h 14}) /5 :=x 9 {{f" 14}. {ho M. M.h) = x' (reduced with respect to {f"h. {ho 14}. EXAMPLE 1. {ft. in our examples. EXAMPLE 1. {h. The theory developed so far is valid for any field k. {f" /5}. M.. Js}. {h M} S(h. and h = y E Ql[x..h}} S(ft..!S} is a Grübner basis for (J" 12. Is}} J4) -S+ 0 Fifth pass through the WHILE loop 9 := {{h. {f. {f" 14}.11. {h.!''!3' J4. {h. {f" Js}. Let us use the lex term order with y > x ta compute a Grübner basis for l = (J" h. BASIC THEORY OF GROBNER BASES 9 :=0 SU" h) -S+ 0 = h The WHILE loop stops.7.10. Zs[x. xz + y. 14} forms a Grübner basis for (f" M ç Zs[x. . y.}} G:= {J" h h. This is not the case as we will see in this exercise. [Answer: x 2y + z..4. Compute the S-polynomials of the following pairs in Q[x.7.{h.7.4) and thus G = {J" 12.1. and thell reducing this basis modulo 5. 2 C.4 to show that the polynomials given in Exercise 1.{J.7. where we assume that the polyaomials in G' have relatively prime integer coefficients. y3 + z3. z] with respect to the lex.6.5.7. y].1.7. 7. b. 1 = 3x 2 yz .] b.{ht. Finish the computation in Examples 1.2xh ---->+ 4y 5 + 3y4 + y2 + Y + 3 (reduced with respect to G) Let 14 := 4 y 5 + 3y4 + y2 + Y + 3 g: {{hh}. xz + y) ç Q[x.2 do form a Grobner basis with respect to lex with x > y > z > w. 14}' The third through the sixth executions of the WHILE loop give S-polynomials that reduce to zero (Exercise 1.2. Use Theorem 1. x 3 y . 7. The reader might think that G could also be obtained by first computing a Grübner basis G' for l = (f" hl viewed as an ideal in Q[x.3. Compute the Grübner basis G' for l = (f" hl ç Q[x.y3 z3 and g = xy2 + Z2.}. In Example 1.7. 1 = 3x 2 yz .] 1. a. -12 =3xy+4x+y3+ y (reduced with respect to G) 13 := 3xy + 4x + y3 + Y g:= {{J"h}'{hh}} G:={J"hh} Second pass through the WHILE loop g:= {{hh}} S(j"h) = yj. 1./4}. Find a Grübner basis for (x 2y + z. y.y.xy3 and g = xy2 + z2. [Answer: z + x 2 y.7. xz + y. z] with respect to lex with x < y < z. y].10 and 1.11 without a Computer Alge bra System.t. You should do the following exercises without a Computer Algebra System. 7. deglex. Exercises 1.7. xy2 . 1 = 3x y . y] with respect . a. and degrevlex orderings with x > y > z : a. 13. xz + y) ç Q[x. y.11 we obtained G using arithmetic modulo 5 throughout the computation.yz and g = xy2 + z" 1.z2. Find a Grübner basis for (x 2y + z. S-POLYNOMIALS AND BUCHBERGER'S ALGORlTHM 45 INITIALIZATION: G := {j" h}' 9 := {{J" h}} First pass through the WHILE loop 9 :=0 S(f" Let 12) =yj. z] with respect to deglex with x > y > z. 1. . . Of course.46 CHAPTER 1.L.. Prove that. BASIC THEORY OF GROBNER BASES to the lex ordering with x > y. 1. Assume that deg(gi) = ai for 1 <: i <: t.12). y] let J = (xZy ...y + X. In lQI[x. and compare with Example 1.} be a Grabner basis for J.. (F) has a Grobner basis consisting of differences of power products. deg(J') is represented by the shaded region in the diagram given below: exponent of y (0. Show how the steps in Gaussian Elimination (see Section 1.7..2) p'--L.6.. the Grabner basis obtained from Buchberger's Algorithm might not be unique. .2) parallel the steps in Buchberger's Algorithm (Algorithm 1. Let G = {g" .8..1). this delinition of deg depends on the term order in use. . . xy2 . Show that.7. Assume that we have lixed a term order on 1['n..{O} (see Exercise 1. In the last section we saw how to compute Grabner bases. Reduced Grobner Bases.x).7. Draw the region which represents deg(J') if we use lex with x < y. 1. with respect to any term order. b..7. b.0) exponent of x c. we obtain uniqueness. Let l be an ideal in k[x" .xnl not containing O.7.1).9. 1. In this section we show that by putting certain conditions on the polynomials in the Grobner hasis. For 0 of f E k[x" .4.11. Deline deg(S) = {deg(f) 1 f E S} for subsets S ç kixJ.xnl deline the multidegree of f by deg(f) = a where Ip(f) = X U with a E !\In.7..fs} ç k[x" . 1. with respect to the deglex ordering with x < y.. Reduce G' modulo 5 to obtain G.2) parallel the steps in Buchberger's Algorithm (Algorithm 1. Prove that t deg(r) = U(ai+!\In) i=l where J' = l . However.g. 1. . Show how the steps in the Euclidean Algorithm (Algorithm 1.( (3...8. Assume that F = {J" .xnl and each fJ is a difference of two power products. .··· ..xn ]. 7.3..7. a. ). For example./3.} are minimal Grobner bases for an ideal l. as the following proposition shows. in Example 1. a minimal Grübner basis for Jean be obtained from {f"h. {f" 12. that is. .1. /3. REDUCED GROBNER BASES 47 In Buchberger's Aigorithm there are two ploces where choices are made... and J4' We could also remove j.. . First. A Grabner basis G = {g"".). Jf Ip(g2) divides Ip(g. COROLLARY 1. we may assume that i = 1. . .gt} be a Grabner basis for the ideal J. since F is a Gr6bner basis for l. J5} for J..8.} is 'caUed minimal iJ Jor aU i.} be a Grabner basis Jor the ideal J. Now 91 is also in 1. there is the arder in which the polynomials are inputed and this affects the application of the Division Algorithm.. . J4. then it is also divisible by Ip(g2)' Therefore.} is a Grübner basis for J. we might end up with a different Grobner basis.8.7. using Definition 1. Second. 0 As a direct consequence of this lemma. To obtain a minimal Grabner basis from G. This leads ta the following definition. then {g2.J5}. Note that Even after we have computed the Grübner basis G = {f" 12.8.f4. It(f. the Spolynomial S(f" 12) would have reduced ta zero. h). and hence.2. . After renumbering if necessary. but. Sa we would have obtained a different Grübner basis. all minimal Grobner bases for an ideal l have the same number of elements. and J4' SA minimal Grabner bases are not unique. . Since ft is in 1 and sinee G is a Gr6bner basis for l. DEFINITION 1. PROPOSITION 1. t. . we now see how a minimal Gr6bner basis can be obtained from a Grübner basis.J3.3. PROOF. by removing j"h. in the WHILE loop of Buchberger's Aigorithm where we compute S-polynomials. we choose {f. LEMMA 1.. This is because any term divisible by !t(f4) = x 2 is also divisible by It(f5) = x. Let G = {g"". eliminate aU gi Jor which there exists j of i such that Ip(gj) divides Ip(gi) and divide each remaining gi by lc(gi)' In Example 1. we can observe that J4 = x 2 can be removed from G.6.4.g. Jf G = {g"". Js} is the Grübner basis for J we would have obtained had we computed S(h. gt} is also a Grabner basis Jor J.. Clearly.8. . and the same leading terms. gt} and F = {f" .1. PROOF.1.g. h. . . /3. {g2. g. .. Sa. then s = t../3) = x first. Js} is also a Grübner basis for (f" 12.). The set {f" 12.) = It(gi) Jor aU i = 1.8. if we were to change either of these choices. J. if a polynomial J is such that Ip(f) is divisible by Ip(g.. and after renumbering if necessary. lc(gi) = 1 and Jor aU i of j. and would not have appeared in the Grübner basis. .10 above. Ip(gi) does not divide Ip(gj). g} Egat random. there exists i such that Ip(gi) divides Ip(j. /3) = x before S(f" 12). Let G = {g" .10 ifwe had computed S(f2.7. . t.} " H..} is a reduced Grobner basis for J. Thus we only need to prove uniqueness. 1 h i .) divides Ip(12)... ht} be redueed Grübner bases for J. (in fact.6. .1..y. !c(Yi) = 1 and Yi is redueed wit. . BASIC THEORY OF GROBNER BASES there exists j sueh that lp(fj) divides Ip(YI)' Therefore lp(fj) divides Ip(J..) for each i = 1. H is also a Grübner basis for 1. Thus s = t and after renumbering lp(fi) = Ip(Yi) for ail i = 1.. . Now 12 is in J... His a redueed Grübner basis.8. and sinee lp(hj) = lp(gj). . Fix a term order.. it is a minimal Grobner basis). t.4.. sinee G is a Grübner basis.8. .." .ht-I}' Then H = {h" .). after renumbering if necessary. lp(h i _ l ). We now prove that redueed Gr6bner bases exist. 0 THEOREM 1. Consider the foUowing reduetion proeess: H. minimal Gr6bner bases are nat unique.. PROOF.} is caUed a redueed Grobner basis if. sinee F is a minimal Grübner basis. no non-zero term in Yi is divisible by any Ip(Yj) for any j of i. Let G = {YI. and henee j = 1. .h respect to G .} 93 ---++ 3. 0 As we mentioned after Corollary 1.hi E J implies that there exists j such that lp(hj ) divides lp(gi . lt(g. . where h.{Yi}' That is. COROLLARY 1. This proeess continues until al! f's and y's are used up.. and. If gi of hi. .. and henee there exists i sueh that lp(y. PROOF. h where h3 is redueed with respect ta H3 = {h" h 2 . Therefore. sinee a redueed Grübner basis is minimal.8. i ::..8. But then . . ). Let i be given. Sinee the division of 9i by hl. We proved in the previous result that every ideal has a redueed Grübner basis. 19t is done by eliminating tenns of gi using lp(h . . then gi .g.48 CHAPTER 1. g. Thus Ip(J.) = lp(y.8.. g. .} H.hi) < lp(h i ). Note that. Let G = {g" . g4. .) imply that i i=. . . 9t ~+ ht. . Ta get uniqueness. ...) = lt(h i ). t.. we have that lp(hi ) = lp(g. = {g2.} and H = {h" . lp(y. we need ~o add a stronger condition on the polynomials in the Grübner basis. The minimality of F and the fact that Ip(YI) = Ip(J.). we see that j of i. for aU i. As above we get that Ip(Y2) = Ip(12). .5.b 9i+1. both G and H have the same number of elements and we may assume that. h 92 ---++ 2. where ht is reduced with respect ta H t = {hll h 2. 1 ::. A Grobner basis G = {YI. . is reduced with respect ta H. h..).7 (BUCHBERGER).. for all j.. Then every non-zero ideal J has a unique reduœd Grobner basis with respect to this term order. for aU i. Note that a redueed Grübner basis is also minimal.. Ip(gi+1)"" .hi)' Sinee lp(gi . . where h 2 is redueed with respect ta H 2 = {h g3. .. DEFINITION 1. we may assume that i = 2. g.Yt} be a minimal Grobner basis for the ideal J. for each i. . We note that by Proposition 1. 91 ~+ hl.3. sinee G is a minimal Grübner basis... .8.8.8.8. We may assume that the polynomials in G are ail hOillogeneous.Js} is not a reduced Grübner basis. (g. Prove that a subset G of a homogeneous ideal I is a reduced Grübner basis with respect to the lex ordering if and only if G is a reduced Grübner basis with respect to the deglex ordering.8. Js} is a reduced Grübner basis for I. Prave thal 8((gi)h... We use the same notation as in Exercise 1. we see that {h.6. g. 1. Since h = y + x.8.. . . .) where each Ji is homogeneous (see Exercise 1. We go back to Example 1.) h.xy+3x+2y3+2y.) be an ideal of k[XI. . Let F ç: k[Xl.} be a Grübner basis for (fi'. This contradicts the fact that G and H are reduced Grübner bases.9.. J.19. .9. . . .4. J!:) with respect to <h . . We cali I a homogeneous ideal provided that I = (fI. Let I = (f" . Show that I is homogeneous if and only if for all J E I. h} is a Grübner basis for I. d. .. Show that any hOillogeneous ideal has a Grübner basis consisting of hOillogeneous polynomials. Since h = y and Js = x. x n ] be an ideal. Prave that I is a homogeneous ideal if and only if the reduced Grübner basis for I consists of homogeneous polynomials.8. Then note thal It<. . x n ]. Let G = {g" . ExercÎses 1.11.1... Compute the reduced Grübner basis for the ideals in Exercise 1. yS+2y4+4y2+4y+2} is a minimal Grübner basis for I. x n ]. Js} and {h Js} are both minimal Grübner bases for the ideal I. Js} is in fact {h.10.7...2.... Let I ç: k[Xl. Js}. J!:)) = J. 4ys+3y4+ y2+y+3}. .19. x 2y+2xy+x.1..8. So gi = hi· 0 EXAMPLE 1. 7. Find an algorithm that will determine a subset F' ç: F such that I = (F') and for which no proper subset of F' generates I.. 3xy+4x+y3+ y . REDUCED GROBNER BASES 49 Ip(h j ) = Ip(gj) divides a term of gi or hi.8. Fix an arbitrary term order on k[Xl.9). .5.3. (f)h = It(A) for every homogeneaus polynomial in k[Xl. since the reduced Grübner basis is unique. .gj)h.. Compute the reduced Grübner basis for the ideal in Example 1. By Corollary 1. 1.6.. Let us go back to Example 1. x n ]. 1. We let <h be as in Exercise 1.. y] with respect to the lex ordering with x> y is {x 2+y2+1. a. J. h. In fact it is easy to see that it is the reduced Grübner basis for I..xn. (gj)h) = 8(gi. Of course. . .8. EXAMPLE 1. . Prave that given .3 that {h.7. [Hint: First note that even without the assumption that Gis a Grübner basis we have (Gh) = (A 1 J E (fi'.3. 1. .7.. . {x 2+y2+1.w]. . . each haillogeneous component of f is also in 1.. There we showed that a Grübner basis for I = (x 2 + y2 + 1. c. x n ] and let I = (F). Prave that G h = {(g.. . Let < be any order on k[Xl' . We have seen Tight after Corollary 1.3.. x 2y + 2xy + x) ç: Zs [x.4.8. the reduced Grübner basis obtained from {h.. we see that h can be reduced to y using Js· So {h. xy-x) ç Q[x.x}.. [Hint: At each stage of Buchberger's Algorithm. = Lt(I). y3 _ y2.17 and in the second add a polynomial to JI. . (*) As mentioned before. Consider the two possible cases: either (ml. .. . G is a Grobner basis.. y]. Let T be tbe (infinite) set of ail possible term orderings on k[XI. b.. . Use c to show that every ideal has a universal Grobner basis. .] (Answer: {x .50 CHAPTER 1. . choose a term ordering which gives this leading term ideal.ms) is the leading term ideal for l with respect to one term order in 11. that is. In this exercise we show that for a given ideal tbere are only finitely many possible reduced Grübner bases.. let l be generated by {JI.. THEOREM 1.9.7. Summary. . then!h ~+ gh. (iii) f E l if and only if f --"->+ o.g E k[XI. and let 1:.xn..ms and an infinite set 'li ç Ta such that It(Ji) = mi for each term order in Ti. Conclude that tbere are only finitely many reduced Grübner bases for a given ideal. there exist terrns ml. Let l be an ideal of k[XI..9. x n ]. We conclude this chapter by giving a summary of the most important results that we have seen so far. (i) For aU f (ii) Lt(G) E l.. 1.w]. BASIC THEORY OF GROBNER BASES f. A set F which is a Grübner basis for an ideal l with respect to every term order is called a universal Grobner basis.1. d.8. Find a universal Grübner basis for the ideal (x_y2. a. is finite.] 1.. For each leading term ideal in 1:. In the first case. or it is not. .f. a Grübner basis with respect to one term ordering might not be a Grübner basis with respect to another term ordering. if f --"->+ g. = {leading term ideals of l with respect to the term orders in T}.. and repeat the argument.. there exists i such that Ip(gi) divides Ip(J). xy .6.7. use Exercise 1. is infinite.6.8.... . The following statements are equivalent for a set of non-zero polynomials G = {gl' . Prove that 1:.g. Let Ta ç T be the infinite set of these chosen term orderings.8. [Sketch of the proof ([MoRo]): Suppose to the contrary that 1:. Prove that there is a one to one correspondence between Rand c. . .x.] c. .f. . .}· Since there are only finitely many terms which appear in the fi 's. . . The first theorem lists ail the equivaient conditions that we now have for a set G = {gl. An example of such a basis is given in Exercise 1...xnJ. . Also.) 1.y2. gt} to be a Gr6bner basis.. .} and l = (G). x 2 . consider ail possible choices of leading terms.. Let R = {reduced Grübner bases for l with respect to tbe term orders in T}. . + .. ... Then every ideal l has a reduced GTobner basis with respect to this term order.7..9.Xn ].. .. 1. ·+hijtgt such that lp(8(gi.g .. (v) For aU i # j.9. SUMMARY 51 (iv) ForallfE k[XI' . we have 8(gi.2.. x n ].v:<..2.6. (vi) For all f E l.. .4..7. andr"r2 arereduced with respect ta G. then Tl = T2.. 1.. there exists h" . 8(gi. (vii) For aUi #j (1 <:: i. iff5'. ..gj) = hij. E k[XI.t(lp(hijv) lp(gv)).1..xn] such that f = h. .. This GTobner basis is effectively computable once l has been given as generated by a finite set of polynomials.6. Moreover this reduced Grobner basis for l (with respect to the given term arder) is unique.+ r2.. THEOREM ..gj) 5'. + . gj)) = max'..+ Tl.. Fix a term order on k[XI' .j <:: t). ht The proofs that ail of these conditions are equivalent are contained in Theorems 1...6..g. and Corollary 1.. and 1. + htgt and lpU) = max'<v<t(lp(hv) lp(gv)).7. f 5'.+ o. . . In this section we want to show how to perform effectively! the following tasks: (i) Given 1 E k[x!. .) be an ideal of k[x!. . (iii) Find cos~t representatives for every element of k[x!..l. for example.. Applications of Grobner Bases This chapter is devoted to giving a number of applications of the theory developed in Chapter 1 to computations in polynomial rings.. In Section 2. J of k[x!.Xn]. .. We also give sorne applications to computations which use polynomial rings. . .. e.. determining whether graphs can be colored by three colors. and if so.1 we give methods for doing basic computations in k[x!. In Section 2. determining if one polynomial is a member of an explicitly given ideal.. and ta a parametrically given variety.. The last three sections present applications of Grübner bases to problems in computational mathematics: deterrnining minimal polynomials of elements in field extensions. .xn]. .xn]/I. (iv) Find a basis of the k-vector space k[x!. . . e. ta compute generators for the intersection of ideals. we show how to find the ideal corresponding ta the projection of a variety. and finding solutions to integer programming problems. find V!. .Chapter 2..xnl/1.xn]/ I. Elementary Applications of Grobner Bases. ..1.4 we study homomorphisms between polynomial rings.. . In the next section we give a method for eliminating variables in systems of polynomial equations and use it..Xn] such that 1 = vd! + . In Section 2.g. .xn ] are equal.··· . + vsI. .. 2. In the next section we give more applications to algebraic geometry.vs E k[x!..g.2 we introduce the Hilbert Nullstellensatz and use it to conHect Grübner bases ta sorne elementary questions in alge braie geometry. . 1We Temind the readeT that by "perform effectively" or by "determine" we mean that one can give an algorithm than can he programmed on a computer. 53 .. . Let I = (h. determine whether 1 is in l (this is the ideal membership problem).. . (ii) Determine whether two ideals I. ... ..xn] and k[x!. We then generalize these results to the case of polynomial rings modulo an ideal (affine algebras). In particular we determine generators for the kernel of such a homomorphism and we give a method to determine whether it is onto. 54 CHAPTER 2. + v. First pass through the WHILE loop g :=0 S(f"h) =yh -xh = _y2+xy+x 2 (reduced with respect to G) .2) Thus Equation (2. . . . Moreover.. . .1) can be transformed to give the polynomial combination of the original polynomials h.. 80 the ideal membership question is answered. xn]/I. .g.} and let G = {g" .. 2. . ... (vi) Find inverses in k[XI' .h}. . .. We begin with Task (i).9.gt} be a Grobner basis for l = (h..1) J= U. . This procedure is illustrated in Exarnple 2. f} and sorne polynomials Wi which are exp!icitly computed in the Division Algorithm....Js.1. Thatis. i=l for sorne v. . say {h". h = xy2 . So we can obtain as an output of Buchberger's Algorithm not only the Gr6bner basis {g" . h. g.1.} but also a t x s matrix M with polynomial entries such that (2.1.. x n ]/ l when they exist. We follow Algorithm 1. . ..7.y + x. + . '" E {1.1. + .1. We use the deglex term ordering with x < y.. . EXAMPLE 2.7..1.1) for the computation of a Grobner basis. Buchberger's Algorithm can be implemented so as to keep track of the !inear combinations of the Ji'S that give rise to the gj 's.. 9 = S(hv.. We have already seen in Theorem 1. but we keep track of the !inear combinations that give rise to the new polynomials in the generating set. .7.1 below. Let h = x2y .}.g:= {{J"h}}.1 that J El<==? J --->+ O. and l = (f" h).. + Utgt· AIso. . .: J as a !inear J = Vd.. This can be seen as follows: during Buchberger's Algorithm (Algorithm 1. Let F = {h. J.. a new polynomial 9 is added to the basis if it is the non-zero remainder ofthe division of an S-polynomial by the current basis. . applying the Division Algorithm to J E l yields Ul . J. L Wihi. Ut such that G (2.x. J.) with respect to a fixed term ordering. APPLICATIONS OF GROBNER BASES (v) Determine the operations in k[XI. h~) - ..1 given in Section 1. INITIALIZATION: G:= {h. In this example we consider k = <Ql.. {h 14}} G:= {f. + x 2fa = x 3 y + x 4 ~x4_y2+2xy_x2 - y2 + xy ~ x 4 +xy-2x 2 (reduced with respect to G) Let 14 := x 4 + xy . + (_x 2 + 1)12 {{f" M. in the above computation we kept track of the !inear combinations of j. and 12 giving rise to fa and 15 and this gives us the following: (2. fa}} S(f" fa) = yj. {h 15}. {h M. {h M} S(h.xh {{f" h}. . ELEMENTARY APPLICATIONS OF GROBNER BASES 55 Let fa:= _y2 +xy+x2 Note that fa = yj. {h. M.2x Note that 15 = (h + xfa) . . M.3) N ow eonsider the polynomial .. 15}.xj. 14. fa}} G := {f" h. {h.fa = (x 2y .2x (reduced with respect to G) Let 15 := x 3 + y .2.fa) = h +xfa = x 2y+x 3 -x ~ x 3 +y.. fs}} G:= {f. M. fa.!2.1. {h. 15} ç := The reader can verify that ail the remaining S-polynomials reduce to zero. {f2. fa} := ç Second pass through the WHILE loop ç := {{h. {f" 14}. 15} is a Grübner basis for J. It is also easy to see that {f"h/s} is a Grübner basis for J. + x 2fa) . {f4. = (xy . + (_x 3 + x)h ç := {{h. and hence G = {f" 12.l. sinee It(h) is divisible by It(fa) and !t(f4) is divisible by It(f5)' In fact {f" h Id is the reduced Grübner basis for J with respect to the deglex term ordering with x < y. h 14. h14} Third pass through the WHILE loop ç := {{f" M.x)j. {fd5}.!2. {h.1)j.j. h}.2x2 Note that 14 = (yj. Moreover. + (2y . We give another illustration ofthis for k = 1':5' We go back to Example 1. with respect to the lex ordering with x > y.2x 2/5 + 2yJ.1. + (_x 2 + 1)12) (_2x 3y . y]. when we are trying to keep track of the linear cornbinations in the reduction process. and X is a term.4x 2 ---> 2y.2y + 2x .h ---> ---> + 2x -2y.3x3 + 2y2 .2xl5 + 2yh . 2It is convenient. .2x -1)((xy -1)J.4)(yJ.2x 4 . Using Equation (2.2x .2x2y + 2y2 . XnJ.3x 3 + 4y2 . 2yJ.f5 ---> ~ -2X'/3 ---> -1.2xh . where g. APPLICATIONS OF GROBNER BASES We show2 that 1 E 1: 1 :. That is.xh) +(_2x 2 .3) we have 1 = (x 2 + 2y)J.2xy + x 2 .c2.15 . . So we see that 1 x 2J. and g3 = 4/4 = y5 + 2y 4 + 4y2 + 4y + 2.y + 2x _2 y 3 +4xy2 -3x 3 +4y2 -4xy-4x 2 -y+2x 2xy2 . +(2x4 + 2x2y + 2x 3 .y + 2x 4y2 _ 4xy . and we get g2 g3 + 312 4(y + 3)g.3xy + 3x 2 + 2x + 1)J.8. It is easy to keep track of how g2 and g3 are generated during the algorithm. then f ~ h means that h = f .2y 3 + 4 xy 2 .4x2 . R.13 ---> O. h E k[Xl) . is {g"g2.2yl5 . EXAMPLE 2.15 -2x 5 + 2x2y2 _ 2x3y _ 2x4 _ 2 y 3 + 4 xy 2 .2xy2 .2x2y + x 3 .y + 2x 2x2y2 _ 2x 3 y _ 2x 4 .413 (x 2 + 2y)J. = x 2 + y2 + 1 and 12 = x 2 y + 2xy + x.4xy .Xg.2x .fl ---> _2x2.2. + 2(3x + 4y2 + 3y)g2 (2xy + y3 + 2y2 + 4y + 2)J. = Il> g2 = 213 = xy + 3x + 2y 3 + 2y.6xy .2y . . + (3x + 4y2 + 3y)/2.fs -2X. + (2y . g.2y 3 + 4 xy 2 .9.56 CHAPTER 2.1)/2.3x 3 .1)fs.4xy . to include in the notation the terrn by which we multiply the polynomial we are using for reduction.y + 2x -2x 3y .2y .4x 2 ..2x .ecall that the reduced Grübner basis for l = (f" 12) c:: Z5[X.g3}.2x2y .4)13 + (-2x 2 . where J.y + 2x _x3 + 4y2 . .y _2x 4 _ 2y 3 + 4 xy 2 .3x 3 + 4y2 . . if J.1.2xy .]5 ---> -4. . .1.. .] = [ ~ g3 [1 ][ft] ~ 3 f' 2xy+y3+2y2+4y+23x+4y2+3y 2 0 The second task.15) and so the map Ne: k[x" . .. Altematively.h E k[x" .9 = (f . The clement r above is called the normal form oJ J with respect ta G. I = (f" . 9 E k[x" . such that J -S+ r (Theorem 1...4.xnJ/I. there exists q E I such that J = q + Ne(f). J. g.1. . Since -y + 2x is reduced.1.. In particular.3.7).Ne(ft) + c2Nc(h)) E I and c. .8.xnl if and only if the reduced Gr6bner basis for I is {1}.Ne(f)) . determining whether two ideals 1. PROPOSITION 2. cd. Thus J + I = Ne(f) + I in k[x" . We know that for all J E k[x" . PROOF.xnl/I.. xnl ----. + c2h) = c.... . since q E I. we have Ne(x 3 ) = -y + 2x.xnl ----. Allio. 2y ..1. we may simply check whether I ç J and J ç I.xn ]} is a set oJ coset representatives Jor k[x" .. . xnl is k-linear...xnl/I. . finding caset representatives for every element of k[x" . Conversely.} is a Gr6bner basis for I. so that J . .xnl. and is denoted Ne(f)..Ne(ft) + c2Ne(h) (see Exercise 1. X n ]. . . so Ne(f) = Ne(g). ELEMENTARY APPLICATIONS OF GROBNER BASES and 80 57 g. reduced with respect to G. I = J if and only if I and J have the same reduced Gr6bner basis. Thus Ne(f) = Ne(q) + Ne(g).... J are equal. Also. xnJ there exists a unique element r E k[x" .. Let J. is a consequence of Theorem 1.. We now consider Task (iii)) that is. Therefore Nc(cd.x. . and for any ft.Ne(ft) + c2Ne(h) is reduced with respect to G. .6.7. .. We keep the notation from the beginning of the section: I = (G). Then J == 9 (mod I) iJ and only iJ Ne(f) = Ne(g).. .6. if Ne(f) = Ne(g). From the Division Algorithm. Now J == 9 (mod I) if and only if there exists q E I such that J = q + g. . Moreover. ..Ne(f) E I. k[x" . to determine if l = J..) ç J if and only if ft. 80. xnl. where G = {g" . then J . k[x" ... E J. ThereJore {Ne(f) 1 J E k[x" .. We go back to Example 2. That is. we note that for a given ideal I.. for any C"C2 E k.5. we have that I = k[x" . .. . and we know how ta determine whether this is tIue. J.Nc(g)) E I and hence J == 9 (mod I). DEFINITION 2..1.2. But Ne(q) = 0. 2 j. 0 EXAMPLE 2.1.xnJ is klinear.. x Y + Y ----.(g .. We note that x 3 ~ -y+2x. + c2h(c. the map Ne: k[x" . . 93}.x. PROPOSITION 2. x.. We are now able to complete Task (v). we have Nc(xy') = x and so (y + I)(xy + I) = x + I. the cosets of such power products are linearly independent by the uniqueness of the normal form. So a basis for Zs[x. .92.y + X. and 93 = yS +2y4 +4y2 +4y+ 2. y'. We have seen that for any f E k[x" .1. (2x' +y)(3xy . t. x n!/ I.9.2x}. = x 2y + 2xy + x.5y 0= 6(xy . x 3 + Y . . . and x is reduced with respect to C. . A Grübner basis for I with respect to deglex with x < y is G = {x'y .) c:: Zs[x.5y + 3x + I..2.16x 2 .. yl/ I. We go back to Example 1. We keep the same notation as above.6. .xy and so diIllQ(IQJ[x.1... PROOF. and 50 dimz. .xnl/ I consists of the cosets of aU the power products X E 1rn such that Ip(9i) does not divide X for aU i = 1. So a basis for lQJ[x. for example. We go back to Example 2. x 1 x y x' xy 1 1 X y x' xy x x x2 xy -y+2x y-x Y Y xy xy+x' y-x x x2 x' -y+2x y-x -xy+2x' xy ..9. The other products are computed in a similar fashion and we obtain the following multiplication table for the representatives 1. 2.x'.. y3. by the definition of reduced. Since Nc(J) is reduced with respect to G. that is. we go back to Example 2. yl/ I) = 6. f + I = Ne(J) + I in k[x" . where fI = x' + y' + 1.xy of the Q-basis {l+I.7.1. y! / I. Xn].5) = 6x 3y -lOx 2 + 3xy: . EXAMPLE 2.1. that is we can now give a multiplication table for k[x" .y+I. since Ne (x 3) # Nc(x'y + y). APPLICATIONS OF GROBNER BASES Sinee 2y-x is reduced.xn!/I.1. a k-linear combination of power products X E 1r n such that lp(gi) does not divide X for all i = 1. . The representative of the coset y + I times the coset xy + I is the normal form of xy2.xy+I} for lQJ[x.x 2 )_ lOx'+3x-5y = 6xy-16x 2 -5y+3x (mod I) andso (2x'+y+I)(3xy-5+I) = 6xy . Moreover.yl/I consists of the CQsets of I. it is..x 2 xy xy y-x x xy . Again. 7 and give a multiplication table for IQJ [x. we wish to find a basis of the k-vector space k[x" . yl/ I) = 5. t.x'+I.8.x+I. 0 EXAMPLE 2. .y. xnl/ I.1. . with respect to the lex ordering with x > y. where 9' = fI. _y' + xy + x 2 . Finally. . is {g. y4. . (Zs[x.. 9' = xy + 3x +2y 3 +2y.y].58 CHAPTER 2. .x.. we have Ne(x'y+y) = 2y-x.. and j.j. The next task we want to consider is Task (iv). Since xy2 ~+ x.y.. yl/ I consists of the cosets of l.1.. A basis for the k-vector space k[x" .8.x' x2 So. y. Recall that the reduced Grübner basis for I = (fI. The representative of the coset of f times the coset of 9 will be the normal form of f 9· EXAMPLE 2.x 2. we see that x 3 ot x'y + y (mod I). 10. We illustrate this in the fol!owing example.. + h. sinee the cosets of 1. and. xnJ/ I and to compute that inverse. EXAMPLE 2..x) + axy + b(y . we first find a reduced Grübner basis H for the ideal (J" . . . in fact. if so. Hence (-2xy ... since Jg .2.1. f) is. we consider Task (vi). x.1 E I if and only if 1 E (I. given a k-basis and the multiplication table. and e = 1.. + cy + dx + e)(y + x + 1) '" 1 (mod I).xn ].xnJ/I must have a finite k-basis. (axy + bx' + cy + dx + e)(y + x + 1) + bx 2 + cy' + cxy 2 +cy + dxy + dx + dx + ey + ex + e ax + a(y . we want ta compute that inverse. we ean express 1 as a !inear combination of h. al! of k[XI'''' .xnJ/ I has an inverse and. . .b = -l.xnJ/I. y. d.x) + b( -y + 2x) + bx' +c(xy + x') + cxy + cy + dxy + dx' + dx + ey + ex + e (mod I) (a + 2c + d)xy + (b + c + d)x' + (a + c + e)y + (b + d + e)x + e.x' + y + 1) + I is an inverse of y +x + 1 +I in Q[x. . to determine if J + I has an inverse in k[XI" .J.'!). b. . Of course if we had started with an element of Q[x.d = 0. x'. these equations would have had no solution.1 again.xn ]/ I if and only if the ideal (I. in the case when f + l has an inverse. + gf. Of course. yJ/I. is to recognize that J + I has an inverse in k[XI"" . yJ/ I that did not have an inverse. . we want ta determine whether an element J + I of k[XI...e = 1.7). given an ideaJ I = (J" . If H oJ {1}.xn ]. as in the solution to Task (i). that is. f).. If H = {1}. So we need to find a. determine its inverse. yJ/ I. then J+I does not have an inverse in k[XI"" ..1. An alternative approach to the method used in Example 2. e E Q sueh that (axy + bx' Now.··· . Using Example 2. + .. These equations are easily solved to yield a = -2. . this problem translates into an exercise in linear algebra provided the k-basis is finite (see Theorem 2. then. 1= hd. 80 (axy + bx' + cy + dx + e)(y + x + 1) '" 1 (mod I) if and oruy if axy' + ax'y + axy + bx'y + bx3 j a+ a +2c+d + c + d + c b + d b + + e e e o o o o 1. Thus. .) and a polynomial J E k[XI. which does not suffer from the defect that k[XI'" . c.J.. xy form a basis of the Q-veetor space Q[x.J" J.1.1. ELEMENTARY APPLICATIONS OF GROBNER BASES 59 Ta conclude this section. we would !ike to determine whether y + x + 1 + l is invertible.10.J.2. + (2x 2 .9.6. 16). Let I ç k[Xl"" .4. In Q[x. 15. 2. for a E Q.8.4/6 .3. We go back to Example 2. a. S(f" 17). Show that diffiQ(Q[x.1. The S-polynomials S(f5.5x 2 .1.1.1. xz . Using the multiplication table of Example 2.y]/I.4x 3 .y 2 +x). . S(h. y. Exercises 2. Show that in Q[x. 2.1. and S(h. Je l.5) ç Q(x )[Yl' Y2]' Note that {yr .1. 2. 2.1.3.2yz. giving us the inverse (2x 4 + 2x2y .yz). In Example 2.1 show that 1 E J and write 1 as a linear combination of fI. z]/ (y4 + 3y2 Z + z2. y] such that g(O) = we have Ig E l. keeping in mind that the field is Q(x). x+I.217 + (2x 2 . 1 + J is a zero divisor in ° . S(f6. We fust compute a Grübner basis for the ideal (f" 13. z]. y2 + J. 2..3.1. Compute the multiplication table for Example 2.1. For those that do have an inverse find it. 17). 15 and also as a linear combination of f" h.10. 18) ail At this point we stop.8. Let 1 = -1 +x2 +xy.y.1. y. where I = (X 2 +y. [Hint: Note that it suffices to show that xl.5)/8 (-2cC + 3)f.1. keeping track of the multipliers as we did in Example 2.4x 3 . sinee all other reduee to zero.x 3 .C' [Hint: Consider the ideal 2.5.1.10.2. y~ . Rationalize the denominator of x+v 3 +?25 I = (yr .1.1..11.1. Let 1 = xy4 + 2x 3y 2 .1. or I = J. 13. S(f" 18).xn ] be an ideal.1. Letting 16 = y+x+ 1 we compute that S(f. x 2 + Z.5} is a Grübner basis for l.f6) --->+ -x 2 -x = 17.60 CHAPTER 2. S(f5' 18) --->+ S-polynomials must reduce to zero using the polynomial Working backwards we compute 1.5)13 + (-2x 2 + 7)/5 +(2x 4 + 2x2y . yi El. let I = (x 2 + Z. the coset xy+y+a+I. Show that for ail 9 E k[x.5y + lOx + 1) + 1.7.9. Consider Example 2..9 determine which of the cosets.y3 . APPLICATIONS OF CROBNER BASES The polynomial 9 is then the inverse of 1 modulo J. S(f5' 17). EXAMPLE 2. x 3 . y~ .5x 2 . 2+X+y2 +I has an inverse. xy + y2 + Z. .xn ].. 1. 1 4fs . 1 M . xy + y2 + z)) = 00.] 2. y4 + 3y2 Z + z2) and J = (x 2 + z.1. has an inverse if and only if a of 0.] 2.1. Determine which of the following (if any) are true: le J.8 we readily see that this is the same answer we obtained in Example 2. xy + y2 + Z. Devise a method similar to that used in the first solution of Task (vi) for determining whether for 1 E k[Xl' ' .5y + lOx + I)Js. y+x+ 1) with respect to the deglex order with x < y. Finally.xy2 + 2x2y .1. You should do this exercise without the aid of a Computer Algebra System. 16) --->+ -2x-l = fs. In Example 2.10.1. Follow the technique used in Example 2. z].y) = {(O. .1. Showthat l isaprime ideal (that is.x 2y+z).xn ]} ---+ ---+ S { Subsets of Kn} V { Subsets of Kn} VK(S) ---+ >---> {Ideals of klxl. EXAMPLE 2.'.. as in Section 1. the field K will affect the properties of the maps above. in klxl.. that is. The purpose of this section is ta analyze this correspondence further. we saw that there was a correspondence between subsets of klxl. we define the variety. In QIX. J(V). (It makes sense to evaluate f E k[xI..O)} ç JR2 On the other hand for K = <C we have that Vc( X2 + y2) is the union of the two lines y = ±ix. n. .2.. ' .2.2) { Subsets of klxl.. We need to expand somewhat the notions given there.1. the generators for l form a Grübner basis.. Qlx..an) E K n sinee k ç K) Also.. . where i = while Vc(x...2. . . then J is the set of all multiples of -1 + X2 + xy + l by elements of k. . . given a subset V ç Kn we define the ideal.zl/I contains no zero divisors).) ç klxl.l 1 2. Hilbert Nullstellensatz. y. . In Section 1.1. . Let K be an extension field of k. (Recall that in a commutative ring A. VK(S). HILBERT NULLSTELLENSATZ 61 kIXI. .Xnl/J. For K = Rwe have V.xnl then We emphasize that the variety is in Kn and the ideal is in klxl. 2. . Xn] at a point (al)'" . ..xnl.1... . Show in Example 2.. in' Kn by We note that.(x. K is a field such that k ç K Given a subset S ç klxl.xn ]} I(V).2.xnl by l 80 now we have the correspondences (2.10. IHint: Note that for lex and z > x > y. . Show in Example 2.. a E A is called a zero divis or provided that a oF 0 and there is a (J oF 0 in A such that a(J = O.2. .. .1. This is illustrated in the following two examples..J. c.(x 2 +y2) = V.y.8 that if J = {g + J E klxl.) b.xnl and subsets of kn .1. ...2.xnl.1) and (2.. let 1= (x+y2. The reason for introducing this extended notion of a variety is that the set of solutions of a system of equations depends on the field K That is. .xnl/l (g + 1)(X2 + 1) = O}. y) is still {(O..O)}.8 that xy + J is a zero divisor. . . if l = (f" . x n ] we define the radical of l.xn ]. but are "essentially different".. the equation f = 0 has a solution in K.. The situation is sirnilar for f = x 4 + y4 + 1. VK(I) = VK(. for ail fields K ::2 k.2) with K = k. "essentially different" ideals will give rise to different varieties.Ji = {f E k[x" .Ji. as we see in Theorem 2.1 and 2.. Recall that a field K is algebraicaUy closed if for every polynomial f E K[x] in one variable. For example.xn ].2. . denoted .(I) =0 if and only if I= k[x" .. We will not include the proofs of these theorems.2. THEOREM 2. by .2. APPLICATIONS OF GROBNER BASES EXAMPLE 2.e. we consider the algebraic dosure of the field k.2.3 and 2. Examples 2.2.2. the algebraic closure of 1ft is <C. . x n ]. It is easily checked that . . By enlarging the field IR to <C.Ji give rise to the same variety. The same situation occurs in Example 2. Then for K = IR we see that Vil< (f) = 0. This field is unique up to isomorphism and is called the algebraic closure of k (see [Hun.. xn]1 there exists e E ft such that rEl}.2. Moreover. that is. In order to state the N nllstellensatz. Lan]).1. For an ideal I of k[x" .5 (STRONG HILBERT NULLSTELLENSATZ).2. Consider the polynomial f = x 2 + y2 + 1.. of an ideal. Before we go to the next forrn of the Nullstellensatz..5. .2. Let I be an ideal containedink[x" .3 (WEAK HILBERT NULLSTELLENSATZ).2 show that a system of equations may have "too few" solutions in kn to give us insight into the algebraic and geometric properties of land VK(I). Note that the result is clear for one variable sinee the field li is algebraically closed. Then V. the variety.2..Ji is an ideal in k[x" . y] whose corresponding variety in IR2 is empty. Every field k is contained in a field k which is algebraically closed and such that every element of li is the root of a non-zero polynomial in one variable with coefficients in k.2.1) and (2.3. denoted k.2. In fact there are infinitely many ideals of IR[x.Ji). Hun]. The interested reader can find thern in [AtMD.. The Hilbert Nullstellensatz has many forms and we will present two of them below in Theorems 2. it gives information about the zero set. For the remainder ofthis section we will consider the correspondences (2. we have that I and .(I)) = . .2..2. The theorem is given this name because.xn]. we need a definition. . In Example 2.4... I(V . whereas Vc (f) has an infinite number of points.. This will be clarified later and the key result for this is the Hilbert Nullstellensatz 3 given below.62 CHAPTER 2. the ideals (x 2 + y2) and (x..Ji for aU ideals I of k[x" . 3The ward "Nullstellensatz" is a German word for "zero point theorem". y) give rise to the same variety over lR.2. THEOREM 2. DEFINITION 2. i. 2.2. HILBERT NULLSTELLENSATZ 63 Theorem 2.2.5 implies that two ideals J and J correspond to the same variety, i.e. Vk-(I) = V:;;:(J), if and only if their radicals are equal, i.e. ..fi = ..(J. This allows us to make more precise what we meant earlier by "essentially different" ideals. Two ideals are "essentially different" if and only if they have different radicals. In Example 2.2.1, note that ..((X:Y) = (x, y) and (X2 + y2) = (x2 + J y2) and 80 are "essentially different" and correspond to different varieties in te 2. We now consider some applications of the above results. Let J = (f"", , f,) be an ideal of k[XI,'" ,xn], and let G = {g" ... ,g,} be the reduced Grübner basis for l with respect to a term ordering. THEOREM 2.2.6. V:;;:(I) = 0 if and only if 1 E G. (i.e., given polynomials ft, ... , f" then there are no solutions to the system ft = 0,12 = 0, ... , f, = 0 in lin if and only ifG = {l}.) PROOF. By Theorem 2.2.3, V:;;:(I) = 0 if and only if 1 E J. But the last condition is equivalent to G = {l}, since G is the reduced Grübner basis. 0 THEOREM 2.2.7. The following statements are equivalent. (i) The variety V:;;:(I) is finite. (ii) For each i = l, ... , n, there exists j E {l, ... , t} such that Ip(gj) for some vEN. (iii) The dimension of the k-vector space k[xJ, ... , xnl/J is finite. = xi PROOF. (i) =? (ii). Let V:;;:(J) be finite. If V:;;:(I) is empty, then, by Theorem 2.2.3, J = k[xJ, ... , x n ] and hence G = {1} and (ii) is trivially satisfied. So we may assume that V:;;:(I) is not empty. Fix i E {l, ... ,n}. Let aij,j = l, ... ,f be the distinct ith coordinates of the points in V:;;:(I). For each j, 1 ::; j ::; R, let 0"" 1; E k[x;] be such that 1;(aij) = 0 (this can be done by the definition of Ii). Let f = fth···f, E k[Xi] ç k[xJ, ... ,xn ]. Then we see that f E J(V:;;:(J)), and hence, by Theorem 2.2.5, there exists e such that E J. Since Ip(fe) = for r x;= sorne natural number m, and sinee the leading power product of every element of J is divisible by the leading power product of sorne element of G, there exists a polynomial in G whose leading power product i8 a power of true for every i = 1, ... ,n. Xi alone. This is (ii) =?(iii). We saw in Section 2.1 that a k-basis of k[xJ, ... , xnl/ J is the set of cosets of power products reduced with respect ta G. Since for every i E {1, ... ,n} a power of Xi is a leading power product of sorne 9j' there are only fiilitely many power products which are reduced with respect to G, and hence dimk k[xJ, ... , xnl/ J is finite. (iii)=?(i). We will show that for any i = l, ... , n, there are only finitely many distinct values for the ith coordinate of points in Vk(I). Fix i E {l, ... , n}. Since, byassumption, k[Xl'" . ,xnl/ lis a finite dimensional k-vector spac€, the powers 1, Xi, of Xi are linearly dependent modulo J. Therefore there is an integer xr, ... 64 CHAPTER 2. APPLICATIONS OF GROBNER BASES Cj m and constants E k, 0 :::; j :::; m, not aU zero, Bueh that :Lcjxi E J. j=O m Since the above polynomial can only have finitely many roots in k, there are only finitely many values for the ith coordinates of the points of V.(I). D An ideal loi k[x" ... ,xnl that satisfies any one of the equivalent conditions in Theorem 2.2.7 is called zero-dimensional. This terminology is adopted because V.( I) consists of only finitely many points. EXAMPLE 2.2.8. In Example 2.1.1 we saw that a Grübner basis for the ideal l = (x 2 y - y + X, xy2 - x) in <Qi[x, yi with respect to the deglex terrn ordering with x < y is G = {x2y - y + x, _y2 + xy + x 2,X 3 + y - 2x}. We see that x 3 and y2 appear as a leading power product of elements of G, and hence V.(I) is finite. In fact it is easy to solve the equations to get Vij"(I) = {(O, 0), (a, -1), (-a, 1), (a', -1), (-a', 1)}, where a and a' are the roots of the equatjon z2 - Z - 1 = O. We note that a Grübner basis for l with respect to the lex term ordering with x < y is G' = {x5 - 3x 3 + x, y + x 3 - 2x}, and again we have that sorne power of x and sorne power of y appear as leading power products of elements of G'. These equations may easily be solved to yield the same answer as above. EXAMPLE 2.2.9. We go back to Example 2.1.9. Recal! that the reduced Grübner basis for l = (f"h) ç Z5[X, y], where j, = x 2 + y2 + 1 and 12 = x 2y + 2xy + x, with respect to the lex ordering with x > y, is {9"g2, 93}, where 9' = j" 92 = xy + 3x + 2y 3 + 2y, and 93 = y5 + 2y 4 + 4y2 + 4y + 2. We see that x 2 and y5 appear as leading power products and hence Vz, (I) is finite. We note that not aU of the solutions are in Z5; sorne are in the algebraic closure 25 of Z5. EXAMPLE 2.2.10. As a third example we again let k = Ql and consider the intersection of the circle j, = (x - 1)2 + y2 - 1 = 0 and the ellipse 12 = 4(x 1)2 + y2 + xy - 2 = O. Using the lex term ordering with x > y we see that the Grübner basis for the ideal (f" 12) is {9"g2}, where g, = 5y4 _3 y 3 -6y2+2y+2 and 92 = X - 5y 3 + 3y2 + 3y - 2. Since Ip(9') = y4 and Ip(92) = x, we see that Theorem 2.2.7 implies that the number of points in the intersection is finite. Also, clearly 9' = 0 has at most four solutions and for each solution of g, = 0 we get precisely one solution of 92 = O. Thus we see in this case the geometrically obvious fact that the intersection of a circle and an ellipse can consist of at most four points. We note that in the last example the form of the Grübner basis was particularly convenient for determining the points in the variety. That is, the first polynomial contained only the y variable, and the leading power product of the second polynomial was a power of x. We will nQW show that, in the case of zerodimensional ideals, this type of structure in the Gr6bner basis is always present when the lex term ordering is used. 2.2. HILBERT NULLSTELLENSATZ 65 COROLLARY 2.2.11. Let l be a zero-dimensional ideal and G be the reduced Grobner basis for l with respect ta the lex term arder with Xl < X2 < ... < xn· Then we can arder 91, ... ,gt such that 91 contains only the variable Xl, 92 contains only the variables Xl and X2 and Ip(g2) is a power of X2, g3 eontains only the variables X" X2 and X3 and Ip(g3) is a power of X3, and sa forth until gnPROOF. This follows immediately from Part (ii) in Theorem 2.2.7. That is, we may reorder the gj sueh that lp(gj) is a power of Xj. It then follows, because of the lex ordering, that the only variables that may appear in gj are Xl,X2,··· ,Xj. 0 We see that the Grobner basis for a zero-dimensional ideal l is in "triangular" form (this is similar ta the row echelon form in the linear case). Thus, in order ta solve the system of equations determined by a zero-dimensional ideal l, it suffices ta have an algorithm ta find the roots of polynomials in one variable. That is, we first solve the equation in one variable 91 = O. For each solution a of g, = 0, we solve the equation g2(0<, X2) = o. We continue in this manner all the way until gn = O. The solutions obtained in this way are the only possible solutions. We still have to test them in the equations 9n+l = 0, ... ,9t = 0 (in the case when t > il) in arder ta obtain the set of solutions of the full system of equations. The techniques for finding the roots of polynomials in one vaxiable are not part of the theory of Grübner bases. The interested reader should consult [Cohl. These ideas will be illustrated in the following example. EXAMPLE 2.2.12. Consider the ideal l = (Z2Y+Z2, x 3y+x+y+ 1, z+X 2 +y3) in lQi[x, y, z].We computé the reduced Grübner basis G for l with respect ta the lex ordering with X > Y > z. We get G = {z4 - z3, y" + 3y 8 Z - 2y 7 4y4z +y3 + y2 + 2y+ z3 _ z2 + z + 1,x 2 +y3 + z,yz2 +z2,xy +x +y7 + 2y4z _ y3 _ Z2 _ z,XZ+y'O _ y9 +y8 + 3y 7z _ y7 _ 2y 6z _ y6 +2 y 5z +y5_ 2y4z - y4 _ 2y 3z +y3 + y2z _ yz +y - z3 + 5z2 + z + 1}. Sa using the notation of Corollary 2.2.11, we have g, = Z4 - z3 is a polynomial in z alone. Also, g2 = y" +3 y 8 z - 2y7 _4y4z + y 3 +y2 +2y+z3 _Z2 +z+ 1 is a polynomial in y and z alone whose leading power p~oduct is Ip(g2) = yll Finally g3 = x 2+y3+ z is a polynomial in x, y, z whose leading power product is Ip(g3) = X2. SO ta find the solutions of the original set of equations, we first note that z = 0 or z = l. Then in order to find the correspondîng y values, we would have ta solve the llth degree equations g2(y, 0) = 0 and g2(y, 1) = o. We continue this way as described above. 4For the remainder ofthis chapter and following chapters, Grobner basis computations will, most often, not be done explicitly in the text and will often require the use of a Computer Algebra System. The reader who wants to verify the computations stated in the text should avail themselves of such a system. The authors usually used CoCoA, but other systems could have been used for sorne of the computations. This is discussed in the Appendix. 66 CHAPTER 2. APPLICATIONS OF GROBNER BASES We have seen in Theorem 2.2.5 the importance of computing VI. For zerodimensional ideals we will show how to do this in Exercises 2.3.23 and 2.3.24. But this is a difficult task, in general, which is beyond the scope of this book. The interested reader should consult [EHV, GTZ]. However we can now give an easy criterion for membership in 0. THEOREM 2.2.13. Let J = (J" ... , f,) be an ideal in k[x" .. . , x n ]. Then f E VI if and only if1 E (J" ... ,f"l-wf) indeterminate. ç k[x" ... ,xn,w], where w is a new PROOF. By Theorem 2.2.5, VI = J(V.(I)), and hence f E VI if and only if f(a" ... ,an) = 0 for aU (a" ... ,an) E V.(I). Let f E VI. If (a" ... ,an,b) E V.( (J" ... ,f" 1 - w f) ), then fi(a" ... ,an) =Ofor all i= 1,2, ... ,8 and 1-bf(a" ... ,an) =0. But then (a" ... , an) E V.(I), and hence f(a" ... , an) = 0, which is a contradiction. Therefore V.( (J" ... , f" 1 - w f)) = 0, and by Theorem 2.2.3, we have 1 E (J" ... ,f"l-wf). Conversely, let 1 E (J" ... ,f"l-wf). Then 1= L hdi + h(l - wf), i=l , for sorne hi,h E k[x" ... ,xn,w]. Then for every (a" ... ,an) E V.(I), we have l=(l-wf(a" ... ,an))h(a" ... ,an,w). Note that the right-hand side is a polynomial in w. If f(a" ... , an) # 0, then we can set w = f(al,:' ,a,.,J ta obtain a contradiction. Therefore f(al, ... ,an) = 0, and so f E VI. D Sa the radical membership question can be answered by deciding whether 1 is in an ideal. Thus, as we showed in Section 2.1, to decide whether f is in VI, we first compute a reduced Grübner basis G for the ideal (fI, ... , f" 1- wf). If 1 E G, then f E VI, otherwise, f 'le VI. EXAMPLE 2.2.14. Let J = (xy2 + 2y2, x 4 - 2X2 + 1) be an ideal of lQI[x, y]. We would like ta determine whether f = y - x 2 + 1 is in VI. SO let us consider the ideal (xy2 +2y2,x 4 _2X2 + 1, 1-w(y-x2 +1)) in the ring lQI[x, y,w]. A Grübner basis for this ideal with respect to the deglex term ordering with x < y < w can be computed to be {1}, sa that f is indeed in VI. Since f E VI, we know that E J for sorne e, and we may want to determine the smallest such e. To do this we first compute a Grübner basis for J. For example, with respect to the deglex term ordering with x < y, wc have G = {y2,x 4 - 2x2 + 1}. We use this Grübner basis to compute the normal form of J' for i = 1,2, ... until the first time that normal form is zero. For example, we can compute that Nc(J) = f # 0, Nc(F) = -2yx 2 + 2y # 0, but NC(J3) = 0, so that f3 E J. r 2.2. HILBERT NULLSTELLENSATZ 67 Theorem 2.2.13 gives a method for determining whether two ideals I and J have the same radical and therefore correspond ta the same variety in li n. Let I = (h, ... , J,) and J = (91, ... ,9')· Using Theorem 2.2.13 we can decide whether each Ji is in ,fj. If so, then I ç ,fj and hence .,fi ç ,fj. The reverse inclusion is checked similarly. EXAMPLE 2.2.15. LetI = (x2z2+x3,xz4+2x2Z2+x3,y2z_2yz2+z3,x 2y+y3) and J = (XZ 2 +X 2 , yz2 _Z3, x 2 y - x 2 z, y4 _x 3 , X 4 Z_X 3 Z, z6+ x 4, X 5 _X 4 ) be ideals in Q[x, y, zr. The reader can easily verify by the method above that .,fi = ,fj. Exercises 2.2.1. Consider the system of equations over <C X'Q - x 'Q { X'2 _ + 51x 4 + 51x 4 'Q 9x + 32x8 22x" 22x" 48x 2 + 18 30x 2 + 18 57x 6 + 51x4 -18y 18z - 18x 2 O. 2.2.2. 2.2.3. 2.2.4. 2.2."5. 2.2.6. 2.2.7. Obtain an explicit solution involving ( and (2 where 1, (, and (2 are the three cube roots of unity. [Hint: Use lex with x > y > z. There are 11 solutions.] Use Lagrange Multipliers to maximize the function J = x 2 + y2 + xy subject to the constraint x 2 + 2y 2 = 1. (At least find explicitly the 4 points where the maximum could accur.) Show (using a Computer Algebra System) that the function J(x, y, z) = (x 2 + y2)(X 2 + y2 - l)z + z3 + X + Y has no real critical points (i.e. places where the three partial derivatives vanish simultaneously). In Q[x,y,z], let I = (X 4 y2 + Z2 - 4xy3z - 2y5 z ,x2 + 2xy2 + y4). Let J = yz - x 3 • First show that J E .,fi. Then find the least power of f which lies in I. Verify the assertions made in Exarnple 2.2.15. Show that the following are equivalent for an ideal I ç k[Xl, . .. , Xn]. a. l is zero-dimensional. b. For ail i, 1 ::; i ::; n, there is a polynomial f E I such that J contains only the variable Xi' Let I be a zero-dimensional ideal of k[Xl' ... ,Xn]. Corollary 2.2.11 gives one way ta compute the monie generator of 1 n k[Xi]' In this exercise we present a more efficient way ta compute this polynomial, for i = 1, ... 1 n. We will assume that we have a Grobner basis G for I with respect ta sorne term arder (any order will do). This method will not require any new Gr6bner basis computation, instead it will use simple techniques of linear algebra applied ta the vector space k[xj, ... , xnl/ I. a. Let m be least such that {l + I, Xi + I, . .. , Xi" + I} is a set of linearly dependent vectors in k[xj, ... , xnl/ I. Let L;;'~o av xi '" 0 (mod I). Prove that f = a~n (L;;'~o av xi) is the manie generator of I n k[x,j. 68 CHAPTER 2. APPLICATIONS OF GROBNER BASES In view of this, we need a method for determining whether {1+ J, Xi + + I} is linearly dependent in k[XI, ... , xn ]/ J and: if so, for finding a linear combination 2:::=0 avxi in 1. Consider m + 1 new variables Yo, YI,··· , Y= and the polynomial 9 = "L,':~o YvNc(x'[) E k[Xl1'" 'X n1 Yo,··· ,Yrn]. We view 9 as a polynomial in k[Xl1'" ,xn ] with coefficients in k[yo, ... , y=] (which we note are alilinear), and we let J be the ideal in k[yo, ... , y=] generated by the coefficients of g. b. Prove that {l + J, Xi + J, ... , xi + J} is a linearly dependent subset of k[XI, ... , xnl/J if and only if Vk(J) "" {O}. c. Prove that if (ao, ... , a=) E V.(J), then f = "L,':~o av xi E J. d. Use the above to give an algorithm that inputs a Griibnerbasis G for J and outputs the monic generator of J n k[Xi]' e. Use the algorithm above to find Jnk[x] and Ink[y] in Examples 2.2.8, 2.2.9, and 2.2.lO. Verify your answers by computing the appropriate Griibner bases. 2.2.8. (Faugère, Gianni, Lazard, Mora [FGLM]) Let l be a zero-dimensional ideal of k[XI, ... , x n ]. Let G I be a Griibner basis for l with respect to a term order <1> and let <2 be another term order. The technique presented in Exercise 2.2.7 can be used to compute a Grobner basis, G2 , for l with respect ta <2 using only linear algebra. a. LetXI , ... ,Xr bepowerproductsink[xl, ... ,xn]. Use the technique of Exercise 2.2.7 to give a method for determining whether {XI + 1, . .. , Xc + I} is linearly dependent or not. b. Assume that all the the power products in k[XI, .. ' , x n ] are ordered using <2 as follows J, ... , xi Modify a to give a method for deciding whether there exists a polynomial f in l whose leading term with respect to <2 is Xc. c. Use the above to find an algorithm that inputs a Grübner basis, G I , for l with respect to <1 and outputs a Griibner basis, G 2 , for l with respect to <2 . (Note that any power product X E k[XI,'" , x n ] is either reduced with respect to G 21 a leading power product of sorne polynomial in G2 , or a multiple of the leading power product of sorne polynomial in G 2 • Moreover, eventually in the algorithm, every power product not yet examined will be a multiple of a leading power product already generated.) d. Why does the method of this exercise not work for ideals which are not zero-dimensional? [Hint: Think about the stopping condition in c.] e. Use this algorithm to compute a Griibner basis for l in Example 2.2.8 with respect to degrevlex with x < y. . . for example. X" . 69 = l..3. Moreover. In the previous section we saw the advantage of computing Grübner bases with respect to the lex term order. In this section we present a far reaching generalization of this idea.xnl/I)::.1 is a term arder. let /-li 2.Ym}.. If the orders <x and <y are lex term orderings. Provethat dimk(k[xI. . For i be the degree of the monie generator of 1 n k[Xi].. Consider two sets of variables {Xl. [Bu83.2. the order within the two sets is unimportant. and let < be an elimination arder with the x variables larger than the Y variables. a... DEFINITION 2..Ym] is a Grobner basis for the ideal 1 n k[YI. . . .3...l. Elimination. ELIMINATION 2.2. EXAMPLE 2. y variables such that one of the Xi appears ta a positive power in Z. . ..2. It is a fact that computations using the lexicographie term ordering are slow (see...Ym. 1'11'2···l'n. = X 2 and Y.3. <y Y 2 • X This term arder is called an elimination order with the the Y variables.. . ·l'n elements.4.Ym]. .2).1 is the lex term ordering on all the variables with the Y variables smaller than the X variables (Exercise 2. The advantage of this order is that when one is interested in properties that the lexicographie term ordering between the two sets of variables is advantageous for..3. Then G n k[YI.3.g.Y.3.xn } and {YI. we define X variables and YI.9. We define a term order < on the power products in the x. J n. variables larger than Elimination orders have the following fundamental property whose proof we leave to the exercises (Exercise 2. . The elimination order is "like" a lexicographie term ordering between the x and y variables.3). LEMMA 2. Prove that Vk(I) bas at IDost 1'11'2 . There is a great advantage to the elimination order as the following result shows. GMNRT]) and it is better to have as "little lexicographie ordering as possible".. Let l be a zero-dimensional ideal in k[Xl1'" .3. <x X 2 or { X. then the elimination order defined in Definition 2.3.3.xn ]. Assume that the power products in the X variables and the power products in the Y variables are ordered by term orders <Xl <y respectively. ..3.. b.. Let 1 be a non-zero ideal of k[YI. if Y is a power product in the y variables and Z is a power product in the X. then Y < Z. Y variables as follows. THEOREM 2.. . . For X" X 2 power productB in the power products in the y variables. . < X 2 Y2 = X.xn ]. .} be a Grobner basis for this ideal. The elimination arder defined in Definition 2.3. . Y 2 X. Let G = {g" . .. xn]. Let j E I n J. every term in 9i involves only y variables. .. Conversely. .w)J) ç k[Xl..Xn.). J be ideals in k[Xl' . .xn. .w)J) is {wh. 0# As a first application of Theorem 2. . Clearly G n k[Yl"" . PROPOSITION 2.8). (1.. APPLICATIONS OF GROBNER BASES The ideal In k[Yl. .4 we now present a method for finding generators for the intersection of two ideals.w)j.70 CHAPTER 2.Ym].(1. Ym] is called an elimination ideal.. Yi E G n k[Yl' .. .. .. frOID Lemma 2. i.'" .w)J) in k[Xl"" . since j has only y variables. . and thenletw=OandgetjEJ.. .. .w)j. n )hi (Xl 1 ••• 1 Xn...).'" .5... .Ym] is contained in In k[Yl. since the x variables have been "eliminated". .. which îs done simply by inspection. . . (l-w)J)nk[Xl. Since j = wj + (1..w[ using an elimination order with Xl..Ym) E Ink[Yl.3.. suppose that j E (wI. there exists i such that lp(gi) divides lp(f). .. . . .. (1.x i=l p .xn ) = L wfi(xl. Moreover. Consider the ideal (wI. Since w does not appear in j(Xl. .w)J) n k[Xl'''' . Xn' W]. we have f(xl.'" . (1 .. .Ym].2.(Xl. . Now let j(Yl. we can let w = 1 and get j E I. . wj" (1. REMARK: If I = (fI.3. First we compute a Grübner basis G for the ideal (wI.W). and let w be a new variable. Xn].'" . . . We then obtain a Grübner basis for In J by computing G n k[Xl l • • • .. Xn].X n ].j.w)jL . .. Ym] is a Grübner basis for I n k[Yl' . we see that lp(gi) involves only the y variables and so.w)J) ç k[Xl"" .thereexistsgi E Gnk[Yl. Then In J = (wI. .'" ....Ym]. sinee j E (wI. Since G is a Grübner basis for I.Ym]' Thus. (1w)J) nk[Xl. (1..3.. Ym]' D PROOF. and hence G n k[Yl. we have jE (wI.}.. A similar technique can be used to compute the intersection of more than two ideals (see Exercise 2.xn ]. D As a consequence of the above result we obtain a method for computing generators for the ideal In J.X n smaller than w. Xn. for every jE Ink[Yl. and J = (f.3. . PROOF.. Let I..'" . Xn).xn)hj(xl. then a set ofgenerators for the ideal (wI.'" j=1 .Ym] such thatlp(gi) divides lp(f). (1. w]. Then.. w) + ~)l-W)j.j.e. . (1 .xn]. w(x . by the definition of f. (iii) lc(f) = lc(f)lc(g).3y2 . g). g). .3. g)). Conversely. We get G = {x3 y 2 _ x 3y _ 3x 2y _ xy + x 12853w + 118x4y + 3. denoted by lcm(f. Dually.. We compute a Gri:ibner basis G for the ideal (w(x 2 + y3 . PROOF. .5 can be used to compute least common multiples and greatest common divisors of polynomials in k[Xl"" . then h = aJ = bg for sorne a.. x 5 y + 3x 2y3 + 3x 2y2 - x 3 + 3x 2y .g E k[Xl. to be the polynomial d such that (i) d divides both J and g. We wish to compute In J.x2 + l. to be the polynomial f such that (i) J and 9 both divide f.g).2152. so that hE (f).xn]. Hence J divides h and 9 divides h. 80 a Gri:ibner basis for the ideal l nJ is x5y + 3x 2y3 + 3x 2y2 _ x 3 + 3x2y . y.3}. bath non-zero.'" .xn ]. ELIMINATION EXAMPLE 2.XnJ is a unique factorization domain (UFD) (see [Hun].3. if hE (J) n (g).xn].g). w] using the deglex term ordering on the variables x and y with x > y and an elimination order with w greater than x.3y . b E k[Xl. LEMMA 2.g)gcd(f.3y2 .yx + 3) and J = (x 2y .Xn].7. denoted gcd(f. y]: 1= (x 2 + y3 . x 2y4 + x 4y _ x2y _ y3 . Let f = lcm(f.1)) ç QI[x. both non-zero.g E k[Xl. (ii) if J and 9 both divide a polynomial h. x .g) and gcd(f.972x2y2 + 2152x2y - 118x2 - 9y2 +972y .'" .'" . We now show that Proposition 2. 0 5The reader should recognize that this is due ta the fact that k[Xl. then f divides h. we define the least common multiple of J and g.1). we have (f) n (g) = (lcm(f.g) exist 5 and that Jg = lcm(f.3y . and thus f divides h by the definition of lcm(f.w)(x 2y . (iii) lc(d) = 1.l. Consider the following ideals in 71 QI[x.3. .1). Then f E (f) n (g). we define the greatest common divisor of J and g. For J.6. Fix a term order on k[Xl.2. For J.3. . g). . y. + 357x + 9X2 y3 - 357x 3y .3}.yx + 3). (ii) if h divides both J and 9 then h divides d.) . It may be shown that lcm(f. 1.. 80 in particular gJi E J for i = 1. Then J:l=n J :(J..72 CHAPTER 2. g).h. ..3. .y2 . .'" . Let J = x2y2 . . 0 We have seen a method for computing intersections of ideals.. 80.y2 + x 2 . and gCd(f"h. To compute gcd(f.3. x n ] 1 gl ç J}.. .5. Let l and J be ideals in k[XI' . g) is the polynomial in G in which w does not appear. s.J. Then 1 J (f) = y(J n (f)) . (1. we first compute the reduced Grübner bMis G for the ideal (wJ.8).)· i=l PROOF. . (1 W)(xy2 . Then lem(f. .w)g) with respect to an elimination order with Xl.11. g) = ~~) lem(ft. 12. we use the Division Algorithm to compute gcd(f.g To compute the lem and gcd of more than 2 polynomials. and hence 9 E n:~l J: (fi)..g) = l_x 2 _y4+x 2y4. LEMMA 2. s. LEMMA 2.1 and 9 = xy2 .. Xn smaller than w.3. Xn].8. we use the Division Algorithm to divide J9 by lem(f.y2 . 13) = lem(f"lem(h. To compute lem(f. y].... g). so that 9 E J: 1. y.y2 + x 2 . Proposition 2.w)g) = (W(X 2y2 . 13)). Letl= (f" .xn]. DEFINITION 2. we use the above method repeatedly and the fact that gcd(f.3. Thereforelem(f. To obtain lem .X + 1)) ç lQI[x.g). We will discuss the geometric significance of these ideals after Proposition 2. w] using the lex term ordering with w > x > Y to get ci = {x 2y4_x2_y4+1. g).2wx 2 -2w+ x 2y2_x 2 _y2+1}.) and J be ideals in k[x1..10. and hence gl ç J.3. in view of the aliove lemma. EXAMPLE 2. we first compute the reduced Grübner basis G for the ideal (wJ.. APPLICATIONS OF GROBNER BASES Therefore to compute the lem and then the gcd of two non-zero polynomials J and 9 in k[XI' . . . g) to get a quotient equal to x .9._wxy2+wx+wy2 -w+xy2-x-y2+1. xn].h))· (Alternatively we could note that (ft) n (hl n (hl = (lem(ft. if 9 E n:~l J: (fi).1. The ideal quotient J: l is defined to be J: l = {g E k[XI. If 9 E J: 1. . and this is gcd(f.5 has another application: the computation of ideal quotients. Conversely. then gl ç J.3. 13)) and apply Exercise 2. 13) = gcd(f"gcd(h. Xn]. then g(fi) ç J for i = l. . (1 . WB only need to concentrate on computing J: (f) for a single polynomial f.1).. Let J be an ideal and J # 0 be a polynomial in k[XI.X + 1 be polynomials in lQI[x.'" .. 3. y. sa that h (J n (h)) = (x. y. wy. h = x 2.w)h) w > x > Y ta obtain 1 n (h) by computing a Grübner basis G l for the ideal ç lQl[x.xy + y2.y).3. Verify the assertion in Example 2. Exercises 2. sa that 12 (J n (12)) = (x 2 . u. Conversely. Y2 = y. Assume that we have the lex ordering on x.2.x. y.2. then yf E J.11 First we compute J (WYl.1. y. v. w with u > v > w. ELIMINATION 73 PROOF.w)fû order as above.x . and hence yf E Jn (1). y).3. WY2. (1.w)y) ç lQl[x. (1. 1 .3.3.xy + y').3.y[.w)(x 2 . y larger than the variables u.xy.wy.y. Fmally we compute (x.3. (1 .12) and J = (gl. If . Use the elimination order with the variables X. v. . Let YI = x(x + y)'.). Second we compute J n (12) by computing a Grübner basis G 2 for the ideal (WYl. Consider the ideals J = (11. and henee y E J: 1 ~f). We wish ta compute J: J. and sa by Lemma 2. 2.wy. y). w] using the same G.10 we have J: J= (J: (J. 2. (1 .xy + y2. y with x> y and the degrevlex ordering on u. WY2.x 3 + y3.)) n(J: (h)). w] with respect ta the lex term ordering with 2 2 32} G 1= { xw-x. y E J: (1).3. w] with respect ta the lex ordering with w > x > y. 0 EXAMPLE 2.3. = {wx . Prove Lemma 2. and we obtain ç lQl[x. wl in orcler of descending terrns.xy+y2}. and 12 = x + y in lQl[x. sa that y E y(Jn (1)). w to write the polynomial f = xu2vw+3y3u2vw-6xuvw-100x2yuv2+2xu3w-x2yu2w+9xuw2 E k[x. y) by computing a Grübner basis G for the ideal (wx. ta obtain Therefore J: J = (x 2 .3. By Lemma 2. v.2.3. y) n (x 2 . then yf E J.12. If 9 E y(Jn (1)).y. z}. Prove that there is a term order < which extends ~p . c. m.. . y) n (x .Xn } (S and T may overlap).3. APPLICATIONS OF GROBNER BASES 2.y. X <:'p Y or Y <:'p X. Let <s be deglex with x < y. compute (x. a.p Y if and only if 2:::7~1 Ui"i <:.8.ES Xf' (resp.··· . . • ~p is reflexive and transitive.Z . Prove that h n 12 n··· n lm = J n k[X1"" .2. In Q[x. For X = .1.5. Without the use of a Computer Algebra System. . Use b to compute generators for the ideaJ (x.3. Show that <:'p extends to a term order on 1['n. For a power product X = X~l •.x3y+xy2+2x2+2y).5. In this exercise.X n ].x) c:: Q[x.wmlm ) c:: k[X1.3.lm be ideaJs of k[Xl1'" ) x n ]. y) c:: Q[x. y]. z]. . if X < Y.wm ].. xr 2. In this exercise we generalize the concept of elimînation orders. .3. n lm· c.3. . Let S and T be subsets of {Xl"" . where l = (x 2y .. For each i = 1. Let J = (1.6. Ler u = (Ul l . Let <:.. y) n (x 1.10. X T ) to be fIx. and let <T be lex with X > Y > z.un) E ]Rn be a vector with non-negative coordinates. xy + y + 1) and J = (x . Y E 1I'n. Order the following power products according to <: 2.9.'" . consider a new variable Wi. b.". for ail X. + w m ). Compute the intersection.y.74 CHAPTER 2. X <:. E 1I'n. • X <:'p Y implies X Z <:'p y Z. a. .xn ]. 2.. y] compute gcd(x3-x2y-3x2+xy-y2-3y.. Z. Y. b. Let S = {x. .(W1 + W2 + . [Answer: x 2 + y. • .3. Use a to give a method for computing generators of 1. We deline a new order < as follows: for power products X and Y in k[X1. Construct the deglex ordering using b.y} and T = {x.z. X <y = Xs <s Ys or (Xs = Ys and X T <T YT ).] 1 . y1) c:: Q[x. 2.7.3.. Verify the assertions in Example 2. . w. Let I" h. y) n (x -1.3.. we extend Proposition 2. Z2 .Xn } such that SuT = {Xl.p be a relation on 1I'n satisfying • for ail X. a. fIX. ' x~n E k[Xl1'" l xnl we deline X s (resp. 2:::7~1 Ui{3i. • 1 <:'p X for ail X E 1I'n.4. That is. 2.y.. Let n = 3 and the variables be x. y]. X~n E 1I'n. In J. X~n 1 -and Y = X~l .xn . y.8.3. n 12 n ... then X <:'p Y [Hint: Use the idea of an elimination order to combine <:'p with any term ordeL] b.ET xf')· Let <s and <T be term orders on the variables in S and T respectively. Prove that < is a term order.··.. W" . 1 Define ~p on 'JI'n as follows.h. 2. gt} be a generating set for Jnk[x" . .. For a polynomial 1 E k[x" .. Find the radical of the ideal (_x 2y2 2x 5y 2 _ x 6y2 + x 7y2 _2xy3+4x2y3_4x4y3+2x5y3_y4+3xy4_3x2y4+X3y4) ç Q[x.] 2.9t. ~8 ' .y2 + xy2 . . Let 1 be an ideal of k[x" . Pm. 2.9t homogeneous and with no 9i divisible by X n1 then {g" .13..8. In Q[x..xy2-x) and J = (x 2 +y2.m 1 l' - gcd(J. . .x 3 +y). z] compute generators for 1: J and J: 1 where 1 = (x 2y . > X2 > .) 2.13..3. then g" .4.. Let k be a field of characteristic zero. .14.. .3..:O.9. ...1. .....) d. y]. P2.. Consider degrevlex with Xl > X2 > ..gt is a Griibner basis for the ideal 1: (x n ).· ... '~8 ) Xl X n 8f 8f' c.] 2. (See Exercise 2.. where land f are as in Exercise 2. Prove that if {x~'g" ..2.9vXn.3. Compute generators for 1: f=. 19t is a Griibner basis for the ideal l.l. Conclude Ihat 1: 100 is an ideal of klx" .. We define 1: r = U1: (Ji).xn ].1. [Hint: Recall Exercise 1.xn . .3.+1.) a. .1.2. . every power product occurring in each gi has the sarne total degree) where X n does not divide any of 9".xn ].... b.. Show that 1 = 1: (1) ç 1: (J) ç 1: (J2) ç . compute generators for 1: (1 . . > X n with 9b .9.x 2 +xy-l)..9v+l. Let 91XnJ'" . Pm are irreducible. Prove that 1: 100 = 1: (Jm) for sorne m. b. .t.5.2.. Show Ihat = (J').] 2.xy). > Xn· Prove that if 91Xn. e.··· . .3.xn ] 1 . where 1 is the ideal defined in Example 2. Define m = rnaxij(degw(uij))' Prove thal 1:/00 = 1: (r)..11.1 . a.xn ].xn ].12. [Hint: See Exercise 1..4. see the discussion following Proposition 2. .15. Now write gi = (l-wf)hi + L.] + x 3y2 + 2x 4y2 - ... In tbis exercise we show how to compute the generator for the radical of a principal ideal.gt} is a Griibner basis for 1: x.xn ] wrile 1 = P~'p~"" p~m where the polynomials Pl.. ..9vxnl 91/+1.~.. Show that in k[x" . c. called the square free part of f. (Compare withTheorem 2. Set f* = PIP2 . y. Provethatl: 100 = Jnk[x" .. [Ànswer: (x 3y ..4.x~'gt} is a Griibner basis for 1 with respect to the degrevlex ordering with x. In Q[x... ... .. uijlj for i = l. [Answer: 1: J = (y2-l.. i=l 00 (The ideal 1: 100 can be thought of as the ideal defining the points in V(I) which do not lie on the hyper-surface defined by 1 = 0.3.x 2 .} be a generating set for 1 and {g" . Let {J" .. and let J = (1.3.xy)..16.wf) ç k[x" . . w]. y]. .xn ]. Let w be a new variable.'" .Y + x.9t be homogeneous polynomials (that is.3. .xn]. ELIMINATION 75 2. Let 1 be an ideal of klx" .... and let 1 E k[x" .. . .3.3.8. where hE k[X1"" . let 111'" 1 lm be distinct ideals in k[Xl 1 ' " ) xnL and let ft.··. --"->+ n::..1) f == fi (mod I. Sa we may assume that no power of any single variable Xi appears in J.17. Give a method for computing such an f.] .1) has a solution if and only if 9 h..w=I=).1) is given by h + h In particular.. then the set of solutions of System (2. . In this exercis€. Use Exercise 2.y)) x (mod (x -l.19.3. .18.J. [Hint: Use Exercise 2. 0) = l. .(W1 + . Let al.17. prove that u E k[X1"" . a. then the system above has no solution. 1 SiS m.2. . then X = Y Then show that if xi appears as a term in f. then any term in f with Xi is in ft. we compute the solutions ta a system of congruence equations... and let al. . +w=). f f f x -1 (mod (x.. 2. . and f(2.am E k be given. . As in Exercise 2. c. .xn ].'" .xn ] of the system (2.xn ] such that f(ai) = ai for i = 1. .3... c.3. We wish to compute the set of solutions f E k[xl. Prove that if we replace (x.. 2. W1I" . Prove that if h = Nc(g) and hE k[X1. Find a polynomial f in Q>[x.. let W" . a. Use the same argument for terms in f of the form XiXj. Let f E k[X1"" .". b. Show that the following system of congruence equations has a solution and compute the set of solutions.] b. and if X divides Y.xn ] is a solution of System (2. Prove that f E ft.. [Hint: First show that if X and Y have the same total degree. 0) = -l. .. y] which satisfies f(O. .y -1)).m.3. In this exercise we use the Chinese Remainder Theorem (see [Hun]) which states that if ail the ideals Ii are maximal then System (2.. let 9 = 2::::. J = (1 .76 CHAPTER 2.4. More precisely.3. 1) = 1..xn ] be a homogeneous polynomial. Prove that there exists f E k[X1. Consider the ideal 1 ç k[X1. f(l... .. Finally.3.f= E k[X1...xn ] generated by all power products X which appear in f and such that IpU) = X with respect to some term order. and G be a Grübner basis for J with respect to an elimination order with the w variables larger than the x variables.3. l am E k n be distinct.3. . .1) if and only if h = N Cnk[x.W= be new variables.y)) y (mod (x . .y). APPLICATIONS OF GROBNER BASES 2. Xn ]. Prove that System (2.3. Wij..y) by (x 2 .1) always has a solution.xn ]. .xnl (u).. c.1: . Use a to give an algorithm that inputs a Gr6bner basis Ci for each Qi with respect to < and outputs a Gr6bner basis for l with respect to <. (Lakshman [Lak]) Let 10: where aij k[XI.} is linearly dependent for each i. x2 . X r + I} is linearly dependent if and only if {1 + Qi. y).y... y3 ..YnJ. . Xlh i + J.. [Answer: for lex x > y.2x2 -. .2x _ y2 + y. We wish to compute a Grobner basis for 1: J using linear algebra techniques only.J 2....3.'" • h. . yJ. . x 2 . X + y). (Lakshman [Lak]) Let l be a zero-dimensional ideal of k[XI.. Y + x) ç lQI[x.J b...3.21.. . a. . . [Hint: {1 + (1: J)... Q2 = (x 2 + y.... c. . Use a to give an algorithm that inputs a Grobner basis G for l with respect to < and a generating set for J and outputs a Gr6bner basis for 1: J with respect to < . Let Ci be a Gr6bner basis for Qi. .. We wish .y2 + 3y)..y2 .. xnJ (1 <: i <: r) and let < be a term order on k[XI.y2) ç lQI[x. (y3 . ELIMINATION 77 2.3y2 + 2y. Xrh i + I} is linearly dependent for each i. .2.8 to give a method for deciding whether there exists a polynomial f in 1: J whose leading term with respect to < is Xr..2.2. y) ç lQI[x. be a Gr6bner basis for 10(1). n Qr..8. Let G. [Hint: {1 + 1.20. As in Exercise 2.X2.. E k be a non-singular linear transformation. yJ.. .YnJ 'L.. yJ. . write all the power products in k[XI.8.Xr + (1: J)} is linearly dependent if and only if {hi + J. Let J = (hl.1. For an ideal l of k[XI. . XI + 1.2. Modify the method of Exercise 2.) be an ideal of k[XI"" .. r.3y + 2)... Modify the method of Exercise 2. . . . Use b on the following example: J = (2xy ..2. (ii) QI = (xy + 1. . XI + (1: J).7=1 aijYj. . a. . . . . x 3 . . XI + Qi. .8 to give a method for deciding whether there exists a polynomial f in l whose leading term with respect to < is Xr. (Lakshman [Lak]) Let Qi be a zero-dimensional ideal of k[XI.J 2. As in Exercise 2.1).XnJ. 1: J = (x +y . . xnJ ordered by <: 1 < XI < X 2 < . . write aH the power products in k[xl. xnJ.3. note that 10(1) is an ideal of k[YI. .22. We will compute a Grobner basis for I with respect to < using linear algebra techniques only.Xn].xnJ. Use b to compute generators for the intersection QI n Q2 in the following cases: (i) QI = (x. . . Q2 = (x . and let C be a Gr6bner basis for l with respect to sorne term order < . . X r + Q. . [Answer: For lex with x > y. Let l = QI n Q2 n .3.3.XnJ Xi --> !-------)o k[YI... J = (y . ·.J b.. i = 1.xnJ ordered by <: 1 < XI < X 2 < . We will consider a zero-dimensional ideal J. ..r. .8 to give a method for deciding whether there exists a polynomial f in 1 whose leading term with respect to < is Xr. Modify the method of Exercise 2.xn ] ordered by <: 1 < X. and so it suffices to show that . Prove that 1 = n...3.3. y... Use the above ta determine whether 1 = . a.yz. Give an example that shows that part a is not neeessarily true if ehar(k) # O..Xr+I} is !inearly dependent if and on1y if {1>(1) + 1>(1).pj) for j = 1.23. ln.j(I. e..j(I.fi = 1 if and on1y if it contains a univariate square-free polynomial in each of the variables. Assume that for each i = l. Conclude that a zero-dimensional ideal 1 of k[Xl' .2. l cantains a squarefree polynomial gi E k[Xi].xn ]. a. n. ..' . Pro with Pj E k[Xl] irreducible and pairwise nonassociate. .] b. .. let g. . Use this to give an algorithm which inputs a zero-dimensional ideal 1 and outputs TRUE if . and let gi be its square-free part. ... 2.In]. 2... . ) n are square-free in (k[xd/(Pj))[X2' . .pj)). < X 2 < . (*)(Seidenberg [Se]) Let char(k) = 0 and let 1 be a zero-dimensional ideal of k[Xl1'" . i = 2. By induction hypothesis. . . Prove that 1 = ... 1>(Xr ) + 1>(I)} is !inearly dependent.y] X -4 f-.pj) = (I. For n > 1. Consider the map Use Exercise 2.X y.X . = Pl . ) + 1>(1)..v] 2u-v -u+v and let 1 = (X3 + 3x2y + 3xy2 + y3 + x2 + 3xy + 2y2.78 CHAPTER 2. . .xn ] satisfies.pj). c. .. Use a to give an algorithm that inputs a Gr6bner basis G for 1>(1) and outputs a Gr6bner basis for 1 with respect to < .pj) = (I.:~l (I. Conclude that .fi. let gi be the monie generator of 1 n k[Xi]. 1>(X. z].2.8.3.. Z2 ..] b.fi where 1 = (X2y . d.pj)) = 1>j((I. .16 to observe that a polynomial f is square-free if and only if gcd(f.pj). [Hint: {l+I..24. +1.. write all the power products in k[Xl'. c.'.fi = 1 and FALSE otherwise. Let ehar(k) = 0 and let 1 be a zero-dimensional ideal of k[Xl' .. As in Exercise 2.xn].--. Use b on the following example: Consider the map 1>: Q[x.. x 2 + 4xy + 4y2 + 2x + 3y). xy + x + z . . APPLICATIONS OF GROBNER BASES ta compute a Grobner basis for l with respect to sorne term order < using !inear algebra techniques only.t y ~ Q[u.j1>j((I.. 1') = 1 and the fact tbat gcd's are invariant under field extension ta conclude that the images of gi.y) ç: Q[x.. [Sketch of the proof: Use induction on n. Use deglex with x > y. For i = 1. . Use Exercise 2. Also. a. gi. Use a to give an algorithm for computing generators for v'I.3.xnJ! Ja 9! k. xnJ. . y.Xn ..14 to prove that J is zero-dimensional and that dimdk[XI'''' . x2 +xy + 2x 2 +y+y2).2." . . .'" . xnJ. g.J 2. Assume that v'I = 1. . [Sketch of the proof: Let G be a Griibner basis for 1 and let J be the ideal of k[xI.y.J b. g. POLYNOMIAL MAPS 79 a. show that <jJ induces a map 1/1: k[xI.an) ç k[xI. Prave that J = naEV.xnJ/J ---> II k[xI.'" . (*) Let char(k) = 0 and let 1 be a zero-dimensional ideal of k[XI"" .3. We recal! that such a k-algebra homomorphism is a ring homomorphism . [Hint: Use Exercises 2. xy + x). .XnJ. In this section we are interested in k-algebra homomorphisms between the polynomial rings k[YI.xnJ/I) =dim. xnJ! I).(I) . Let J be the ideal of K[XI.an) E V. consider the ideal Ja (Xl .xnJ generated by G. yJ! 1 and the number of points in Vc(I) where 1 = (x'y .. .3.fJ = J by Exercise 2. [Answer: dim = 6 and there are 3 points.fJ = J.y.6.. zJ.18 show that <jJ is surjective.xnJ.. Compute generators for the radical of 1 = (x2y .4. let K be an extension field of k..xnJjJa.3.(k[xI'''' ..4. Compute the dimension of lQI[x. Prove that .26.. Prove that the number of points in V.X . 2.xy + z ..xnJ/J). .(I) is less than or equal to dimk(k[XI.').16 and 2.a" . Consider the map <jJ: k[xI'''' .X . where 1 is a zero-dimensional ideal. then ..J d. xnJ.(J) J a .3.2yz' + yz + y2 . .. y=J and k[XI.J c. yJ! 1 and the number of points in Vc(I) where 1 = (x 2y . Give an example that shows that this is not necessarily the case if char( k) "1 O.y. and condude that 1/1 above is a bijection. xy + x + z xy.'" . . Prove that f + Ja = f(a) + Ja for al! f E k[XI'''' . Prove that if v'I = 1 and char( k) = 0. xnJ f Using Exercise 2.. Use Exercise l. [Hint: If v'I = 1.23. Compute the dimension of lQI[x. Let char(k) = 0 and let 1 be a zero-dimensional ideal of k[XI"" .25. z4 .. aEVk" Since k[xI"" . c..x . b.3. Polynomial Maps. [Answer: dim = 3 and there are 3 points. .. .xnJ generated by 1. For each point a = (a" . then equality holds in a.'" . we see that 1/1 is a k-vector space homomorphism..x) ç lQI[x.3.25. J 2.23 to prove that v'I = (1. •..2) v Also. Such a map is uniquely determined by (2.an-bn )· . (ii) The image of 1> or more precisely. .4...1) where fi E k[XI.. . conversely.• .b1 . we get a k-algebra homomorphism from formula (2. Recall that the kernel of 1> is the ideal ker(1)) = {h E k[YI.. if we let h E k[YI..an . This map is. i ::. given any such assignment (2. This subalgebra is denoted by k[fI. as is easily seen. Xn ]. ~ b a2 ~ " b2 . Letal.1.bn be elements oia commutative ring R... Then the element a.. where Cv E k. a2 . . fm) = D and so ker(1)) is often called the ideal of relations among the polynomials fI. That is.. . and only finitely many Cv '8 are non-zero. LEMMA 2.4. Before we give a characterization of the kernel of the map 1>.'" . m.4. . Ym]. We know from the theory of abelian groups that as abelian groups under the map defined by g + ker(1)) >---> 1>(9). a k-algebra homomorphism. fml. then we have (2.. APPLICATIONS OF GROBNER BASES which is also a k-vector space linear transformation. and the image of 1> is the k-subalgebra of k[XI.3 to determine the following: (i) The kemel of 1> or more precisely. an algorithm to decide whether a polynomial f is in the image of 1> and an algorithm to decide whether 1> is anto. y~rn.. xnl.. . an ~ b. v = (Vl..a21'" ..4. b2 .4.80 CHAPTER 2. . in fact. a Gr6bner basis for the kemel of 1>. im(1)) = {J E k[XI"" . ."" v rn ) E Nm .. . . . say h = 2:vcvy~l .··· .. Another way to think of ker( 1» is that h E ker( 1» if and only if hU" .xnll there exists h E k[YI. . fm· We will use the theory of elimination presented in Section 2.Yml with f = 1>(h)}. bn is in the ideal (a. and thus is a k-algebra isomorphism... Ymll1>(h) = D}.1). This is called the First Isomorphism Theorem for k-algebras.2). we need a technical lemma. .b21 . 1 ::.. '" m i=l . .xu.y 2u- . let 9 E ker{ 1».2. Y= . and hence gEKnk[YI. V.ft. y variables larger than the T. x n ].fV' 9-9-9 -L-CVYl 1 v . . .4.Jm) and hence 9 Eker{1». an -b2 . order between them with the x.y4 .4. v = (VIl'" .·· 1 Xn]. and only finitely many cv's are non-zero. Y variables with Y > X and the degrevlex term ordering on the T.w . .. EXAMPLE 2. The polynomials in G without any X variables fOTIn a Grübner basis for the kernel of 1>. Let 9 E K n k[YI..x.'" .··. x n ). c:: k[YI.yr .. 81 The proof is easily done by induction using the fact that ).u .x3 y . Therefore 9 is zero when evaluated at (YI.yv ..XI.··.x4.. each term in the smn above is in the ideal K..w.Xl" .1. fV mm).'" . .Ym]' D We now have an algorithm for computing a Grübner basis for the kernel of 1>. POLYNOMIAL MAPS PROOF. Conversely.v.·bn =al{a2· . · . Then g{YI"" .Ym]' PROOF.u.u.3.. Y=.Ym..xn ] with respect to an elimination order in which the x variables are larger than the Y variables. Xl.··· . .v .· ... w] ----> lQI[x.·. .xy3. . we have (f l.4. Therefore. ala2"'an -b l b2 . since g{ft. y=]. V. . We can write where Cv E k. By Lemma 2... u.2.. We get G = {x 4 . Y=) = L{Yi . w variables. .r. ·.. . Let 1>: lQI[r. with an elimînatiop. u.w.x3 y . . · . xn))hi{YI.y4) c:: lQI[r. We first compute a Grübner basis G for the ideal K = (YI . y] be the map defined by r U ----> ----> ----> ----> V W X4 X3 y xy3 y" We first compute a Grübner basis G for the ideal K = (r ... f m ) " (V' "'Ym Vm . Ym.ft.xW.Y=)={ft. .v.y] with respect to the deglex term ordering on the x. fm) = 0.Vm ) E Nm. u. Let K = (YI . Ym ..4. v.. ·bn ) +b2 ···bn {al -b l D THEOREM 2. w variables with r > u > v > w. xy3 . where hi E k[Yl1'" 1 YmJ Xl..fm) in k[Yll'" .f=) Thenker{1» =Knk[YI.fi{XI. 0 The following Corollary shows that the result of the preceding theorem gives an algorithmie method for determining whether f is in the image of <1>.. ..'" ....xn ] be the ideal considered in Theorem 2.Ym) = Lgi(Yl"" . ..g(Yl.4. Therefore.'" .Xl.'" .'" . if f E im( <1». . and f is in the image of <1>. .. sinee f -S+ Nc (f) .f=) ç k[Yl. f E k[Xl' .xv 2 1u 3 .Ym]' Therefore h E k[Yl.y=].4. . Then f .Ym.Xn)).'" i=l .4.y= . x 2 v . .Xn)(Yi .X" .xn ) .g(Yl"" .ft. Shannon and M.Y=]. . This material has been adapted from D.'" . THEOREM 2.Xn]. ' .fm)' PROOF. where h = Nc(g) = Nc(f).·· . Conversely. Sinee the x variables are larger than the y variables in our elimination order the polynornials used to reduce 9 are in k[Yl.4.··· .y=. . 9 -S+ h. .. In this case.'" .'" .Ym]..f=) forsome g E k'[Yl' . With the notation of Theorem 2.y=. But..J.4..y=). Then f=g(J"". c Conversely. If we substitute J. . by Proposition 2.Ym] such that f -S+ h.xn ] bein im(<I».xn ) h(Yl..5.'" . . .'" . we see that f(Xl.(Xl.v 2 .Y=) is in K. let f --->+ h. .'" . Sweedler [ShSw].xn ) ..h E K..y=].y=]. where h E k[Yl. and let G be the reduced Grobner basis for K with respect ta an elimination arder with the x variables larger than the y variables. .X" .xn ] is in the image of <1> if and only if there exists hE k[Yl'''' .. namely..1... sinee 9 E k[Yl'''' .g(Yl. Note that f(Xl. and henee.r 2 v .4. for Yi.g(Yl. then. v 3 . ..1. using Lemma 2. Then f E k[Xl.4.2. UV . we have f E im( <1» by the Theorem. COROLLARY 2. so 1 f(Xl.Ym) = g(J" ..fm) .. ..Y=]. Let K = (Yl . Consider the polynomial f(Xl.Xn) .uw 2. X 2 y 2 w . APPLICATIONS OF GROBNER BASES U 2W .xn ] is in the image of the map <1> and an algorithm to determine whether <1> is onto. ' .4. we see that f = h(f" . the question of finding an algorithm to determine whether an element f E k[Xl"" . then by the Theorem f -S + h . If Nc(J) E k[Yl. f = <I>(h) = h(f" . 9 can only be redueed by polynomials in G which have leading terms in the y variables alone.xn ] is in the image of <1> if and only if Nc(f) E k[Yl'''' . rv 2 Therefore a Grübner basis for ker( <1» is yuw . Let f E k[Xl.rw .fm) = <I>(h)..82 CHAPTER 2.Y=) E k[Yl"" . PROOF. yu 2 - xrv} • We now tUTn our attention to the second question posed at the beginning of the section. and f -S+ h. and _x 2 +xu is reduced with respect to G. THEOREM 2. x 2 u ~ v 2 -v.fm). .7.6.. we have. Ne(f) = Ne(h) and as in the proofaboveweseethat hE k[YI... . by Corollary 2. . Let 1.. let us assume that the order is sueh that Xl < . is onto if and only if for each i = 1.Ym. and henee there exists g2 E G sueh that lp(g2) divides lp(x2 . . without loss of generality. . v. and since G is the reduced Gr6bner basis and any term involving Xl could be reduced using 91 = Xl . V - x 3 ) ç: <Ql[u..hl. is onto. . E k[Yll'" .3v 2 .Ym].4. Then 1. and henee there exists gl E G sueh that lp(g.Ym] implies Ne(h) E k[YI. -hl for sorne h. .4..X n] be the ideal considered in Theorem 2.'" .. there exists 9i E G Buck that 9i = Xi . Therefore.h. E k[YI. that x 5 is not in the image of 1.4..4.h 2 for sorne h 2 E k[y!.h~) = X2. x] with respect to the lex term ordering with x > u > v to get G = {u 3 - v4 - 3v 3 .Ym]' By Proposition 2.4. .h\) = x. Now that we have an algorithm to determine whether a polynomial 1 is in the image of 1. xu 2 _v 3 - 2v 2 ..x 4 - X.5. . Therefore Xl . Then by Theorem 2.1.n. .Ym]' Similarly. x 3 -v}. is onto.. .lm) ç: k[YI.V. .Ym] such that Xl ~+ h~.2. there exists h~ E k[Yll ..2.). We simply check whether x" ....4. xv +x - U. and let G be the reduced Grobner basis for K with respect ta an elimination arder with the x variables larger than the y variables. . ..··· . Sinee the only terms strieUy smaller than X2 are terms involving Xl and the y variables only. Xl.h il where hi E k[Yl1' .Ym] such that X2-S+ h~. v]. POLYNOMIAL MAPS 83 with h E k[YI. we can determine whether 1. Also. since Xl is in the image of <p. gl = X..xn E im( 1....h\ E K.Ym . D EXAMPLE 2. .) divides lp(XI .. v] ---+ <Ql[x] be the map defined by u v We would like to determine if x 5 is in the image of 1. In the next result we see that we can do this simply by inspecting the Grübner basis.4. . Ym]. there exists h.. in this case. Let us first assume that 1. .Yml We proceed in a similar fashion for the remaining Xi'S. sinee the only terms strickly smaller than Xl are terms in the Y variables alone. since X2 is in the image of 4>.... . Since N e (x 5 ) = _x 2 +xu rt lQI[u.. We now reduce x 5 using G to get X 5 x _v 2 xv+x-u ---+ X V ---+ 3 -x 2 + XU. Let K = (YI .V.: <Ql[u. < Xn. . Xi= hi (f. Moreover. 1 PROOF. we must have 92 = X2 . We first compute a Grübner basis G for the ideal K = (u . .4... Moreover.. . . We first compute the reduced Gr6bner basis G for the ideal K. the Gr6bner basis G did not have a polynomial of the form X ~ h(u. Since Xi . xnl.Yml--->k[xl.". . we have that Xi is in the image of 4> by Theorem 2. D The above result gives an algorithm for determining whether the map 4> is onto or not.uv 2 + uv . Again by Theorem 2.xnl/I..hi E C...4. .u + w 2 E· G*. and hence 4> is onto.4. with respect to the lex term ordering with > u > v > w to get ~U2V + v 2 w + 2vw + w.mt to determine whether <jJ* is anto. v).. fm E k[Xl. then.4. ." .4. .9. of the map 4>: k[Yl. i ::. .Yml/J ---> k[Xl. we have Xi -S+ hi_ Since hi is a polynomial in the y variables alone. . V - x3 )W X x5 ) ç Q[u.8. We have seen in Example 2. xnl is an affine k-algebra.n. . as we saw at the beginning of this section. . fm]. . the map 1>* is anto. Yml/ ker( 4» and hence is an affine k-algebra. .4.. w. Now consider 4>' : Q[u. xnl/ l for sorne ideal l of k[Xl. . Let J be an ideal of k[Yl.'" . . We first compute a Gr6bner basis G* for the ideal K* = (u . xnl.'" . the image.x 4 - X. In fact we have X = 4>'(uv 2 ~ uv + u ~ w 2 ) = (X4 + X)(X3)2 ~ (x4 + x)x3 + x4 + X ~ (x5)2 We now extend the preceding resnIts to quotient rings of polynomial rings. Xn ].. fm). EXAMPLE 2. . we check whether there exists Yi = Xi ... with hi E k[Yl. DEFINITION 2. . . wl u v w ---> Q[xl x 4 +x ~ We Wf.xnl which sends Yi to fi is isomorphic to k[Yl. v. An k-algebra is called an affine k-algebra if it is isomorphic as a k-algebra to k[Xl. since x5 was not in the image of 4>...6 that the map 4> was not onto. We now want to study k-algebra homomorphisms between affine k-algebras.Yml.4.. Clearly the polynomial ring k[Xl. Consider a k-alge brahomomorphism 4>: k[Yl. n.. x].'" . by inspection. Yml and let l be an ideal of k[Xl'''. v. and.4. .. if fI. u 3 ~ v 4 ~ 3v 3 ~ 3v 2 ~ v} Since we have x . APPLICATIONS OF GROBNER BASES For the converse we first note that 4> is onto if and only if Xi E im( 4» for 1 ::..'" .84 CHAPTER 2. . we see that Xi = hi(f"". .hi E G for each i = 1. AlBo. . k[fI.. . Then we can write m f'(YI. .fm)) + f'(fI. .f.fm) = L v Cv (Yr' . Thenf'(fI. . That is.YI. Then ker(</J) = K n k[YI.fm. . .Ym) = I)Yi-f. . .. .XI..." . . . y~""'.4. POLYNOMIAL MAPS 85 Let us assume that </J: Yi+Jr-->fi+ J · We note that the rnap </J is well-defined if and only if the following condition is satisfied: if J = (91..... Let f' E Knk[YI.(XI. Generalizing Theorem 2.. .. .Ym) = L:vcvYrl .. .. 9i(fI.. ..Vm ) E N m . Let f'(Yb. v is in the ideal (YI . . ..I El.·· ..6.Ym .fm) = K.fI. then..··· . where v = (VIl'" . .. .'" v .fm) E J. . This condition can he checked easily using a Gr6bner basis for J and Theorem 1..2.Ym) E (J.Xn) </J(f'+J)=f'(fI.Ym. .xn ) i=l where W(YIJ··.Ym) = (f'(YI.xnl whose generators are those of J together with the polynomials Yi .9'). if Knk[YI.. ..··· . ..Xl. XI.Ym . . let f' E k[YI. Then f'(YI.4. . . ..4. thenker(</J) = (f{ +J. .fi. . .xn))hi(YI. sînce each Vj.10...'" .fm) E J. . . ..Xn}Vv(Xll'" .··· . for i = l.2.Xn) = l:::Uv {YIJ .. .Ym) .··· .Ym..2 we have THEOREM 2.XI..'=) + f'(fI. .Yml = (f{. . .. ...fm). .. LetK be the ideal ofk[YI. ..Xn ) El. Cv E k. t. and only finitely many Cv are non-zero. By Lemrna 2.= - g' . YI .)..··· .Xn)+J=O.fI.. ...fm.. . ..f.f'(fI..Ym...Ym - f m) and hence f'(YI.Ym.··· . K = (J.. sinee w{JI.1. .. ..··· .··· . .fI.XI.Xl.Yml (mod J).Ym] with </J(f'+J) = o.lm. . Conversely.fm)+J=w(fI....+J)· PROOF. y:.Xn}Vv(Xl..Ym]. that is. . .Xn ) = L v Uv(!I... f:.... .4. . XI. ..fm). 1 SiS m. .. xnl/ 1 is in the image of q.hi E G .Ym].4.. ..13..Yml Xl.fm) + I.Ym) = g(f" .xnl/1 is in the image of q.Xn) = LUV(Yll'" . .. THEOREM 2.Ym.Ym] sueh that f .Xl. ...xn )g(y" .ft..fm) .Xn)Vv (Xl. .!m) El.g(y" .. .. . .fi(Xll . we have.7. D .. we see that f ...fm) be the ideal as in Theorem 2.h(f" .xn ]. .fm) E l. The argnment proceeds as in the proof of Theorem 2. .··· ...xn ) .. y.Ym.xn ]. ....Xl. We now prove the analog of Theorem 2. . ..Ym) + (f(x" . we have that f +I E k[x" . and hence f -S + h with h E k[y" . . Let f +1 be in the image of q.(h+ J) = h(f" ..Ym]. .10. ..g(f" . . where W(Yll'" . ..11.Ym) E Knk[y" ....Ym. and let G be a Griibner basis for K with respect ta an elimination arder with the x variables larger than the y variables.12... i=l .8).Ym. Therefore J'(Y" . . Then q.. .g(f" . PROOF. Ym) E k[y.' .4. . that f(x" . .. where hi E k[Yl. Then there exists g E k[y" . Xl. let f E k[x" ..Xn ).. .7 and is left to the reader (Exercise 2. Let K be the ideal as in Theorem 2... . . Then f +1 E k[x" . .h E K.Ym..... .fm)). .n there exists a polynomial 9i = Xi . . D THEOREM 2.. again in a fashion similar to Theorem 2. . Xl.4.4. using Lemma 2... .. ..(h + J). If we substitute fi for Yi. We finally determine whether the map q.x n ] be such that Then.4.4. PROOF..4 (Exercise 2. .9). . and let G be the reduced Grobner basis for K with respect ta an elimination arder with the x variables larger than the Y variables..Ym.4.10.xn ) ..fm) El... if and only if Nc(f) E k[y" .··· .Xn) with V v Eland where 9i.4.4..4. Conversely.] . .. . Let K = (1. APPLICATIONS OF GROBNER BASES since J'(ft. .4.Ym) is in K.xn )) + W(Yll'" .g(y" .. is onto.··· . .86 CHAPTER 2. Continuing the notation of Theorem 2.xn ) -g(y" .. The proof is similar to the one of Theorem 2.Ym]. . m Lgi(Yb'" .. Uv E k[Yll'" . if and only if there exists h E k[y" .4. D As before we have COROLLARY 2.'" v .. .11.4.. Since f(x" . We consider the polynomial f(x" .. -S + h..Ym].. and henee f + 1 = q.Xn)(Yi .. x" .4. Ym] such that f In this case f + 1 = q.1. is onto if and only if for each i = 1.... and hence <QI[u. sinee x . . v ~ (x ~ y).y. By Theorem 2.4. 80 let K = (x 2 ~ 2xy.14. by the First Isomorphism Theorem. - 2w. 1 To conclude.W . v.vw + J. UW. uv . VW.4. and I = (x 2 ~ 2xy.-u 1 . uw + J. since = (uv ~ (x+y)(x~y) ~2xy+I x 2 ~ y2 ~ 2xy + I and = 0. We compute the reduced Grobner basis G for K with respect to the lex order with x > y > u > v > w to get G= {u 2- 2w v " 2Y - 1 + -v 1 x . w]/ L "" <QI [x.w 2 + J) = L/J. where J w. note that. wl/L. v 2 + J. yl/ I. vw + J. w 2 . v 2 . uv ~ w + J.v 2 . (x We now compute the kemel of 1> as in Theorem 2. Consider the following <QI-algebra homomorphism 1>: <QI[u.w]!J u+J v+J w+J ----> ----> ----> ----> <QI[x. y]. y2) ç <QI[x. v. we see that the map 1> is onto.~u . w].y]/I x+y+I x~y+I 2xy+I.v 2 + J. We fust note that the map 1> is well-defined.10.w vw uw w -u 2 2' 2 2' '" 2} . 1>(u 2 ~ v 2 ~ w 2 ~ 2w + J) + y)2 ~ (x ~ y)2 ~ 4x 2y2 ~ 4xy + I ~4x2y2 + I = O.2. v. we have ker( 1» (u 2 ~ 2w + J. w]/ J)/(L!J) "" <QI[u. v.4. w 2 + J) (u 2 ~ 2w+ J. where the isomorphism is obtained from <p. u 2 ~ v 2 ~ w 2 ~ 2w) ç <QI[u.13.4.v. In fact we have where L = (u 2 !u 1 x +I = 21>(u + v + J) and y +I = 21>(u ~ v + J). we have (<QI[u.y2.u.v. by Theorem 2.uw+ J. Also.~v and y + ~v are in G.-v 1 uv .2w) (recall that uv ~ w E J).w ~ 2xy) ç <QI[x.4. POLYNOMIAL MAPS EXAMPLE 87 2.u ~ (x + y).10.w 2 .w]. u 2 . .!m) is a minimal polynomial for f over k(fl.xn ]. 1>: v r--+ X + Y and 1>: w r--+ X . . Let G be the reduced Gr6bner basis for K with respect to the lex ordering with Xl > .3. w] ---> lQI[x.' . . . Analyze this reduction. x n ] respectively.. fI. Let f. Xn]..··· .xn ] is a symmetric function. Compute generators for ker(1)). a.xn ) are the fields of fractions of k[Yl. Show that go (y.. APPLICATIONS OF GROBNER BASES Exercises 2. .Ym] and k[Xl.. . YI.f. .fm E k[XI.. Ym. YI .4.6.4.4...... + Xn_lX n Xl Use Theorem 2... where 1>: lQI[u. ..4.eade.Xl.Xn).w] ---> lQI[x. Show that f is algebraic over k(fl... . where 1>: lQI[u. use the method of Example 2. f .Ym)Yv such that h(f. 1>: u r--+ X + y.y] is defined by 1>: u r--+ x 2 + y..'" . A polynomial f E k[Xl.. . and 1>: w r--+ x2 + y2 2. fi}. where k(Yl..Tl.y2. 2.fm) =0 and ho (fI.] b..Ym) to k(Xl. Ym) and k(xl.88 CHAPTER 2.4. > Xn > Y > YI > .. x n ] is called symmetric if for all permutations (Y of {l..4..4. We have h -S+ O. . u.fm) # O.. . . .'" .18. In the case that Go # 0 let go E Go be such that lp(go) is least. .· ... v.4. 2. . > Ym' Let Go be the set of all polynomials in G involving only the variables y. fm).2. .. YI.. .fm). . In Exercise 1. . [Hint: In the case where we assume that fis algebraic over k(fl.. . . . . we can find h= Lvhv(Yl"" . .fm) in k[y. . + XIXn + X2X3 + .ft.·.4. . .18 we saw that the set of symmetric functions is a k-algebra generated by the following n functions: 0"1 0"2 + X2 + . v] ---> lQI[x] is defined by 1>: u r--+ x 2 and 1>: v ---> x"' 2.Ym and in which Y actually appears. y] is defined by 1>: r r--+ X2 + y. Ym .. . Determine whether the map 1> is onto.··· .3 to find ker(1)).1..4 to give a method for deciding whether a given function E k[.4. . ft. part d. .5. + X n XIX2 + XIX3 + .ft.. Use this method to check your answer in Exercise 1. (Shannon and Sweedler [ShSw]) This exercise assumes that the . (Shannon and Sweedler [ShSw]) In this exercise we extend the results of this section tomaps fromsubrings of k(Yl....4..v. 1>: v r--+ X . is familiar with the definition of the minimal polynomial of an algebraic element over a field.. . Consider the ideal K = (y . where 1>: lQI[r.. f m) if and only if Go # 0. Without using a Computer Algebra System.' ..y2 Is 1> onto? 2. . is not onto.: k[YI. .: w+J ----> X 3 +xy2 +J and where J = (uv-w) and J = (xy+y).X n ] generated by Yi . c. Find the kernel of q.Yml Xl. Ym] at P. . w] ----> Q>[x.Ym] andfli E k[Xi+I"" .4. 2.. Choose "Ii such that lp(?.' .g E k[YI. 'P . Let q. a.f3il where Qi rtp. Consider the map q. a. .4.4... .y2. b. yJ! J be defined by q. ... c. is onto.'mXi.. .Ym]P = f. Consider the map q.) is the smallest possible.n there exists Yi E G such -that 9i = (XiIi . .8. Prove that q. b.Ym)Xi -Oi(YI. and sa conclude that gi is of the required farm. v.!m) [Hint: If Xi is in the image of 0/'. > Ym' Then 7/J is onto if and only if for each i = 1. "Ii E k[YI. Prove Theorem 2.] d. It is called the localization of k[YI.9.(f"". Let P = ker( q. "i E k[YI. v] ----> k[x] defined by u ----> X4 + x and v r----+ x 3 . Let K be the ideal of k[YIJ'" .13. Ym] such that "Ii rt P.' . Letq. Note that k[YI..'" .4.11.!m) for some Di. e.4. For the converse.4. 2.7. Prove the following analog of Theorem 2.fil i = 1. first prove that X n is in the image.: u + J ----> x 2 + Y + J. v. Let G be a Grübner basis for K with respect to the lex term arder with Xl > x2 > .. Ym]P is a subring of k(YI..: u ----> x 4 +x 2 +xandq. Show that q.'" . Prove that there erists a polyuomial gi E G such that lp(gi) = yr' .YI." .. . is onto..v]---->Q>[x] bedefined byq.Ym]P ---->k(XI"" . is onto. .m.4..: Q>[u. and v 1-----+ x + y and W J-----+ x .Ym)' Prove that t i E K.. . .7. Show that x3 is not in the image of q. POLYNOMIAL MAPS 89 a. y] defined by u ----> x 2 + y.. etc.Xn.: Q>[u.7 show that q.Xn).··· .4. and q.. then Xi = ~. Complete the proof of Theorem 2.Ym] andg rt P}. Using Theorem 2. then Xn-l. .··· .. Prove that q.Ym]...u. wJ! J ----> Q>[x.: v ----> x 3 -x.: v+J ----> x+y+J.'" .4. can be extended ta a k-algebra homomorphisffi 7/J: k[YI. 2. We consider the following map q.: k[u.... Compute generators of P and determine whether 'if. . Show that the map 7/J. Consider ti = ?i(YI. is well-defined. corresponding ta the one given in Exercise 2.. > x n > YI > .. c.: Q>[u. b. q. . where k[YI. Ym).4. Compute generators of P and determine whether 7/J is onto. 8.)..6. HI 2. Ym] Yi ----> r----+ k[XI"'" Xn] fi.. y:.10...2.. Give a method for coruputing generators of ker( 7/J). .. . . 5.bn ) If we apply this map to a variety V.2.2.W) = I(V): I(W) (Exercise 2. .xn]/I fi+I.W consists of the union of the two lines x = z = and x = y = excluding the origin.am.3. Recall that we showed how to compute the ideal quotient in Lemmas 2. it is easy to see that I(V) = (x(y . Moreover W is the plane y = z which contains just two of these lines. 2. Therefore V(I(S)) c:: W 0 As a simple example of the above proposition. the variety V(xy -1) projects onto the x-axis minus the origin.5. But V(I(W)) = V( v0) = V(J) = W.. APPLICATIONS OF GROBNER BASES 2.1) (al l . If Sc k n .··· .W)) (although it is geometrically obvious that this variety is the union of the two lines including the origin).5. and its Zariski closure is V(I(V . Thus V .3.90 CHAPTER 2. Sorne Applications to Algebraic Geometry. consider tWQ varieties V and W contained in k n. PROPOSITION 2..z).5.z)) and W = V(y-z) in IC"' Then V consists of the four lines y = z = 0. ° ° ° . .1. EXAMPLE 2. then V(I(S)) is the smallest variety contain- ing S. x = z = 0. .bll ··· . We will abbreviate ·this variety more simply by V(I). .. x = y = 0. We use the above method to compute the smallest variety containing V . We note that I(V . Then V ~ W need not be a variety. ifW iB any variety containing S. Throughout tills section. AIso. for 1 an ideal in k[xl. We are interested in finding the smallest variety containing 7f(V). For example.5. Then I(W) c I(S) and V(I(S)) C V(I(W)). Consider the varieties V = V(x(y .. Before we do this we give the following general proposition..2). y(x .z).6 to the case of </J: k[Yj. namely y = z = and x = y = z. That iB. namely V(I(V . . then V(I(S)) called the Zariski closure of S. lare ideals in the appropriate rings and I is a prime ideal. by Theorem 2. .10 and 2.4.( I) c::: k n as we did in Section 2.z)) and I(W) = (y .y(x .W)).11. PROOF. we may not obtain a variety. For those who have the appropriate algebra skills.Ym]/J Yi+J ----> ----> k[xj..W) = I(V): I(W). and this is not a variety. . where J is an ideal in k[xj. Let c:: W This set iB W = V(J) c:: k n be a variety containing S..2.Xn]. generalize ExerciEe 2..4.W. In this section we will apply the results of the previous sections to study maps between varieties.. where we assume that J. We begin by considering projection maps km+n (2.X n ].11.z).5. By the above we have I(V . we will consider the variety V. and x = y = z. am.. . yz). Therefore the smallest variety containing V .1).5.. In view of Proposition 2. Let 1 E I(7r(V)).y(x .Ym].xn ]. ..z) compute I(V): I(W) using Lemma 2. . If " ) = 0. We (X (y . We consider a map r. .a=... The Zariski closure o/7r(V(I)) is V(I n k[YI.bn ) = 0 for all (a" . .z) = (x.jI.bn ) E V...X" .am) = for ail (a" . so that I(a" .X n ) Ym = I=(XI. to complete the proof of the theorem we need to show that c:: V(I(7r(V))).11 to get 91 = (y .p: lin ----+ k'l7\ given by <P(XI..bl. Therefore 7r(V) c:: V(Iy).5. there exists an e such that E I. am.2. . . But.··· .. (We note that V is.4 and 2...3.jI. ..lm).z)) = (x(y .Xn )). . Wc first show that I(7r(V)) c:: .Xl.z). . Let us first prove that 7r(V) c:: V(Iy)..X n ).5. .y=. This process is called impliciiization.z)): (y ..Yrn. .5.V=.b n 0... the union of the two lines x = z = and x = y = O. Now we have V(Iy) = V(.. so that (a" . These equations define a variety in k m+n namely V= V(YI.Y=]).. .1.J=(XI. .z). D V(Iy) ° r r We now turn our attention to an application of Theorems 2. PROOF.xn ]. ... Let (a" .z).3. then I(a" . .3. sinee f involves only YI. .. .. ..3. ... ... .. E Iy. . . .y= .. in fact.a=) E 7r(V).X n ) Y2 = h(XI. One has ta be careful.. ....xn ). .··· .bn ) E V.. where the Ns are in k[XI.. y(x ..(X" .. since parametric equations do not always define a variety and so we will find the Zariski c10sure of im( <p).. .h. . . By Theorem 2...··· .am) E 7r(V).... . . 80ME APPLICATIONS TO ALGEBRAIC GEOMETRY sinee y'(x(y .5.. .. bl . the graph of <p.z). . .jI.am. This complete the proof of the theorem.2...) We want to convert the parametric equations into polynomial equations in the y variables alone. . b .lm: YI = h(xI..". Let V = V(I) and Iy = Ink[YI.W is. .. b . sinee 1 contains only y" variables.. We now return to the projection map (2. .X n ) = (f..z)) and y'(y . and thus I(a" . and hence 1 E .. Let I be an ideal in k[YI. as we observed above. If we view f as an element of k[Yll'" .xnL then f(al.5.y(x . ° THEOREM 2. since 1 E I.) c:: V(I(7r(V))).. 1 . One can think of this as a subset of k= parametrized by h.am) = 1 E Iy. 5. -xv 2 + yuw.an) ---> >---> W (ft(a" . . _x3 y + 1t. y4) r (2.Xn]. .··· . -xu + yr..an)).. where ft.·.Ym]/ I(W) Yi + I(W) ---> e----> k[xl. .3. and so we apply Theorem 2.2 1[.4 (x. Sueh a map Q' gives rise to a k-algebra homomorphism cr'" between the affine k-algebras k[YL. Then the Zariski closure of rr(V) is V(I n k[Yl. .fm E k[Xl' .5. EXAMPLE 2.. The polynomials that do not involve x or y are-those that determine the smallest variety containing the solutions of the parametric equations (2. . .92 CHAPTER 2.2): More generally. .··· .. ...··· . _x4 + r}. -rv 2 + u 2 w.X n]. That is we let l = (Yl - ft.5.5.y4) with respect to the lex term ordering with x > y > r > u > v > w is G = {uw 2 . .. .xnl/ I(V) as follows: a': k[Yl..an).Jm(a" .4. one computes a Grübner basis G for l with respect ta an elimination order with the X variables larger than the y variables. -x 2 v + y 2 u.x 3 y... u ..X3 y . . . xy3... Therefore. V .Yml/I(W) and k[Xl. XTV yu 2 ..Jm) ç k[Yl. . APPLICATIONS OF GROBNER BASES Consider the projection rr: Tim+n Then we can see that rr(V) is the image of 'P. ta find a set of defining equations for that variety. -xw + yv. Xl. .x4. _ xy 3 + V. . Consider the map 'P: 80 1[..2) U v W x4 x3 y xy3 y" The Grübner basis for the ideal l = (r . The polynomials in G which are in the y variables only are the desired polynomials. ' xn]/I(V) Ji + I(V).Ym..Ym . .) a: ç k n and W ç km v (a" .xy" w ..Ym])..v 3 . Note from Theorem 2.2 that the Zariski closure of the image of 'P is defined precisely by the ideal of relations of the polynornials ft. Le. _y4 + w.. y) that the parametrization is given by: (X4.. . .. . we consider maps between two varieties V given by polynomials. .4. -r 2 v + u 3 .Jm. _x 2 y 2 r + u 2 .rwuv. 5.. PROPOSITION 2.. The PROOF.Y=] Yi -+ >--+ k[XI. Let 9 E argument is clearly reversible.. . Let V be the variety in C 2 defined by X2 + y2 -1 (a circle in the x. Consider the map Œ given by the polynomials h = X2.. so that a'(g) = 0 and 9 E ker(a').f=) E I(V). Œ: V (x. In the case when V = k n we did this at the begining of this section. y) . . Note also that if the map a is the identity map of the variety V onto itself. .4. A polynomial 9 E k[YI.'" .xn ]: Ct: : V ------+ li rn (h(a" ..2. i.an) E V.an) E V. . EXAMPLE 2. xnl/ I(V) fi + I(V). g(a(a"" .. Then for any (a"". . and hence as desired.an).f=(a" .. .10 gives us an algorithm for computing the ideal I(im(Œ)). .... we have ..an) E W. and h = xy. then the corresponding map Œ' is the identity of the affine algebra k[XI.. .an)). . D This proposition together with Theorem 2... We can find I(im(Œ)) by considering the corresponding map Œ': k[YI.'" . Thus the study of the map a between varieties might be done by studying the corresponding map Œ' between the corresponding affine k-algebras. we need to show that for ail 9 E I(W).Xn)) E I(V).e. . . We will give two examples illustrating this idea: deterrnining the image of a variety and determining whether a given map is a variety isomorphism.f= E k[XI.xn ]/ I(V) onto itself (this follows since the field k is infinite).. .5. Y plane). SOME APPLICATIONS TO ALGEBRAIC GEOMETRY 93 gU.5. ..5.Y=] is in I(im(Œ)) if and only if 9 E ker(a'). . and hence g(a(xI'''' .... an)) = 0. (a" . But for ail (a"". and hence for determining the smallest variety containing im(Œ). We have then that a' is a k-algebra homomorphism. h = y2. Suppose that we have a variety V in k n and a map a into k = given by polynomials h.an) >--+ We would like to deterrnine the Zariski closure of the image of the map Œ. wehave a(a"". I(im(a)). .··· To see that a' is well-defined.6. ' . PROOF.(a" ..gn(b" . ..v + w 2 = 0 is the equation of a cylinder whose axis is parallel to the u axis.x 2.. v.bm ).Ym]/I(W) and k[x" .. . y]/ (x 2 + y2 .bm ) V (g.. . .1 = 0 is the equation of a plane parallel ta the w axis.. .gn E k[y" .. U + v-l.. APPLICATIONS OF GROBNER BASES 0.5." . and suppose that {J: W ---> ---> (b" . . ((Joa)* =idk[x" .fm(a" .Yw.. Two varieties V ç Tin and W ç Tim are said to be isomorphic over k if there are maps a: V ---> W and {J: W ---> V given by polynomials with coefficients in k such that a 0 {J = idw and (J 0 a = idv . DEFINITION 2. . .1).. v 2 .. W .*: is ---> Ta find I(im(a)).. Similarly.Ym]' Note that the map ({J Xi >--+ gi(J" . The second example illustrates how a* can be used for determining whether two varieties are isomorphic.v. V qu.y2_ v.w] u v w qu. ... Therefore I(im(a)) = ker(a*) = K n qx. . .. u . ..5. y2.xv . where idv and idw are the identity maps of V and W respectively. THEOREM 2...XV) and find a Gr6bner basis G for K with respect ta the lex term ordering with x> y > u > v > w ta get G = {X2 + V . .xw + yv _ y. .Ym]/ I(W) onto itself. .. Therefore 0:* i8 a k-algebra isomorphism ..1) ---> x 2 + (x 2 + y2 . v 2 - V + w2 ). Suppose that a: V (a" .1. ({Joa)* =a*o{J*. xnl/ I(V) onto itself. Now. the equation u + v . 0 where g" .xn ]. .an).W.xy . The varieties V ç Ti n and W ç Ti mare isomorphic over k if and only if there exists a k-algebra isomorphism of the affine k-algebras k[y" .fm E k[x" ...1) y2 + (x2 + y2 _ 1) ---> ---> xy + (x 2 + y2 .94 CHAPTER 2.. ..bm )).. where ft. First let a and {J be inverse polynomial maps.an) ---> ---> W (J...(b" . Geometrically..xnl/I(V).. we consider the ideal K = (x 2 + y2 -1. w] = (u + v-l. ..7.* The corresponding map 0.an)). The equation v 2 . The intersection of these two surfaces is an ellipse.8. . we have that (J* 0 a* is the identity of k[y" ..V + w 2}. since {Joa =idv ... fm) while the map a* 0 {J* is defined by a)* is defined by and sa.xnlII(V) and hence a* 0 (J* is also the identity of k[x" . 4. So let (x.fm(a"".16 part a to see why 1= I(V) and J = J(W)).. 13 maps W into V.u 4 +u3 + 2u2v Thus K + uv +v' + l}. .3.10 how to compute the kernel of Q* and in Theorem 2.u . ..(a" . we assume that we have a k-algebra isomorphism We will see that e = a* for a map a between the varieties V and W which is given by polynomials and such that a. . since (x.bm )). wheregj E k[Yl.an) t---+ W (f.9. since x + u 2 + v and y .5.2. v]/ J f +J --> t---+ qx. Also. SOME APPLICATIONS TO ALGEBRAIC GEOMETRY 95 For the converse. . . consider the variety W ç: IC' defined by the equation u 4 + u 3 + 2u 2 v + v 2 + uv + 1 in the u. y) E V Then if we replace u and v by y and _y' ... y]/ l f(y.l exists and is also given by polynomials. we need ta check whether a* is a k-algebra isomorphism. . y plane.m. " " " It is readily seen that a maps V into W. n k[u. D V (ail'" --> . 13: 13 Therefore.'" ..y .an)). V+y2+X) be the ideal in qu.. Consider the vitriety V ç: IC 2 defined by the equation x' yx + 1 in the x. We have seen in Theorem 2.y) --> t---+ W (y. .Ym] for j = l. and let e-l(xj+I(V)) =gj+I(W). Also. We will show that this gives an isomorphism of the varieties V and W.10.bm )..gn(b .xn ] for i = l.. x. Finally consider the map a: V (x. y) EV Now consider the corresponding map a': qu. . We compute a Grübner basis for K with respect to the lex term ordering with x > y > u > v to get G = {x+u' +v.. where J = (u 4 +u 3+2u 2v+v'+uv+ 1) and l = (x' -yx+l) (see Exercise 2. _y' . y] as in Theorem 2..13 how ta determine whether a* is anto. to determine variety isomorphism... v. and that a and are inverse maps.. .an).. v plane.no Consider the maps a: and --> V W t---+ (b . . .4.x) + l. Let e(Yi + I(W)) = li + I(V).x respectively in the equation defining W we get y4 + y3 + 2y' (-y' _ x) + (-y' _ x)' + y( _y' - x) + 1 = x' - xy + 1 = 0. where fi E k[Xl"" . ..4..5.bm ) (g.(b . u-y.u._y2 -x).10 and + 2u'v + uv + v' + 1) = J. First we show that a maps V into W.4. . and hence ker(a') = 0 a* is one ta one. EXAMPLE 2. Let K = (x' -yx+ l. v] = (u 4 + u 3 80 by Theorem 2. b) f---' (a 2 +b. if you use deglex on the u. Exercises 2.5. Show that the tangent variety of V is parametrized by x = t 3 +3t2 u. Q gives an isomorphism of the varieties V and W. 2.x2y_z2.x 3 . z = t 5 in IC 3. b.5. Find the equation in IC 2 for the curve parametrized by x = t 3 + l. Prove that a defines an isomorphism between V(I) and V(J) (you rnay 2.a+b.96 CHAPTER 2. y = t 4 + 4t3 u. Z = t 5 + 5t4 u. Verify that also I(V) = (xz .] b. 80 by Theorem 2. y.v) = (_u 2 .48y5 .5.x 2 +xy-2x-y+1). a.z2 = 0 and x 3 + y = O. d. However. z variables with an elimination order between thern.4. a.c).Z2) and in qu. you should encounter no difliculties. [Hint: The tangent variety is defined to be the union of ail the tangent lines of V.v) E W If the reader is interested in studying further the ideas presented in this section we strongly recommend the book of Cox. y = u + v 2 + w 2. In qx. Show that in IC 3.x2y .4.y2.the Zariski closure of what remains is an ellipse.x3_yZ). Compute generators for the ideal of the tangent variety of V.] d. 2. v. the map a* is onto by Theorem 2. [Hiot: Use lex with x > y > z. 2.5. for (u.6. Little.3.Z4). 2x-yz. APPLICATIONS OF GROBNER BASES are in G.] Consider the variety V parametrized by x = t 3 . and ü'Shea [CLOS]. Y = t2 • c.Z2). b. 2. 2. Let V and W be varieties in k n.z] let J = (_2y_y2+2z+Z2.yz. Do part c by solving the equations directly. Prove that I(V .16x 5z + 80xy3z .1 (u. . ShowthatI(V) = (y5_z4.5. Conclude that V is the union of this ellipse and the given plane.5.Definethemapa: V(I) ---+ V(J) by (a.y. Find the ideal of the image of a.2. c. if the plane defined by x = 1 is removed frorn the variety V = V(xy2+XZ 2 _xy_y2 _z2 +y.a-b).5. y = t 2 + 1. Therefore a* is a IC-algebra isomorphism and.v] let l = (uv+v). Find the ideal for the intersection of these two varieties and then determine aU points on this intersection.30x 2yz2 .xy3_z3. Find the equation in IC 2 for the curve parametrized by x = t 3 . your computer may llot be able to complete the computation.8.u). Show by the rnethod of this section that the variety io IC 3 parametrized by x = u + uv + w.13.] Let V be the variety in IC 3 defined by x 2 + y2 .c2 +a. Note that the inverse map is given by a.W) = I(V): I(W). Sa this exercise is done using elementary multivariable calculus. y = t 4 . Define a: V ---+ IC 4 by (a.7.v.a+b._y2+xz.1. Z = u 2 + V is all of IC"' [Hint: If you try to compute this example using lex. c) f---' (a 2 . w variables and also on the X.5. [Answer: (15x 4y2 .5. there is a polynomial i E k[x] such that gi '" 1 (mod !. under the map defined by x + (p) r--4 a. by Equation (2. the composition of the affine algebra homomorphisms </J: k[y] y ----> >--> k[xl/ (P) h + !.x+···+bmxm be the eorresponding polynomials in k[x]. .Pl).p.6. The results of this section will not be used in the remainder of the book and may be skipped.1).6. Note that h(a) = (3.e. We will assume that the reader is familiar with the most elementary facts about field extensions [Go. Altematively. that is gf .1.P) ~ >--> k(a) (3.P)). with a algebraic over k. bo+b. then the minimal polynomial of " over k is defined ta be the monic polynomial p in one variable.gy . We assume that we know the minimal polynomial p of Œ.p.4 to find the minimal polynomial of an element algebraic over a field k. + ana n (3= .1 E (p). We note that in arder to compute in k(a) it suffices. MINIMAL POLYNOMIALS OF ELEMENTS IN FIELD EXTENSIONS 97 assume that l (V (J)) = J). y . the map </J is onto.p. THEOREM 2. Let 0 i' (3 E k(a). Let p E k[x] be the minimal polynomial of Œ over k.1) k[x]/(p) ~ k(a)...6.1). considering the k-algebra homomorphism </J: k[x] ----> k(a) defined by x r--4 a.h = y .h) = !. Let k ç K be a field extension.h) nk[y]. of smallest degree such p( a) = O. Then the minimal polynomial of (3 over k is the monie polynomial that generates the ideal J n k[y]. Consider the ideal J = !.O <: i <: n. We first consider the case where K = k(. k[y] is a principal ideal domain).. and g(a) i' 0 (it is the denominator of (3). with coefficients in k.. In this section we will use the results of Section 2.P) is a field. say ao + alQ + . y]. sinee this is true for every ideal in k[y] (i. PROOF.2.P). since a is algebraic. the kernel of the map </J is !. First note that y . Let h = jf.6. and 80 (2. Recall that if a E K is algebraic over k. Note that q is in the kernel of </J if and only if q((3) = O.P).gy . Therefore it suffiees to show that !. to compute in the affine k-algebra k[xl/!.O <: j <: m.x + . Moreover. Now consider </J. Therefore to find the minimal polynomial of (3. + anx n and g(x) = bo+b.ff '" f(gy . 2. and let a E K be algebraie over k.6. we find the generator of the kernel of </J.).y . He]. Let k ç K be a field extension.4. Let f(x) = aD + a. and onr goal is to compute the minimal polynomial of any (3 E K.1) (mod !.. Minimal Polynomials of Elements in Field Extensions. By Theorem 2.p.a+···+bma m where ai.10. we have that ker(</J) = !. Note that sinee k[x]/!.f) of k[x. Note that J n k[y] is generated by a single polynomial. bj E k. . . Then the minimal polynomial of (3 over k is the monie polynomial that generates the ideal J n k[y].. y]. 0 The above result gives an algorithm for finding the minimal polynomial of an element (3 in k( a): given a and (3 as in the theorem.J) of k[x. We wish to find the minimal polynomial of (3. We compute the reduced Grübner basis G for J with respect to the lex term ordering with x > y to obtain y5 11 y4 + 4 y 3 +2 5y2 + 95y + 259 } ..98 CHAPTER 2.h E (p.X n ..Lti-d[xil be the minimal polynomial oiai over k(O!I' .Pn. . y].an). gy . APPLICATIONS OP GROBNER BASES and hence y .. .6.fR) = g(y .an)' where f. which is similar to Theorem 2..J) contained in k[Xl. Let (3 E k(O!l' ..an) using the following result.g(CXl1'" .. say (3 .1. For this we need the following notation which we use for the remainder of this section.Qi-Il Xi) = p. Therefore the minimal polynomial of (3 is y5 + y4 + 4 y 3 .h) (mod (p)). where a is a root of 3 the irreducible polynomial x 5 . THEOREM 2. Therefore gy .n and p E k( a" .6.6.O!i-I).3.Xn].Xi] such that p(Ql..gy-J) ç (p.J) so that (p.J). . Now consider the element f3 = 1_0:.hl. gy .f(O!I' . let Pi E k(al1'" . The polynomial in G which is in y alone is the minimal polynomial of (3. .. we let P be any polynomial in k[Xl' . Consider the extension field «. For i = 2. .. .2.5y2 + 95y + 259. This technique can be extended to a more general setting of field extensions of the form K = k(a" .h) ç (p. y] with respect to the lex term ordering with x > y. and (p. . . . ..x . .y .. . but every ap~ plication we will make of P will not depend on the particular choice we have made..]I[x.]I(a). For i = l.O!n) .gy.. Let K = k(O!l.O!n) be an algebraic extension of k..]I(a) of «. We now determine the minimal polynomial of any element (3 of k(O!l' .ai.. Consider the ideal J = (PI'··· .2.gy ..X .Il [Xi]... We first show that . . We note that p is not uniquely defined. We consider the ideal J = (x 5 .. Y . xy + 2x 3 + x -1) in «..g E k[Xl' .f == g(y . . .2.gy . . ln.. Conversely. 'i PROOF..::-20: E «.O!n). . EXAMPLE 2.y-h).]l.f E (p. we compute the reduced Grübner basis G for the ideal (p. . 4. We wish to find the minimal polynomial of v'2 + .X2.PnRn E (Pl"" . . Now let 1 E k[Xl"" .xn] and write fThen.48y 5 1187x2 - Xl > X2 > Y ta obtain 45y4 + 320y3 + 780y2 . MINIMAL POLYNOMIALS OF ELEMENTS IN FIELD EXTENSIONS 99 using the map We note that 4>n is anto. Consider I-PnRn E k[Xl"" .lO y3 + 12y2 60y+17.y5). x n ) E k(a" ..(Xl +X2)) = (xl -2.Pn-l)' and 1 E (Pl"" . and note that h(Œn) = O. v (f .PnCn = 0. we see that for an v. .û:n-l) kernelof = O. .6. Therefore the minimal polynomial of v'2 + .(Xl +X2)) c::::Q[Xl..'" . Therefore gv(Xl. Œn-l.60y + 17}. Œn-.Y] with respect to the lex term ordering wih G = {1187xl ."p2'Y. Œn_l)[Xn].6. D EXAMPLE 2.1).y5.1.Pn E ker(<Pn) is immediate.y5). Say h = PnCn. .y5 over Q is y6 . The proof now proceeds as in Theorem 2.1820. Œn-l.5 E Q( v'2)[X2]. . Consider the field extension Q c:::: Q( v'2. x n) = h . .PnRn)(Œl.2 E Q[Xl] and the minimal polynomial of .. + 48y 5 + 45y4 - y6 _ 6y4 _ lO y 3 + 12 y 2 .. Œn) = O. forsomeC n E k(a". 320y3 . .780y2 .y5 over Q( v'2) is P2 = x~ .2. We also see that v'2 + ...Œn are algebraic over k. The minimal polynomial of v'2 over Q is Pl = xi . .x~-5.6.735y + 1820.'" . x n] be such that I(Œl.y5 has degree 6 over Q and hence Q( v'2 + .y5) = Q( v'2. .y.)[xn].. It remains to show that ker(<Pn) = (Pl"" . The case n = 1 is Equation (2.Pn)' For this we use induction on n.. since pnfn = :L:gv(Xl1'" . gv(Œl 1 '" .Xn-l) is in the and hence by induction. The fact that P"". Therefore Pn divides h. Thus 1 .6.6y4 . since Œl. We compute the reduced Gr6bner basis G for the ideal J= !J.xn_dX~. Let h(xn) = I(Œl.452y . . . by definition of Pn." .Pn)' Thus ker( <Pn) = (Pl"'" Pn) as desired. We consider the ideal J = (Xf .4y + ~. where i the i complex number such that i 2 = .. Q( ij2.Œn). . APPLICATIONS OF GROBNER BASES EXAMPLE 2. Œi = hi(f3). the minimal polynomial of i over Q( ij2) is P2 = x§ + L Then P.y4 _4y 3 + 6y2 -4Y+~}... .6. The minimal polynomial of ij2 over Q iSPI = Xf-2 E Q[x. Let 1 = (Pl"" .th f..3. X.hl. and note that h(ŒI"" . iij2).4y 3 + 6y2 . and since the degree of Q( ij2.Qn in Theorem 2.hi.xn ] besuch that gR-l E 1.Œn) f3. . Alternatively.4y +~} . We conclude by using Theorem 2. .100 CHAPTER 2. i). PROOF.6.'" .y .xn]/I h+I k(ŒI.L We consider the element f3 = 1 + ij2. for each i = l. x§ + l. X2 + 2y 3 - 6y2 + 6y - 2. y . we used a degree argument for deciding whether k(f3) is equal ta k(al. . The reduced Grübner basis for J with respect ta the lex term ordering with Xl > X2 > Y is computed ta be G = { xi + 2y2 - 4y + 2. The minimal polynomial of iij2 over Q( ij2) is P2 = x§ + v'2. .6. Let f(e".. AU the roots of this polynomial generate the extension Q( ij2.. and leU E k[XI. . . and let G be the reduced Grobner basis for J with respect ta an elimination arder with the X variables larger than y.4 y3 + 6y2 . there is a polynomial gi E G such that gi = Xi . Set h = ff.J.'s in terms of f3.xiy . . .4. .. . Œn) = k(f3) if and only if. w.3. x§ + xi.Pn ). and P2 = P2. Consider e".6. il... Let J be as 9 Ql.Pn be as in Theorem 2. . 1 an and 1511'" . Therefore the minimal polynomial of f3 is y4 . .(Xl + X2)) of Q[XI' X2.6yX2 + 2X2. THEOREM 2. . Then k(a"". y4 . we see that Q(f3) is a subfield of Q( ij2. g E k X" .2. First. We wish ta find the minimal polynomial of f3 over Q.6.(xi + X2)).. Since the degree of f3 over Q is 4. i) over Q is 8. We again want to compute the i minimum polynomial of f3 = 1 + ij2' We consider the ideal J = (xi . 0 .X n .. . .5.Œn) = f3. say f3 = ( ) . n. [ ] f3 E k ( Œn ) . Sa we obtain the same result as before. Moreover.X~ + 1. This algorithm is a consequence of the foUowing theorem. The reduced Grübner basis for J with respect ta the lex term ordering with Xl > X2 > Y is computed ta be G = {Xl . i) = Q( ij2. Then k(a"".an)' We will give another method for determining this which has the added advantage of expressing the a.. in"this case.13 and the fact that J = (1. y]. k[y] y k[XI'''' . but is not equal ta it. In the preceding two examples.. = Pl.. for some hi E k[y].2y3X2 +6y2X2 . Let al. . an) = k(f3) if and only if 1> is onto. Thus P2 = x§ + xi.2. 1820) .6.2) 0= 11~7 (48( 0 + ?'5)5 + 45( 0 + V5)4 - 320( 0 + ?'5)3 -780( 0 + ?'5)2 + 735( 0 + ?'5) . 2. aa+3.48y 5 - 45 y 4 + 320y3 + 780y2 .7.6. The minimal polynomial of QH' over k would not do as the following example shows. E k[y].6. the Gr6bner basis G contained the following polynomials 1187x. where a. b.3.2. 0 + {12 + 5. MINIMAL POLYNOMIALS OF ELEMENTS IN FIELD EXTENSIONS 101 EXAMPLE 2.1 = 0. Compute the minimal polynomial of the following over Q.. .6. ?'5).6 .6. il.6.hl exists in G with h. Let K = .780y2 .3) ?'5 = 11~7 ( -48( 0 + ?'5)5 . Show that Q(a!. This gives another proof that Q( 0 + ?'5) = Q( 0. In Theorem 2.6. and thls gives another proof that Q(. we have (2.1 =0 c.6 + 45887' ExercÏses 2. c. Moreover.6 + 45887.2. Compute the minimal polynomial of the following over Q. Finally. no polynomial of the form Xl . ij2+ «"4+5. . .6. {12 + {15 + 5.6) oF Q( ij2..6) and express a. 4 2183 3 10599 2 8465 101499 a = 45887.0 .a2) = Q(. .45( 0 + ?'5)4 + 320( 0 + ?'5)3 + 780( 0 + ?'5)' + 452( 0 + ?'5) + 1820) . 0 + v'3 + y'5.1.6.735y + 1820..6. However. we also note that in Example 2. a. and (2.6 + 45887. 2.2 we have Q = Q(a) and ('-aa-a' ) 1438 where . a.6 = l-a-a? a .] be the minimal polynomial of "'H' over k(a" .1820.6. in Example 2. ij2+1' 2 +a h 3 .3 we required that Pi+' E k(a" . In Example 2.4. and a2 in terms of.Qi).werea . 2. .6. b 0+7 . "'i)[xH..a.45887. - 1187x2 + 48y 5 + 45 y 4 320y3 .6.5.4. a 2 = 5 and .6 = 3 2 ŒI0!"2+1 01 +02 .452y . the 3 distinct cube foots of unity. they need to have a different color. Each vertex is to be assigned one of the 3 colors 1. Let us first state the problem precisely. explicitly.'" .7.6. ç". and two vertices are connected by an edge if the two corresponding regions are adjacent. the method does not give the correct minimal polynomial of (J over IQI. (The same technique would work for any coloring.7.6. We wish to find the minimal polynomial of (J.3 ta compute the minimal polynomial of (J over k..1. The material in this section will not be used elsewhere in the text and may be skipped.) This material is based on a portion of D. and no two vertices connected by an edge are colored the same way. Xj which are connected by an edge by the polynomials in Equation (2.5.. the quadratic subfield of IQI(() where ( is a primitive 7th root of unity. 2. i VIz) and (J = e (2. Therefore Xi and Xj will have different colors if and only if xy (2. THEOREM en. ç. nQW.6.2) Let l be the ideal of QXl. We represent the 3-colors by 1.e.7. Complete the proof of Theorem 2.2). ç. . . we let ç = e'. 2. a.7. The 3-Color Problem.7. The graph gis 3-colorable if and only ifV(I) cl 0. and see that. Now. note that the minimal polynomial of ( is ~. b. Find. [Hint: .Using the method of the text. 2.' . This can be seen to be the same as the 3-color problem for a map: the vertices represent the regions ta he colored. We want ta color the vertices in Bueh a way that only 3 colors are used.6.] 2. = 1). First. we have (Xi ..6.::-. use the minimal polynomial of i VIz over IQI in the method described in Theorem 2.6.102 CHAPTER 2. APPLICATIONS OF GROBNER BASES VIz + i VIz. then g is called 3-colorable.3.1) AIso. In this section we want to illustrate how one can apply the technique of Grübner bases to solve a well-known problem in graph theory: determining whether a given graph can be 3-colored. we let Xl. ç2.7. We are given a graph g with n vertices with at most one edge between any two vertices.xn ] generated by the polynomials in Equation (2.Xj)(xy + XiXj + xJ) = o. Since = x. if the vertices Xi and Xj are connected by an edge. If g can be colored in this fashian.' E C be a cube root of unit y (i. We will consider the variety V(I) contained in The following theorem is now immediate. Instead of the minimal polynomial of i VIz over IQI( VIz). This can be represented by the following n equations IQI( VIz.' and that this quadratic subfield is the fixed field of the automorphism defined by ( >----> (2.X n be variables representing the distinct vertices of the graph Çj. Use Theorem 2.3.1) and for each pair of vertices Xi. B"yer's thesis [Ba]. 6).2. 7). (5. Let us assume that the 3 colors we are using are bIue. (6.(4.1.. (7.1. we have that V(I) of 0. EXAMPLE 2.X7. THE 3-COLOR PROBLEM 103 We have seen in Section 2.j) E {(l.7). We obtain X6 - Xs. (l. Since 1 't G. (5. The graph The polynomials corresponcling to ç ç are: = 1.11. we use the lex term ordering with Xl > X2 > . by Theorem 2.8 xf and 1. 5). because of the polynomial x~ + X7X8 + x~ E G. We must first choose a color for xs. say red. X6 must be red because . 2). and hence. Keeping in mind Corollary 2.7.2 that we can use Gr6bner bases ta determine if V(J) = 0... (2..7.6).5).7. (2.1.4).(3.8). We can use the Gr6bner basis G to give an explicit coloring.2.8)}. for the pairs (i. Then we have that Xl and X3 must be blue because of the polynomials Xl . > xs. since the only polynomial in one variable in G is x~ . Consider the graph ç of Figure 2.. .1. We then must choose a different color for X7. x~ - 1}. red. say bIue.7). sinee the system of equations represented by G turns out to be easy to solve. x3 .3).2.4). (2. If 1 E G. and green. and X4. for i Xr + XiXj + x. ç is 3colorable.8). .X7 E G. We compute a Griibner basis G for the ideal J corresponding to the above polynomials. then V(J) = 0 and otherwise V(I) of 0 (see Theorem 2.2. (l. We first compute a reduced Griibner basis for J. 6).(3.(4. Xs ~X~3~________________~~X2 '"" ______~X5 FIGURE 2. x~ + X7XS + xâ. and 80 it should not be too surprising that solving the equations determined by the Gri:ibner basis is easy.X8}. this is because the polynomials X2 + T7 + xs. > T8 > Tg ta obtain Q' = {x~ . for the pairs (i.X4X7 + X[jX7 . EXAMPLE 2.2 that there is only one way to color the graph g. (2.2. (5. The graph The polynomials corresponding ta Xi - g' g' . and X5 + T7 + Tg are in G.Xg.4). ' .X4X8 + X[jX8 . (7. Xs Xg FIGURE 2. ~ are: = 1.3. X~ + X3X9 + X~. (1. 6). (X4 . (X5 X9)(X4X7 . We note that it is evident from the Gri:ibner basis in Example 2. + xJ. up to permuting the colors. (6.X7 . + X8X9 + x~.X7XS + X3X9 + X~.7. (3. (-X5 + X7)( -X5 + xs). We compute a Gri:ibner basis G' for the ideal l' corresponding to the above polynomials using the Lex term ordering with Xl > X2 > . (8.2. + X4XS + X7XS + X4X9 + T7X9 + XSX9). (1. x.9). (X7 Xij - x9){X7 + Xs + Tg). 2 X3X4 + X4X[j .X5)(X4 + T7 + TS).104 CHAPTER 2. X6 . (3. 2). (3.Tg E G. 6). 9)}.5). + XiX. + X7 + Xs. X3X7 + X3xs . Finally X2 and X5 have the same coloT. Consider the graph g' in Figure 2. which is a different coloT from the colors assigned ta T7 and TS 1 80 X2 and Xs are green. However 1 if there is more than one possible coloring.9 3 1.j) E {(1.. This is illustrated in the following example.4).1. APPLICATIONS OF GROBNER BASES of the polynomials X4 . 8). (1.6)..X3XU . X2 + X7 + X8l Xl + X[j . (2.8). 5). for and x. ..7.7). (2. the Gr6bner basis may look more complicated.3).x g. . The integer programming problem has the following form: let aij E Z. anlO"l an 20'"2 anm(J"m which minimizes the "cast function" (2. and Cj E R. In fact.2.2. or by observing what was done in Example 2. No use of this section will be made elsewhere in the book.7. Show that if we add one edge between vertices Xl and X3 in the graph Q' of Figure 2. Q' is 3-colorable. Conti and C. then Q' is still 3-colorable..7.8.. each pair of which is connected by an edge. .2.2) c(al 1 0"2. i = 1. Generalize the method given in this section to the case of determining whether graphs are 4-colorable. ExercÏses We note that we have tried to keep these problems small.7. and hence. Traverso [CoTr]. This Gr6bner basis looks much more complicated than the one in Example 2. Show that if we add an edge between X2 and Xs in the graph Q of Figure 2.n.7. .2.1. This reflects the fact that there are many possible colorings of this graph. but your Computer Aigebra System may still have trouble doing the computations.8. 2. bi E Z.3 to show that in the trivial example where there are 4 vertices. the graph is no longer 3-colorable.7.am) in l'lm of the system (2. Use the method of Exercise 2. by Theorem 2. 2. 2. it is easy to 3-color the graph Q' by trial and error. . 2.2. .(Tm) = L j=1 m Cjaj_ . Show that now the 3-coloring is unique except for the permutation of the coloTs.5.7.8.1.7. INTEGER PROGRAMMING 105 Since 1 cf' G'. (This can be done either by computing a Gr6bner basis for the new ideal.4.1) {"'""' a210"1 + + + a120"2 a220"2 + + + + + + almeTm a2m Œm bl b2 bn . without making thern too trivial looking.. the graph is 4-colorable and show that the equations imply that all 4 vertices must be colored a different color.1. j = 1.7.3.7.8. . .. Show that in principle (by this we mean that the computations would probably be too lengthy to make the scheme practical) the method in this section could be generalized ta giving a method ta determine whether a graph is rn-colorable for any positive integer m. .) 2. The material in this section is taken from P.mi we wish to find a solution (0"11 l72.··· . 2. Integer Prograrnming. we have that V(I') of 0. 8.1) as for i = 1.. the image of the power product Y2 2 . x~n is the image under 1> of a power product in k[yl. . There are many books on the subject which the reader may consult. Our strategy is to: (i) Translate the integer progranuning problem into a problem about polynomials..4. "=) E N= is a solution of System (2. .. . Then System (2. . We have presented in Section 2. (iii) Translate the solution of the polynomial problem back into a solution of the integer programming problem. not a polynomiaL But. .4 an algorithmic method for determining whether an element of k[Xl.. say YI. In order to motivate the general technique presented below. We introduce a variable for each linear equation in (2. LEMMA 2.. say Xl. then (Ul.2))..1).3) We note that the left-hand side power product in Equation (2. Y2.8. We will alsa concentrate first on solving System (2. x~n be the image of a power product. Œ2. x n ] is in the image of a polynomial map such as 1> (see Theorem 2.. .·.1..4)..8. we will first start with the special case when ail aij's and bi's are non-negative integers.""Y=] Yj The following lemma is then clear. .1).8. see. . X2.8.. y~m under the polynomial map Yrl k[Yl..106 CHAPTER 2.. (2.Œm) E Mm of System (2. 'X n1 and a variable for each unknown (Jj. However the above lemma requires that the power product X~l X~2 •. [Schri].1) can be written as a single equation of power products or equivalently.c. Our purpose here is to apply the results of Section 2.. n... Then there exists a solution (ŒI. We use the notation set above.8. .4 to indicate a solution method to this problem.1) without taking into account the cost function condition (Equation (2. . APPLICATIONS OF GROBNER BASES This problem occurs often in scientific and engineering applications.. for example.3) can be viewed a. .Ym]' Moreover if X~l x~ .8.'" ..8.8. and we assume that all aij '8 and bi '8 are non-negative.1) if and only if the power product x~' x~' . ..8. U2. x~n = <p(Yrly~2 . . (ii) Use the Grübner bases techniques developed so far to solve the polynomial problem.. y~"')..Ym' We then represent the equations in (2. then ((Ji. lndeed. . If h = Yrly~2 . X2.m} be theideal considered in Theorem 2.4 b.8.j . Moreover.8. . X b -{==:? Xl' 2 n n G h --++ k[Yll'" ·th h E k[ YI. then it is the image of a power product Yrly~2 .8. Consider the system (2. X2 b. INTEGER PROGRAMMING 107 '" 1 because the map <p sends the variables Yj to power products in k[Xl l have Xn].8. . x~n = <jJ(h)..8.1). y~rn. Therefore. x~nj 1 j = 1.8.. If X~l xg 2 • • • x~n is in the image of <P.. . x~nj 1 j = 1.1) has a solution.. Therefore h is a power product and we are done. . .1)). Let G be a Griibner basis for K with respect to an elimination order with the x variables larger than the y variables. one for each equation. We use the notation above.. Yi. the S-polynomial of two polynomials which are both differences of two power products is itself a difference of two power products.Ym] then X~lX~2 . Now if 1 2 ••.. we consider a simple example. We have two x variables. .1) does not have non-negative integer solutions. 0 xt xt The proof of Lemma 2. Y2.3.8. one for each unknoWIl. and for finding a solution: (i) Compute a Griibner basis G for K = (Yj ..2 gives us a method for determining whether System (2.4) + + (J3 (J3 10 5. Then. (J2. X bn xl n ("') · E lm 'f' b. LetK = (Yj_X~ljx~2j .8. To illustrate the ideas presented so far.j x~.Ym]' PROOF.7. . and we assume that aU aij '8 and bi '8 are non-negative. Y3.. EXAMPLE 2. We also have four y variables.. The corresponding polynomial ..2. and the one step reduction of a polynomial which is a difference of two power products by another polynomial of the same fOrIll produces a polynomial which is itself a difference of two power products... X b.Ym]. .. Y4. x~n by G. then it reduces to a polynomial h E k[Yl. (ii) Find the remainder h of the division of the power product xî' x~' ..x~. y~tn E k[Yll'" . Therefore the polynomials in G are all differences of two power products. .. . But the one step reduction of a power product by a polynomial which is a difference of two power products produces a power product. then System (2. Xl. x~n is in the image of cjJ. We first note that the polynomials that generate K are all differences of two power products..4. during Buchberger's Algorithm to compute G (see Algorithm (1.4..... we LEMMA 2.2.m) with respect to an elimination order with the x variables larger than the y variables. by Theorem 2.. (iii) If h rt k[Yl.Ym].··· Wl 1 Ym ] ... if X~lX~2 ···x~n --S+ h with h E .(Jm) is a solution of System (2...4.. . only polynomials which are differences oftwo power products will be generated. b2 ..4).. . Instead. .XIX2. . . .bn ) = (b~. Y3. .8.. n sueh that for each j = 1. .'" .yj.n. and define tJ Therefore we have the following equation which corresponds to Equation (2.2)). X2]' The Grübner basis for K with respect to the lex order with XI > X2 > YI > Y2 > Y3 > Y4 is G = {JI.0..m and i = 1.5) We therefore proceed as before. Xl.5) can be viewed as the image of the power product yf'y. Y4 . APPLICATIONS OF GROBNER BASES map is Q[YJ.8.'" .Y3Y4. -1). -1. We may choose non~negative integers a~j and (Xj. not necessarily non-negative. Then in the a1fine ring k[xl. Y3. where l = (XIX2" 'XnW -l).8.' .-5) = (2.XrX~.. '" c--+ c--+ c--+ c--+ Q[XJ. Y2. h}' where JI = Xl .. . we introduce a new variable w and we work in the affine ring k[Xl1'" .X n . 14 = Y2 . Then and and h = Yh~ is reduced with respect to G.7.. 80 K = (YI .anj) = (a~j. .5) is a solution of System (2. we still ignore the cost function condition (Equation (2.1) are any integers.3) (2.8.0)+5(-1. h. .XIX21 Y3 . Using the exponent..2.8.a~j) +0:)(-1. 12 = X2Y4 . Y2. y':.. of h we get that (0.." under . We still focus our attention on determining whether System (2.108 CHAPTER 2. (b l .-1).. and we note that the left-hand side of Equation (2. where b. . .Xl) ç Ql[yl. and are non-negative integers for i = 1.1) has solutions and on finding solutions. b. ..-1. For example.YI. .. • x~nj + l by defining Similarly..8.•• . We will proceed as before. Y4. wlj l we can give meaning to the coset x~lj x~2j .. (-3. h = X2Y~ . 12. this cannot be done in the polynomial ring k[XI.a~j.. Now we tUTn our attention to the more general case where the aij's and bi's in (2. 15 = YIY4 .m we have 1 (alj. Of course.5. Y4] YI Y2 Y3 Y4 --. .Y4.a2j.xn.wl/I.-1. . . except that we now have negative exponents on the x variables.b~) + tJ( -1.... Y2 . for each j = 1.xn ].8.. ..Y3. 14.. .-1). X2] XrX~ XI X2 XI X 2 Xl. that is. '" . Then there exists a solution . not a polynomial.. the polynomials in G are aU differences of two power products.2. by Theorem 2.13.'x 2' ••. one for each unknown. Y3..'x2'" ·xnnw~+I is in the image of q" then it is the image of a power product Yr' y~' .8. We also have four y variables.(Yr'Y~' .8..'" 2. .O-m) is a solution of System (2. Xnnwf3 + l be the image of a power product. w] be the 1 . the argument used in the praof of Lemma 2. y~=). .8. Ifx . "'XnnJwCl:J a' . As in the first case we considered. . x~~w(3+I is the b' b' b' image underq. Let G be a Grübner basis for K with respect to an elimination order with the x and w variables larger than the y variables. We consider the ideal l = (XIX2W -1) .0-2.... As before we have b' b' b' 2.Ym]' Moreover ifx .Xn.. As in Theorem 2.'" .6) We have two x variables. J = 1. then (0-1...8. . (2. (see Theorem 2.8. of a power product in k[Yl. As in Lemma 2. Y2. . Xl. We have presented in Section 2..wl/I is in the image of an affine algebra homomorphism sueh as q. one for each equation. Then. X" . {Yj-XIa. X2. ]X2] a.2 ean again be applied.4. the polynomials that generate K are ail differences of two power products. Xnnw~+ 1 = q. y~= E k[Yl. . INTEGER PROGRAMMING 109 the algebra homomorphism As before we have LEMMA (0-1.2. Y4. . .1).0-2. therefore.4 an algorithmic method for determining whether an element of k[Xl"" . We consider the following system -1 5.8...x~.4. That is.xn.13.. Wc use the notation set above.Ym.6. We use the notation set above.. IdealgeneratedbyxIX2'··xnw-land PROOF.4..4. let K ç k[Yl.8.. and the reduction by G of a power product produces a power product.13). the above lemma requires that X l l x2 2 .m } .8.O-m) E Hm of System (2.Ym]' LEMMA b' b' b' .1) if and only ifx:. D EXAMPLE 2. . YI.5.8. We return to the original problem. Observing the exponents of the different power products obtained during the reduction.8.!7.17).Jsl --> -4 -4 -4 -4 YhPY~s Vfy§5 y J.1) described above. fg Y2Y3 . J4 = YIytyI . Y4] YI Y2 Y3 Y4 --> >---> >---> >---> " >--> Q[XI' X2. DEFINITION 2.0. . Y=~ ) cr' 0-' al and C(o-ll'" 1 O"m) < C(Œl.(YI 'Y2' . (5. we want to find solutions of System (2.7.24).l.7.10). (4.x~w2. The Gr6bner basis for K with respect to the lex order with XI > X2 > w > YI > Y2 > Y3 > Y4 is G = {fI.yi. Js YIyl l . the only requirement on the term order in the method for obtaining solutions of System (2. J. w] and the algebra homomorphism Q[YI.6) (6. is that we have an elimination order between the x. xi Thus K = (YI . (3.. As we mentioned before. we have the following solutions of System (2.11. ll o yty§ly~4 yfyJyF YÎYhl ~ YIY2Y~.8. .3. XIX2W . J.(yf' y~' .2)).y~.8.!9}. A term order <c on the Y variables is said to be compatible with the cost function c and the map <f.31). Y2. Y2 .1'=1 CjO"j (Equation (2.xiw.jh = YIY3yl° ..15.Y2. APPLICATIONS OF GROBNER BASES of Q[XI' X2.. wlj l xi + l x~w2 + l xIw+I X2W +1.38).(Tm) .0.Y3Y'l.0. h = X2 -YIYh~. J6 = YIY~Y'.0.··· .X2W. where fI = XI -YIYtyl.19. ho J4.110 CHAPTER 2. Y::"~ ) = <f. J5.Y4· We now reduce the power product x~w by G (note that xllx~ + l = x~w + I) = = yi.1) that minimize the cast function C(O"I' 0"2.8..!8. h. iJ <f.. 0. (l. Y3 .0"=) = 2:. Y3.xixi.0. Our strategy for minimizing the cost function is to use the cj's to define such a term order.8. l. x~w {j. J5 = YIYh~ . 13 = w .. J6. w and the y variables with the x and w variables larger. (2. That is. and YIY2yl is reduced with respect to G. Y4 . 2). J.1). and therefore yr y~rn . so that y~'" .4..8.. . and as long as we use an order on the y variables compatible with the cost function e and the map 10. . y~'.. then (0"1) ..= y~l .a:n) of System (2.. Wl . y:..8. yr .• l . Therefore l + yr l l .. PROPOSITION 2.8. EXAMPLE 2. Let G be a Grobner basis for K with respect ta an elimination arder with the x and w variables larger than the y variables.6 but we now consider the cost function We use the lex order on w and the x variables with Xl > X2 > w. y':n'.... By Theorem a' a' l yr It{yfl .5). .?.1) with minimum cast as the next proposition shows...1O.8.... Since l • ..•.. is reduced with respect to C. Consider the correspond- ing power product y~. .9. If XII X2 2 ••• X nn Wf3 + y~l . as long as we have an elimination order with the x and w variables larger than the y variables.."= . ''-' 'Ym~ E K...m.. Hence y~'" .2.. Now assume to the contrary that there exists a solution (Œ~.y.y~l . y~m by assumption.Y.. y~m >c y~l .y~~ . y':n= E ker(cp). the term order <c is easy to obtain. 'y~rn -.j U' b~b~ b~j3G (Tl 0" 'th YI U X 2 .3. Then the following order is a term order compatible with the cost function and the map 10: first order power products using the cost function. We use the notation set above. L et Xl respect to G. however the general case is more involved.. Finally we use 1 .. .'rr..8.8.. y':n= .. For sorne cases. we have 2.-. Then (0"1.. that is Cj 2: 0 for j = 1..'" ...) = V b' b ' XI1X22 .n cannot reduce to 0 by G.. +0"2+0"~+1000"~ or. where l • • • Yr:n"" is reduced with respect ta G.1) such that I:?~1 ejO"j < I:?~1 CjO"j. The power products in y are first ordered using the cast fUI1ctia~ a~d ~ies are broken by lex with YI > Y2 > Y3 > Y4.. INTEGER PROGRAMMING 111 Term orders on the y variables which are compatible with c and 10 are exactly those orders which will give rise to solutions of System (2..• l •. We refer the reader to the original paper [CoTrI. 0 = •. 1O?00".1) whieh minimizes the cast junetion c..~. .8. Note that 1o(y~' .• X n W ---1+ YI .. and an arder <c on the y variables which is compatible with b' b' b' G the cost function c and the map cp... . ymm.8.•.. ' Xnn w f3 J.1 +?"2+0"3 +1000"4 = lOOOO"~ + O"~ + 0"3 + lOOO"~ and Yrly~2y~3y~4 <lex y~ly~2y~3y:4. y:... 1 (Tm) is a solution of System (2.8. O"m) is a solution of System (2.n) yr y~Tn. The following illustrates this idea. We go back to Example 2... k~r(1o) ç We note that a different minimal solution may be obtained if we use a different order.1) by Lemma 2. and break ties by any other order (see Exercise 2.. YntrY> Te d lice d'th Wl P ROOF.y:. y:.'rr. But yr y~m..8."=) ~ 1o(y~. K.. y':nm -S+ O. One particular simple case is when the cost function only involves positive coefficients. y~'Tn.y~~ . That is Yrly~2y~3y~4 < y~ly~2y~3y:4 if and only if 10000"1 +0"2+0"3+1000"4 < 10000"... 2. 2.7.Y2Y3. { 30"1 40"1 20"1 + + + 20"2 30"2 40"2 + + + 0"3 0"3 20"3 + + 20"4 10 0"4 12 25. 4 -3.112 CHAPTER 2.8. Use the method of Lemma 2.1. Use the method of Lemma 2.4. + + 0"3 0"3 10 12.9 replace the cost fnnction by C(O"l' 0"2. Use the method of Lemma 2. + + 0"3 0"3 10 4. Use the method of Lenuna 2. In Example 2. Prove Lemma 2. APPLICATIONS OF GROBNER BASES an elimination order with w and the x variables larger than the y variables. 2. 2. 0"4) = 10000"1 + 0"2 + 0"3 + 0"4· 2.8.8. Use the method of Lemma 2. The reduced Grübner basis for K is 2 3 G = {w .8.8. { 30"1 40"1 20"1 + + + 20"2 30"2 40"2 + + + 0"3 0"3 20"3 + + 20"4 0"4 10 11 10. X2 . This solution is of minimal cast.1 to solve the following system of equations for non-negative integers. 0"4) = 0"1 + 10000"2 + 0"3 + 0"4· . 2.8.YIY2Y3J YIY2Y3 . Xl - 6 10 6 9 7 11 YIY2Y3 .5. 0"3. In Example 2.9 replace the cost fnnction by C(O"I. 0"3.1} .8.Y2Y4J Y4 .8.8.4. Exercises 2.1 to solve the following system of equations for non-negative integers.8. 2.8. 3.6.8. 0"2.8. 2.8. We have x~w -S+ Y1Y~Y§ which gives the solution (l. 0).1 to solve the following system of equations for non-negative integers.8.1 to solve the following system of equations for non-negative integers.8.3. 2.4 to solve the following system of equations for non-negative integers.8. . Although we will always consider the elements of A= as column vectors (which we will enclose in square brac. Then in Section 3.xnJ= and their quotient modules. In Section 3.kets).4 we show how 10 compute an explicit generating set for the syzygy module of a vector of polynomials. In Section 3.. In Section 3.5) and show that the same type of applications we had for ideals are possible in the more general context of submodules of free modules (Section 3. In Section 3. We will consider the cartesian product That is. Finally. In the next to last section we explicitly compute Hom. Am consists of all column vectors with coordinates in A of length m.1. In Section 3. The ring that will be of interest to us in later sections is A = k[Xl'''' . N) given Iwo explicitly presented modcles M and N.1 we briefly review the concepts from the theory of modules that we require (see [Hun]). in the last section we apply the previous material to give results on free resolutions and outline the computation of Ext(M. We then generalize the definition of Grübner bases and the results of the previous two chapters to modules (Section 3. we compute a presentation of Hom(M.4 to systems of linear equations of polynomials. Modules.N).2 we define the module of solutions of a homogeneous linear equation with polynomial coefficients (called the syzygy module) and use it to give yet another equivalenl condition for a set to be a Grübner basis for an ideal.. in the interest of saving space in the book 113 .8 we show how this theory can be applied to give more efficient methods for the computations of Chapter 2 that required elimination. In this chapter we consider submodules of k[Xl' . 3. In this section we let A be any commutative ring.Chapter 3.6). Modules and Grobner Bases Let k be a field.XnJ.3 we follow Buchberger [Bu79J and show how to use this new condition to significantly improve the computations of Grübner bases.7 we generalize the results of Section 3. That is. m' E M.0. (ii) for ail a E A and for al! m.. i=l m Note that an ideal of A is an A-module. as In other words.am) E Am. let M= (al. . . A set Mis called an A-module provided that M is an additive abelian group with a multiplication by elements of A (scalar multiplication) satisfying: (i) for ail a E A and for ail m E M. .aam ). (a + a')m = am + a'm. (al. . al + . = (l.. is done componentwis€.. . a.. . if al. (v) for al! m E M. The module Am..114 CHAPTER 3. that is a generating set of linearly independent vectors.em = (0.0). (iv) for al! a.am) = (aal).1) is a basis which we will calI the standard basis for Am. a(a'm) = (aa')m. (iii) for al! a.1. .as! ç Am.as) ç: Am.0). or using our space saving notation. for a E A and (aIl'" . . except that the set of scalars in the module case is the ring A which is not necessarily a field. The concept of an A-module is similar to the one of a vector space... . For example. where ai E A. .0. . we will often use vectors enclosed in parentheses. a(m + m') = am + am'.} a generating set of M..0.. . + bsa s 1 bi E A. Submodules of Am are used for linear algebra in Am in the same way subspaces of km are used for linear algebrain km. MODULES AND GROBNER BASES rQW we will usually write the elements of Am. The set Am is called airee A-module. .s} ç: Am is a submodule of Am. and in fact a submodule of the Amodule A = Al A submodule of an A-module M is a subset of M which is an A-module in its own right. For example.a... . We denote this submodule by (a" .... Scalar multiplication in Am.. In other words.. am) instead of [ am' ] a The use of column vectors is necessitated by the desire ta have function composition match the lisuai notation used in matrix multiplication. lm = m. am E M. then M = {b ." . . is called fTee because it has a basis.. a' E A and for ail m E M.. . . that is.. and we calI {a" .e2 = (0. . a' E A and for ail m E M..as are vectors in Am. a(al 1 ' " . every element a = (aIl'" 1am) of Am can be written in a unique fashion a = L aiei. . e. . Forexample. i = l.. 1. To conclude. let 1 = {a E A 1 a is the first coordinate of an element of M}. 1 is finitely generated. let ni be the element of Am with 0 in the first coordinate and the coordinates of n~ in the remaining m ~ 1 coordinates..Xn]. .. we will assume that the ring Awe are eonsidering is Noetherian.. For i = 1.). . that is.-l and so. Let ml.. Then 1 is an ideal of A. We use induction on m.. For m > 1. and the result follows from the Hilbert Basis Theorem (Theorem 1. using the Hilbert Basis Theorem.1. Note that M' is a submodule of Am. . the system of m linear equations (with unknown the coordinates of b E A') determined by the matrix equation (3. Other linear algebra problems in Am can also be translated this way into a module theoretic question.3..Now consider M' = {(b 2 .1. . Note that ni E M. then M is the "column space" of the matrix S. b2 .. . Let I=(a" . and System (3.. say by n~ l ' _. PROOF.1) has only one equation. x n ] which is Noetherian by the Hilbert Basis Theorem (Theorem 1. . given a E Am. If m = 1." . Since we are mainly interested in the ring klxl. then M is an ideal of A.1.1) Sb=a can be translated into the question of whether the vector a is in M or not.1. we show that . is finitely generated. We have the following THEOREM 3.1).. Let M be a submodule of Am. In this chapter we will develop algorithmic tools ta answer these questions in the case where A = k[Xl1'" . bm ) E M}. Every submodule M of Am has a finite generating set. by induction. M is an ideal of A. We first go back to the general theory of modules. that is. .1. So.. 1 mt E M be sueh that the first coordinate of mi is ai. 1 C. MODULES 115 If we think of the vectors ai '8 as the colurrms of an m x s matrix S. bm ) 1 (0.a.1. and henee.n~ E MI ç Am-l. then this problem is the ideal membership problem diseussed in Section 2. In the case where m = 1.1..1). . (m) = O}. then there exists no such that Mn = Mno for ail n 2:: no.(m') for all m.m E M.1 states that if A is a Noetherian ring then Am is a Noetherian module for all m ?: 1 and. Now for N any submodule of the A-module M. Then m' E M and its first coordinate is zero. Then the first coordinate m.m' E M. It is an easy exercise to see that this multiplication is weU-defined and gives MIN the structure of an A-module. we define MIN = {m+N 1= E M}. We caU MIN the quotient module of M by N. i=l t i=l m = m' + 2:diffii = i=l LCini + Ldiffiil as desired. = L:~=1 diai· Now consider m' = m .(m) = q. it is easy to see that N is a submodule of M.( M) is a submodule of M'.) = {m E M 1 q. which satisfies aq. that is.: M ---> M' an A-module homomorphism provided that it is an abelian group homomorphism. t .(m+m') = q. MINis the quotient abelian group with the usual addition of cosets. In this case we write M "" M'. any submodule of Am is a Noetherian module. we note that q. mE M. D DEFINITION 3. is one to one and onto. An A-module M is called Noetherian if and only if every submodule of M is finitely generated. Therefore m' = 2:.2 was proved. That these tWQ statements are equivalent is proved in exactly the same way as Theorem 1.2. For M and M' two A-modules. Let N = ker(q. we caU a function q.1.ni. So.1.2 is equivalent to saying that if is an ascending chain of submodules of M.(am).1.116 CHAPTER 3. Definition 3.(M) . Thus Theorem 3. for aU a E A. of m can be written as m. The homomorphism q. for aU a E A. Then. We know from the theory of abelian groups that MIN"" q.(m) + q.=1 ct. Also. in fact.1. MODULES AND GROBNER BASES Let m E M. We make MIN into an A-module by defining a·(m+N) = am + N. q.2:!=1 ~mi. is called an isomorphism provided that q. . .. . The second way ta have an explicitly given module M. the image of q. . the submodule M = ('lnl + N. Since 'Ins generates M we see that q. more generally.. ..1. an A-module homomorphism as is easily seen and thus is au A-module isomorphism.aTn E AS sueh that M ~ AS IN for sorne explicit isomorphism. . + N) of AS IN is explicitly given.(ei) = mi. .3 ensures the existence of such an sand N. 80 the first question we have to answer is what do we mean when we say that we have an explicitly given finitely generated A-module M? The first way is to be given N = (aIl'" .1. . the submodules of MIN are ail of the form LIN. Consider the map q. Lemma 3. . is onto. ..s. then q.m..(M) q. let M be an A-module. for i = 1....ms of the A-module M.. . Let N be the kernel of q. by the First Isomorphism Theorem for modules.1. Now given generators ml.fit E AS such that M = ('lnl.ms E M. .3. As in the theory of abeliau groups. is au A-module homomorphism." . . as) = L: aimi' i=l s It is easy to prove that q. if we have an explicitly given submodule N of AS. ..m. Our purpose in this chapter is ta do explicit computations in finitely generated modules Qver A. This fact is referred to as the First Isomorphism Theorem for modules. MODULES 117 as abelian groups under the map MIN m+N ----> ----> q.3. is the submodule of M generated by ml) . Every finitely generated A-module M is isomorphic to AS IN for sorne positive integer s and sorne submodule N of AS.1. Then. namely by specifying q.ms' Hence if ml. This rnap is. where L is a submodule of M containing N. So we conclude LEMMA 3. DEFINITION 3. Now.(a" .. . If M ~ A' IN. .: AS ----> M defined as follows: q. . provided that M is a submodule of AS.. ' .(ei). we define cp as above. Letting el. is onto. is ta have explicit ml. We will often define homomorphisms q.es denote the standard basis elements in AS 1 we note that cp is uniquely defined by specifying the image of each ei E AS.am) for explict ab . and let 'Inl.. Moreover. in fact. from AS by simply specifying q.ms generate M. then we cali A' IN a presentation of M. . M~AsIN..(m).. ...4.). we have m" . Or... . . . ... .hs) = Fh and SYZ(f" . M ~ AS IN.Js) is the set of aU solutions of the single linear equation with polynomial coefficients (the Ns) (3. L i=l hdi' As we have seen in Section 3..1.1.. We consider the A-module homomorphism 10 defined in Section 3.1).) be an ideal of A. Every finitely generated A-module M is Noetherian. and 3. Since AS is Noetherian (Theorem 3.. The kernel oJ the map 10 is called the syzygy module oJ the 1 x s matrix [ fI Js J. By Lemma 3.2..2. MODULES AND GROBNER BASES tWQ It is also useful sometimes to have a presentation of M in the last we will show how we can obtaîn this in Section 3. and h = [ ~: ] E AS. where the solutions Xi are also ta be polynomials in A. . and hence every submodule of A' IN is finitely generated.. 3. + .... = O. where L is a submodule of AS containing N. An clement (h" . + hsJ. .. . Js) is the set of all solutions h of the linear " equation Fh = O. + JsXs = 0. 0 3..hs) = [ fI " = Lhdi' i=l That is.... cases. Let l = (fI. then lo(h . if F is the 1 x s matrix [ fI Js J. .8.2) fIXI + .3. 10: A' given by ---> l S (h"". The submodules of AS IN are of the form LIN. Grobner Bases and Syzygies. ... as A-modules.1. .1.1) DEFINITION l ~ AS1ker(lo). In this section we let A be the Noetherian ring k[XI"" .xn ]..1.118 CHAPTER 3.1) (3...!s) is called a syzygy oJ [ fI Js J and satisfies hIf. for sorne s and sorne submodule N of AS. we have that every submodule of A' is finitely generated.. hs) ... Therefore A' IN and hence M is Noetherian.J. We note that the map 10 can also be viewed as matrix multiplication: s lo(h .2. . Another way to say this is that Syz(fI.. COROLLARY PROOF...2. . ... .hs) oJ SYZ(f" .Js).5.. It is denoted SYZ(f" . .) is a very important abject in commutative algebra. w) and (-z. .2.. -x... GROBNER BASES AND SYZYGIES 119 EXAMPLE 3. -x)) ç AS.yw) + y(xy . h 3) >---> h (X 2 .). homogeneous systems sueh as Equation (3.2. -x) are both syzygies of [ x2 since - yw xy - WZ y2 - xz ] . xy . For i # jE {l. . . < _ S} .xz) = O.xz) = ((y... bepower " products in A.xy-wz.X .yw) .. Syz(x 2 - yw.. . y. and 1= (x 2 -yw. we can view SYZ(f" . Also. { --ei CiXi CjXj where el.1). l (h" h 2. we define X ij = Icm(Xi. Because of the isomorphlsm given in Equation (3. its use will lead to improvements of Buchberger's Algorithm (see Section 3.j. . The map given by q.. w). We note that SYZ(f" . .xz). y(x 2 ..: A 3 ----.1). .2. similar to the role they play in the usuallinear algebra over fields. mare generally..2. in particular. w].. LetcI. that is. (-z. . such systems in Am play a central role in the theory of rings and modules.wz..1. _ ~ < J.x(xy ..j.2) or..s}.. ..yw. . since it is a submodule of A' (Theorem 3. E k-{O} and letX X 2. The next lernma shows how to compute these generators in a special case (the general case is presented in Section 3. Let A = lQI[x. Syz(fI. ' Then (y.2) that in fact these two syzygies generate Syz(x2 .wz.4.j. One of our goals is to compute generators for SYZ(f" .X j ). y2 . PROPOSITION 3. . .3.) will play a critical role in the theory of Grobner bases. xy .wz) .c. For these reasons and others.j.. y.X(y2 .j.. Mareover.) as the set of alilinear relations among fI. y2 . .e j E A' Il <. .2. . We will show later (see Example 3.) is generated by " ij Xij X ..X.. . is q. the ideal l can be described as a quotient of a free A-module and SYZ(f" . Finally SYZ(f" . Then the module Syz(c.X.wz) + h3(y2 . y.3).)..es form the standard basis for AS.2. .xz).4)..yw) + h 2(xy ..3. . .y2 -xz). z.j. -x..wz) and + W(y2 - xz) = 0 -z(x 2 .c.) is finitely generated.. j. .s.. . .X. Then the coefficient of X in h. . .0. we have Il:.X. CjX CjX --i O.. + hscsXs must be zero... j ith coord ~ jth coord Therefore SYZ(C. let (h" ..1 " . . MODULES AND GROBNER BASES PROOF. +c~tCiJc. ' 't 't ~O 1 Recall that (al.0) = 1(0. Therefore.c. +- +(C~ICil + .. that is. + C~t Cit = O... using the same technique as in Lemma 1.. be the non-zero cj '8.h.. + .0..) be a syzygy of [ C..X. .0... Let X be any power product in 1l'n.... ..i. Then we have ci Cl C2C2 + .. i < j :. + h. . First note that for all i i j we have X ei CjXi ij - CjXj X ij ej is a syzygy of C.---.. + . . . + C~Cs = C~l Cil + . .5. s) ç hi <X:.X.as) stands for the column vector bs as in the lisual matrix multiplication..). ..C~t' with il < i 2 < . = O...X . and where c~ = 0 or XiX~ = X for a fixed power product X.. " To prove the converse.--. ..120 CHAPTER 3. < i t . 1 We have adopted this awkward looking convention instead of the lisuai "dot product" because it will be consistent with the notation of computing syzygies of column vectors of matrices later.7. csX...c.X . Let <1' . . h. Thus it suffices to consider the case for which = i = 1. Xij Xij .c.0. e.)). We say that h E Syz(e.). ... The next theorem presents another equivalent condition for a set ta be a Gr6bner basis.X. ..9. .2.h.. DEFINITION 3. then I::~lhdi has a " leading term strictly smaller than In particular the syzygy ----.2. - ~ jth coord Xi} CjX ' j 0. ith Xi} Xij CiXi' 0. . .0. Then. Let B be a homogeneous generating set of SYZ(lt(g...X .g.X .X.} be a set of non-zero polynomials in A. .h....) E Syz(e.. THEOREM 3. Then G is a Grobner basis for the ideal (g" . .Xs be power products and CI.h. In order to implement this idea we need the following definition. for a power produet X.) if and only if for aU (h" ....) 1 might be relevant to the computation of a Grübner basis for (J" . Let G = {g" .3 consists of a finite set of homogeneous syzygies.:!:Lei CiXi x· x· J X't Cj j ej of [ C1 X 1 CsXs J gives rise to the S-polynomial of fi and fJ. ..2.0.2..csXs are the leading terms of the polynomiais f" ..) E Syz(e. + h. ... 1 Cs be in k . we have h. Let X l.3.) is homogeneous if it is homogeneous of degree " X for some power product x.2. .e.3 seem to indicate that the syzygy module of [ It(f...g ..9' G ----> + O........e..{O}. "It(hi ) = hi for aU i) and Xi Ip(hi ) = X for aU i sueh that hi i' O..-X fJ = S(j"fJ)· Cii Cjj Xij This last observation and Proposition 3.. . + .).. . ..4. D We observe that if C1 X 11 .f" and if (h" . sinee [fI ls ] (0. 10) = coord -X fi .) homogeneous of degree X provided that eaeh hi is a term (that is..) ItU.) E B. . GROBNER BASES AND SYZYGIES 121 as desired. . .X. Note that the generating set given in Proposition 3. .)..It(g.. .5. . .f. . .X . we caU a syzygy h = (h" . .. h. let 9 E (g" ..t} IIp(ui)lp(g.ht. ..he}.. . hj = (hlj) . .t least....g. .. By hypothesis.. Then hE Syz(lt(g.2. J - The latter strict inequality is beeause L:~l hij lt(gi) = Thus.3) for 9 with a smaller value for X.2.j sueh that ajhij # O. Thus by Theorem 1.9 we have for eaeh j = l. . Sinee..3) with X = lp(g).i::.3) 9 = u. we need to show that 9 ean be written as in Equation (3.g . where for each j = 1..g. then by Theorem 1. Let h = LiES lt(ui)ei (where e" . we may asssume that the aj '8 are terms Bueh that lp(a]) lp(hij ) lp(gi) = X for all i. gt).. MODULES AND GROBNER BASES PROOF.2.6.. 80 h = L~=l ajhj.... say (3. o.. + Utgt Llt(ui)gi + L(Ui -lt(ui))gd LUigi iES terms lower than X LLa h j j=li=l e t ij 9i + + terms lower than X L L aj v j=l i=l e t ij9i terms lower than X .2. Let S= {i E {l.) =X}. If G is a Grübner basis.6... 9 U. for eaeh j.. . + . +Utgt. where..et is the standard basis for At).3) with x = max (lp(ui)lp(gi)) l::.. .2. . Choose a representation as in Equation (3.+ O..5. since h is a homogeneous syzygy.f t t L i=l h ij9i = L i=l V ij9il sueh that t l<i<t - max lp(vi·) lp(gi) = lp('" hi ·gi) J L i=l J < l<i<t max lp(hi ·) lp(gi). Conversely. L:~l hijgi . .. l htj ).lt(gt)) and h is homogeneous.2. we may assume that lp(g) < X and show that then we ean obtain an expression sueh as (3. . Now let B = {h" . by Theorem 1. 1 P.. + htgt --7+ 0 G for all h" . iES + . .. Then Llt(ui) lt(gi) = iES o. + .122 CHAPTER 3.2..). 4.2.2.. Consider the set G = {g" g2}. (x. In fact.. .2.2..2. 0 1. In Theorem 3... .2. For the converse.). = x + y. gj) ~+ O. + v.3.xlp(aj) lp(hij ) lp(gi) = ~. 7. G is a Grobner basis.l)g2 --++ 0 and that xg. .. We observe that the above proof uses an argument similar to the one used in the proof of Theorem 1.. PROOF.Xg2 --++ O.2.lt(gt)) was required to be homogeneous..4. Prove that (x + l)g.. We have obtained a representation of 9 as a linear combination of the gi '8 such that the maximum leading power product of any summand is less than X.2. -x .1).. Therefore.xlp(aj) lp(vij) lp(gi) t. Prove that G is a Grobner basis for the ideal (g" . each element of B gives rise to an Spolynomial.J x.t. 8(gi..t. + (-x . . gt) if and only if for all (h"".gt} be a set of non-zero polynomials in A.2. GROBNER BASES AND SYZYGIES 123 We have ma.. . COROLLARY 3. we have 8(gi.5 the generating set B of SYZ(lt(g.1. Prove that G is not a Grobner basis for the ideal it generates. + . a. Exercises 3. 3.3.. . 3.t } ÇA t is a homogeneous generating set of the syzygy module of [ lt(g.. g2 = X + 1 E <Q[x.. Let B be a homogeneous generating set of Syz(lt(g.2. which reduces to zero by hypothesis. Let G be a Grobner basis.j = 1.J < ma.g" G G .. as a corollary to the above result.) E B. .).g. yi· We will use lex with x > y.).2. . Let G = {g" .3 ta see that the set Xij Xi.5 is necessary.1 we will give an example which shows that the hypothesis that the generating set of syzygies be homogeneous is necessary. .3. .gt).. B= { lt(gi)ei-lt(gj)ejl'<J.5. Then.7. Thus the theorem is proved. Let G = {g" . Then G is a Grobner basis if and only iffor aU i. since 8(gi.lt(gt)). gj) is an element of the ideal (g" . . Prove that the set {(x + 1. ...2. by Theorem 3.. we can recover Theorem 1. -xl) is a generating set for Syz(lt(g. 0 In Exercise 3.g. + . . b. .J=I. lt(g2))' c.) lt(gt) As we have noted after Proposition 3. + htgt = V. Give another example that shows that the hypothesis that B be homogeneous in Theorem 3. where g.gt} be a set of non-zero polynomials in A.6. we first use Proposition 3..gj) ~+ O. we have h. In thls exercise we show that this hypothesis is necessary. .h. . . Give a method for fimling elements of Syz(fI. [Hint: Think of 3 x 3 matrices whose determinant must be zero. Sorne algebraic results. Huy. GMNRT. y] generated by fI. the desired Grobner basis is obtained but we do not "know" it. We note that the results of this section are rather technical and will only be used occasionally in sorne of the exercises and examples in the remainder of the book. Recall that the algorithm has two steps: the computation of S-polynomials and their reduction. the scope of the applications of the theory.] 3. So we will now discuss improvements in the algorithm for computing Grobner bases. where 1.4.2. . Indeed. h hl. Generalize this idea to the case of m x (m + 1) matrices with entries in k[Xl"" .. Since the algorithm does terminate. + .2. in practice. Xn]. such as the results presented below. y a. the number of polynomials in the basis gets larger.] b.1). at some point before the algorithm terminates. the proportion of S-polynomials which reduce ta zero eventually increases as we get far in the algorithm. Bu83. S=[x 1 x y xl. However the computational complexity of Buchberger's Algorithm often makes it difficult to actually compute a Grôbner basis for even small problems and . lp( V.this limits. in certain instances we can easily find sorne elements of SYZ(f" .3.g . MODULES AND GROBNER BASES where lp(h. We will see in Section 3.. h h the 2 x 2 minors of the 2 x 3 matrix S.)).7.). So far we have not discussed the computational aspects of Buchberger's Algorithm presented in Section 1. At this point we do not know yet how to compute generators for the syzygy module of [ fI J. [BaSt88. .) lp(g.. .. Improvernents on Buchberger's Algorithm.). for example.4 how to do this. since few new polynomials are added to the basis. + h. A carefullook at this issue is outside the scope of this book (see. as the computation progresses.. 3. MaMe.g. J.5.x(lp( vIl lp(g. and obtain 2 syzygies this way. In fact. However. Laz91])..124 CHAPTER 3. [Hint: See the proof of Theorem 3. the number of S-polynomials to compute also increases. each time a new polynomial is added to the basis.7 (Algorithm 1. It can also be shown (but we shall not do it here) that certain algebraic (respectively geometric) properties of the ideal l (respectively of the variety V (I)) have a direct influence on the difficulty of the computation of a Grôbner basis for l (see the references above). and therefore.) = ma. Let l be the ideal of k[x. A problem that arises is the potentially very large number of S-polynomials that have ta be computed. A huge amount of computation IDight be performed for very little gain. can be used in COlljunction with sorne simple heuristic observations ta improve significantly Buchberger's Algorithm (see [Bu'T9]). . {J. lt(g) = bY.4) S(j.g) = O. g} is a Grobner basis if and only iflp( ~) and lp( prime. where lt(j) = aX. Then.'bg'j.g')i".'.5. LEMMA 3.3. Then f and 9 are both terms and S(j.(j .g) = ~Yf .I. In parlicular.'bfg = . .g'J).'. (ii) S(j. bath non-zero. b E k.(j'g .3. we see that the leading term of ~ ((j' . Using an argument similar to the one above.g) ~+ O..J'lg = . g) .3.. (i) =? (ii). Therefore this reduction process continues using only f Or g. At each stage of the reduction the remainder has a leading term which is a multiple of lp(J) or lp(g). .g) = ~Yf -l. we have lp(j) divides lp(j') and lp(g) divides lp(g').g')f _. since gcd(lp(j). products. IMPROVEMENTS ON BUCHBERGER'S ALGORITHM 125 At that point. l' = 0 and g' # O.lt(g')) J) is a multiple of lp(j) or lp(g).J'l..Xg = '. This is a contradiction. S(j. a. 9 = bY + g'. the computation of S-polynomials and their reductions are ail together useless except for the fact that they verify that we do have a Grübner basis.(g . Then X = ~(j .I. The following statements are equivalent: (i) lp(~) andlp(~) are relatively prime.~Xg = . and let d gcd(j.(g' .lp(g)) = l. We write f = aX + 1'. Y are power . and the leading term of '. we have S(j.!.'b(g . l' = g' = O. we have S(j. If lp(j' g) > lp(g' J).. Let f.g E A = klx" .1. that is. CASE 1..lt(j'))g . and Y = l. a) are relatively PROOF.g)"!"" :b(j'g.9').lp(g)) = l.'b (j' 9 . g) ~ + O.+ O. Then.xn ]. We see that (Exercise 1. 9' # O. and X. l' # 0 and CASE 4. Assume first that d = gcd(j. then S(j. The following results are the basis for two criteria for a priori zero reduction of S-polynomials.g' f) S(j. then --"--> :b ((j' -lt(j'))g .(g . lp(g)) = l. and since gcd(lp(j). a ab ab ab If lp(j'g) = lp(g'J).. CASE 3. l' # 0 and 9' = O. Therefore lp(j' g) # lp(g' J).(g' -lt(g'))J). then lp(j')lp(g) = lp(j'g) = lp(g'J) = lp(g')lp(j). We see that we can continue this process until we obtain 0. since gcd(lp(j). g) = 1.g' J) or ~ (j' 9 . b This is the same as Case 2. g) If lp(j' g) < lp(g' J). One way to improve this situation is to ((predict" that sorne S-polynomials reduce to zero without actually computing these S-polynomials or reducing them.I.9' f) appears in l'9 or g' f and hence is a multiple oflp(j) or lp(g). CASE 2.g). since lp(j') < lp(j) and lp(g') < lp(g). Now assume that d = gcd(J.I) are relatively prime as desired.126 CHAPTER 3. ged(X.13).1). We note that if Ip(J) and lp(g) are relatively prime. This is the form in which we will use the criterion below (see critl).g) will not create a new polynomial in the basis.g)) < Xlp(g) = Ylp(J) = DXY.. d~} is also a Grobner basis (Exercise 1. since S(J.u) = Y. Then {J. From Equation (3. . By Theorem 1.4. X.3).g) of 1. g) = 1. The last statement of the lemma is an immediate consequence of Theorem 1.1 gives a criterion for a priori zero reduction: during Buchberger's Algorithm. by the above.g) of 1. It is easy to prove that if S(J. .u).6. then it is not necessary to compute S(J. . Now we tUTn OUT attention to another criterion that turns out to be remarkably effective in improving the performance of Buehberger's Algorithm. Y) = 1. where lp(uf) S. It is also easy to prove that âS(J. we have lp(~) and Ip(. we obtain. Also. (ii) =? (i). g) will reduee to zero using 1 and 9 alone.I) = 1.g) ~+ O. Let lp(J) = DX and lp(g) = DY. Then ged(~.g} Y By assumption we have S(J. then 2S(J.1) X S(J. and hence.~} is a Griibner basis.xnl Y X S(J.3. 1 and 9 are + v). Thus {d~. 0 Lemma 3.4.3.9) Ill} + 0 (see Exercise 3.g) ~+ 0. (lc%) Therefore 1 divides (let) relatively prime. and hence lp(lerf) . Then gcd(~.7. By assumption We have gcd(lp(~). v E k[Xl"" . lp(u) < Y. We need to show that lp(J) and lp(g) are relatively prime. lp(S(J. and hence S(J. g) = S(~. g) = lc(J/ . g). and 9 divides (lerf) .g)) and lp(vg) S. H. Therefore Ip(J) and lp(g) are relatively prime.u). But 9 divides Cerf) . Let us first assume that gcd(J.g)).lp(~)) = 1.3.u) = Y. and Y are power products in A. lp(S(J. sinee lp(u)DX = lp(uf) S.g) = lc(J/ . by the case above. Thus.lc(g)g = u1 + vg.1). S ( 1. and hence there exist u. +v) 9 = Ceyf) -u) f.g ) ---->+ O. so lp(g) = DY divides lp( lerf) . since d is monic (see Exercise 3. then d = 1 and so S(J.3.3.lc(g)g· ~+ 0.:1) = 1. and hence D = 1. whenever 1 and 9 are such that lp(~) and lp(:'l) are relatively prime. lp(S(J.3).7. Therefore. g) such that (3. MODULES AND GROBNER BASES Now assume that d = gcd(J. where D. .. + .s+1 Il ::. 'Tjl.. and such pose we have three distinct indices i.Cs E k ..3. corresponding to one of the Tij remaining in Band reduce it as far as possible. . X j ). .cs X s ) ÇA. we add the reduction to the set {h.3...e j qXi CjXj Xiji Xiji .. For i.e .xn ] and let Cl. We have Xijl! --Tij Xij Xiji Xiji +-TjE + --7fi = Xii XCi ij X ij ..).)).j.Xj ). Let B ç {Tij 1 1::. be power products in A = k[Xl.--ei = ClXi CR..{Tij} is also a generating set for SYZ(C.. For eachi.f = 1. Tjf Hence Tij is in the submodule of generated by and TRi' 0 COROLLARY 3. Let {f" . Tjf Xii Moreover....3 to improve Buchberger's Algorithm in the following way." {O}. c.XR. .) CeXi X ij ' ('Xli +-e. " We will use Corollary 3.j...3 ta eliminate as many of the Tij E B as possible obtaining a possibly smaller set of generators for Syz(lt(h). Xi).e j E Syz (C1Xl.X.3. Let X X 2 .j = l. . .X .csX. . We again apply Corollary 3.3.T b Xji Xii AS Xij Xij = O. . s} to obtain a new set B which now generates Syz(lt(h).3. i < j ::.lt(f. CjXj Xiji +-ej CjXj p Xe. Xije Xli) --ei CiXi = --ei .3. . . .ij. .xn].8 letXijR = lcm(Xi .. " E B.2. . CiXi Xiii! Xiji 0 .i.Js} be a set ofgenerators for an ideal l in k[Xl.. calling it Js+l. then and TR. CiXi 1 CjXj where el. . es is the standard basisfor AS. . lt(fs+l)). . . . Ii).. s} be a generating set Jar SYZ(C. then Xiji = Xij.. lt(fs). . and we have Tij + --TjR + .. Tij that X.i CeXp. S(Ii. We begin by letting B = {Tij 1 1 ::. . and let ij X ij Xs Tij = --ei . We continue this process until . Sup. s}.. i ::. i < j ::. ( --ei X.s. lt(f.. IMPROVEMENTS ON BUCHBERGER'S ALGORITHM 127 LEMMA 3. We apply Corollary 3. .... It(fs+1))' We again compute an Spolynomial corresponding to a Tij remaining in B.3. divides X ij = lcm(Xi .. . Xj.X . Of course.3 to eliminate as many of the Tij E B as possible obtaining a possibly smaller set of generators for Syz(lt(h).)... f such that 'Til. . Now if Xe divides Xi).. We continue ta use the notation oJ Lemma 3. XP. Js} if it does not reduce to zero. B generates Syz(lt(h). is in the submodule of AS generated by PROOF. lt(fs))' We then compute the S-polynomial.3. We enlarge B by the set {Ti.'" .3. Then we have --Ti) Xi)1! Xij Xiji Xiji +-Tjf + --iei = XjE Tij O.. if Xi divides Xij. define X ij = lcm(Xi...Xej ) Xij CiXt CjXj X ij ' (Xji +-ej XjR. ..2.. .. Then B .e . Let CiXi = lt(fi) and use the notation above. . .Xs ). . we only check.. These two cases are the two main WHILE loops in Aigorithm 3. we know without eomputing it.}. . In order ta keep track. Sinee p" p are interchangeable. PROPOSITION 3. j} is in N C sinee we are only interested in eliminating it fromNC. Either the S-polynomial reduces ta zero. s} cannot lie in C (this explains why checking mernbership in NC U C was often just done by checking membership in N C). We note that the purpose of the first WHILE loop is to initialize NC..3. Mareover. So we continue the algorithm until NC = 0.. MODULES AND GRbBNER BASES all of the S-polynomials eorresponding to elements in B have been computed and redueed to zero. always maintaining B as a basis of the current syzygy module. Because of the way we cali this procedure in the main algorithm (Algorithm 3. Given a set of non-zero polynomials F = {fI.{".2.j) = TRUE. of those Tij E B whose corresponding S-polynomial has been computed and reduced we break up the basis B into two parts. that SUi.. We nQW make a few technical points about this algorithm.1.3.3. At each stage of the algorithm there is one of two possibilities.4). This is because the cases of aH triples with v. j} of Tij EBat any given time for which the S-polynomial has been computed. j} E N C U C} is a generating set for the syzygy module of cuuent leading terms. The command crit2(NC. The command.2. We note that at any time in the algorithm after N Chas been initialized. by Lemma 3. j} E C} is a generating set for Syz(lt(f.3. and . {Tij 1 {i. critI (i.) and IpUj) are relatively prime.3. We now give an improved version of Buchberger's Aigorithm as Aigorithm 3. then. PROOF. .1) will produce a Grôbner basis Jar the ideal l = (F).p"p sueh that {v. returns "TRUE" if and only if IpU.3 and the ensuing discussion. Let G = {f" .1). p ail less than s were checked before. whether {i. The basic idea is to find triples of indices ". We first note that the S-polynamials carresponding to every pair in C reduce to zero. Jj) reduces to zero.p}.p. We also use just the indices. .4. j} E N C u C} is a generating set of the syzygy module of the current set of leading tenns (see the proof of Proposition 3. It then suffices ta show that {Tij I{ i. j} of Tij EBat any given time for which the S-polynomial has not been computed and use C for the set of all indices {i.Jt} (t 2': s) be the output of Aigorithm 3. We see this as follows. X p ). M. it is enough to consider the cases p = sand " = s. J-L.C. So we will use N C for the set of ail indices {i. and for this it suffices ta show that at any stage of the algorithm {Tij 1{i. Finally. Aigorithm 3.3. .)).3. the Improved Buchberger's Algarithm (Algarithm 3. we need only consider the cases where one of v.2 is an implementation of the ideas in Corollary 3. in the algorithm. The commands used in the algorithm are defined as follows. j). in the second main WHILE loop. p is s. we note that a pair of the form {i.. Nevertheless Tij must be added to C.{p"p} are in NC U C and Xv divides lcm(X~. If critI(i.3..lt(f.128 CHAPTER 3.3.J.3.3.).1. .1. s) is given as Aigorithm 3.}. 1 stopped. D We note that in Algorithm 3.. Improved Buchberger's Algorithm Neuc does not change..... IMPROVEMENTS ON BUCHBERGER'S ALGORlTHM 129 INPUT: F = {fi. The algorithm stops for the same reason Algorithm 1.j}} IF crit1(i. or a polynomial is added to G. .i+ 1} Il ~j ~ i} Ne := crit2(Ne.) -'"->+ h. where h is reduced with respect to G := IF hT' 0 THEN Is+1 h G:= Gu {fs+i} s:= s + 1 Ne:=Neu{{i. i + 1) i:= i +1 0 DO WHILE Ne T' Choose {i.1 we do not give a rule for choosing the pair .3. 1. C.fs} ç k[Xi.j) = FALSE THEN S(fi..3.j} E Ne Ne:=Ne~{{i.C.3. . Then we apply crit2 to the new Ne which does not alter this last statement by Corollary 3.Is) INITIALIZATION: G:= F :=0 Ne:= {{l. . XnJ with fi T' 0 (1 ~ i ~ s) OUTPUT: A Grübner basis G for (fi. .j}} C :=CU{{i.s}11~i~s~1} Ne:= crit2(Ne. and the relevant pairs are added to Ne u C and so the set of Tij'S corresponding to this updated Ne u C is a generating set for the syzygy module of the new set of leading terms. .3.3. s) ALGORITHM 3.7. .3.1. 2}} i:= 2 C WHILE i < s DO Ne :=Neu{{j. crit2(NC.130 CHAPTER 3.j} ENC THEN IF X.2.e} ENCuC AND {i. Often the S-polynamials are computed in such a way that S(fi. fJ) is computed first if lcm(lp(fi)' lp(fj)) is least (with respect ta the current term order) among the lcm(lp(fv).s} E NC THEN := 1 i:= 1 WHILEi < s DO IF {i.s}} i:= i + 1 f:= f +1 i:= 1 WHILE i< s DO IF {i. Sorne studies have shawn that this choiee is of vital importance.3 INITIALIZATION: f WHILE f < s DO IF {f.X j ) THEN NC:=NC-{{i.s from Algorithm 3. Experimental results show that this warks very well for degree compatible term orders. divides lcm(Xi .3. This procedure is called the normal selection strategy.3.s} ENC AND {i.C.1 OUTPUT: NC with pairs delete" using Corollary 3.3.j}} j := j +1 i:= i +1 ALGORITHM 3. It is not sa .) THEN NC:=NC-{{i. s) {i.s} E NC THEN j := i +1 WHILEj <s DO IF {j. X.j} E NC far which we compute the corresponding S-polynomial.C.s} ENC THEN IF X.lp(f~)). MODULES AND GROBNER BASES INPUT: NC. divides lcm(Xi. 5}}. {l. h}. J5) = xz 2 ta G. Thus we add J5 = yz3 . and we compute S(h. a Grobner basis for (fI. . {2.3. 2}. 2}. {3. 5}. Thus we add xz. and wc compute S(h.3) = {{l. There is another technique that is used ta improve the performance of Buchberger's Algorithm. changing Ne ta Ne = {{2. 3. then the normal selection strategy will first consider only fI) .xz21 in Qlx. 3}}. we have Ne =crit2(Ne. ft disregarding the Orres in which x n appears. that is. changing Ne ta {{2. There are techniques to get around this problem but they are beyond the scope of this book. We follow Algorithm 3.6}.3. the palynomials Ns are inter-reduced at the beginning.3. 2}}.. and compute SU" 12) = x'yz . {2. 4}.. it helps to avoid an unmanageable growth in the number of polynomials in the basis. 5}. {l.4}}.3}. {3. {3. 2}. update Ne and C = {{l. 6}}. zJ. 4}} and C to C = {{l. we will not trace the algorithm crit2 except for one significant example at the end. IMPROVEMENTS ON BUCHBERGER'S ALGORlTHM 131 good for the lex term ordering. {3. we have put boxes around these polynomials for easier reference. . {3. 2}. 3}}. 1 and = xyz3 .5 . 6}}.xz 2 which is reduced with respect to G. 4}} and C = {{l.xz 2 which is reduced with respect ta G. 4}. We update Ne to include the pairs with 4 in them and then we use crit2 to compute the new Ne = {{2. 3}. This can be explained as follows: if at sorne point the basis contains polynomials fI. In fact. After applying crit2 again we obtain Ne = {{2. 3}.5. 3}. We conclude by noting that the use of critI and crit2 help to reduce the number of S-polynomials that have to be computed. 3}. sa that now Ne = {{2.3.z3 which is reduced with respect to G..4}. 3}. {3.z31 ta G. in effect. 4}}. Experimental and theoretical results show that this Grübner basis subcomputation can be worse than the original Grübner basis computation.3}} and C to {{l.z3 which is reduced with respect ta G.3. We see that we do Ilot have to consider {l. for example.z3 ta G. Now we add 113 116 = xz 3 - 1 .. Applying crit2 again we get Ne = {{2.1.1 we consider the following example. {3. 1 1 EXAMPLE 3. y. and the basis is kept inter-reduced as the algorithm progresses. and Ne = {{l. Consider polynomials 1fI = x2y2 .C. lez]). 4}. empirical evidence shows that if N is the number of S-polynomials that would be computed without these criteria. 3}. Wc start with G = {J" h. {3. Je) will be computed. {l.. 4}.Je in which the largest variable (say x n ) does not appear.2}. JG) = xyz2 . We choose the pair {3. Il 12 = xy2 Z - xyz.{ . We choose the pair {1. Since the polynomials generated by the algorithm are scattered throughout the text of the example. and compute SU" J4) = yz3 . However. Even though this may require a lot of computation.{3.z2. Thus we add 1 J4 = x2yz . and in the number of divisions necessary throughout the computation.. We now choose the pair {1.{1. . . C = 0. We use the deglex term order with x < y < z. {l. Namely. Ta illustrate Algorithm 3. the use of these two criteria reduces that number to about VJii (see. 2}. Next we choose the pair {3.4}. This method is discussed in Buchberger IBu85J. 4}.7}.{3.xz 3 which is reduced with re= X 3 z 2 . {3.{4.7} and see that S(h. We consider the situation right after adding J6 to G.8.6}. {5.{5. We may not eliminate {3. we update NC = {{2. After applying crit2 we have NC={{2. 7}. h) = O. {3. Thus G = {ft.h.2}.6} because {l. Note that the computations in crit2 are always trivial.4}.5. namely.7}. {3. After applying crit2 we have NC = {{2. {2. 6.6} we see that Ip(ft) = x 2 y2 does not divide lcm(lp(12). Had we not used crit2 we would have had to compute and reduce 9 8 = 36 S-polynomials.6}.6} E NC and we only consider {1.8}.6} because the pair {1.6} and all three of thern are eliminated giving us NC = {{2. {4.4}.{1. 8}. For 1 we first note that {1. 2}. 7} and compute S(f4. There we started with NC = {{2.xz 3 to G. Thus we add J8 = _Z4 + X2z2 to G.7}.6} rte NC. 6 of which generated elements of the Grübner basis. and we do not consider the pair {5. 8}. {4. Doing computations with large coefficients can be very spect to G.6}.6} we see that Ip(ft) = x 2y2 does not divide lcm(lp(fs). 5. the most time consuming part of Buchberger's Algorithm is the reduction of S-polynomials and we have avoided most of these reductions by using the CUITent algorithm.lp(f6)) = x 2y2 z 3 and so we eliminate the pair {1.6}. and for {5. we computed a total of 13 S-polynomials. 8}} and C = {{l.5}. {3.{3. We choose next the pair {3. 2 To conclude this section. For e = 3 we consider {2.132 CHAPTER 3.XZ21 to G. {3. 4. 7}}. J4.6}. and compute S(f6. Now we choose the pair {2. the possible rapid growth of the degrees and coefficients of the S-polynomials. Now NC = {{2. {3. 3}. {4. Ip(f6)) = xyz".3} rte NC u C. the intermediary polynomials generated by the S-polynornial computations and reductions can become quite large.9. Js. For {1.9 .6}}. h) = + X2 z2 which is reduced with respect to G. This can dramatically slow down the computation.4}.TheS-polynomialscorrespondingto the remaining pairs in NC aU reduce to zero. _Z4 Now we choose the pair {4.So we eliminate no pairs from NC. 3}. J8. we mention two other difficulties that arise during the computation of Grübner bases. {3.6} and {5. 7}. hl· We nQw give one example of crit2. we recall that. For {2.6} we note that Ip(12) = xy2z divides lcrn(lp(ft).{5. Jo.{3.8}. Although it might appear in this example that no real saving in computation time has been gained in using crit2.6}} andC= {{1. {2. since {4.lp(f6)) = xy2 z3. For e = 2 we note that {2. 8}. {l.6} and {4.{5.3}. 2. MODULES AND GROBNER BASES 1 h = XYZ2 .7} and observe that S(12. 3}. {4. We do not need to consider e = 4. 4.6} and {3.6}}. 12 .6} is in NC and we must consider the pairs {2. J8) = x3 Z2 . 7}. Then we choose the pair {6.6} from NC. Indeed.6}. in practice.6}.6}.6}.{4.{4. h. 8}.{5.6}}. h) -S+ O. Jg} is a Grübner basis for 1= (J" h. 6}.8. 8}}. 5} rte NC U Cl.{3. {6. Applying crit2 again we get NC = {{2. {3. {3. We do not need to consider e = 6. 8}}.{2.5}}. {3. So we finaUy arrive at NC = {{2. 4}. Even though the degree and/or size of the coefficients of the original polynomials and the Grübner basis may be of modest size. 6} (we do not consider the pair {3. 6}.4}. since there is only one pair with 6 in it left in NC. {1. Thus we add Jo e= . 6}. 7}.8.{6. Compute the reduced Grübner basis for l with respect to lex with z > y > x. the reduced Grübner ba. Exercises 3. 3.3. consider the ideal l = (x 7 + xy + y. Comparewith Example 2. z] using lex with x > y > z. xz + y) ç lQI[x. a. z2 +z+ 1) ç lQI[x. y. y3 + 2xy.1. Find the reduced Grübner basis for l with respect to lex with x > y> z. try to make up your own more modest example that illustrates the same point. 12 = y3 + 2xy. Again. a large number of terms make any computation very costly. However the reduced Grübner basis for l with respect to lex with z > y > x contains a polynomial of degree 70. If so. xz + y. while the coefficients during the computation grow as large as 10 8 . and 35 terms respectively.1. lQI[x.56428623 Y . a.3. 70. y. The generators ofthis ideal have maximum total degree 7.3. (x'y + z. 7x 3 + 6y) with respect to the lex order with x > y is {x. The problem with polynomials with large total degree is the fact that they can have a very large number of terms.7. y}. y5 +yz+z.56428623 Y .xy2-x) ç. but again they are beyond the scope of this book. y. Compare the two bases.~is for the ideal l above has 3 polynomials with 58. the reduced Grübner basis for the ideal (4x'y2 + 3x.2. b. y. and Jo = 7x 3 + 6y: 401408 10 1835008 9 9604 8 43904 7 7 2 5 x = ( 54 x y . Let 1= (x 7 + xy + y.1 without using a Computer Algebra System. (x'y + z. y5 + yz + Z. As an example of this situation.3. Compare with Exercise 1.1. y' z + 1) ç lQI[x. Compute a Grübner bases for the following ideal using Algorithm 3. This can be seen by expressing x as a linear combination of the three original polynomials fI = 4x'y2 + 3x. IMPROVEMENTS ON BUCHBERGER'S ALGORlTHM 133 costly because of the large amount of arithmetic that becomes necessary.y] usingdeglexwithx < y.3. The following exercise may tax your Computer Algebra System. Z2 + z + 1) ç lQI[x.18809541 Y 200704 6 18809541 Y 38416 + 3 fI + 175616 1) (7 3 6 27 x y 2 6 + 56428623 x 401408 7 1605632 2 9 y + 56428623 x 917504 7340032 2 8 y 8 2 7 + 18809541 x y + 18809541 x y + 6269847 Y + 6269847 Y + 6269847 Y + 3Y 4802 7 + 18809541 xy - 21952 6 100352 5 3xy + 18809541 Y 1 2 3) 12 1128~7246y5(917504y5 + 14406y 4 + 65856 y 3 + 301056 y 2 + 6269847)fs.18809541 Y . z]. . c. (x 2y-y+x.3. There are techniques to try to get around this problem. (No computation is needed!) b. To illustrate the growth in total degree. z]. z] using deglex with x > y > z. For example. For convenience.J.....) and compute SYZ(gl. such that 1 "::. we let Ip(gi) = Xi and for i of j E {l. . We will assume that we have a fixed term order on A.g) 1. Then the S-polynomial of gi and gj is given by By Theorem 1..v$. [Hint: In the given example.9t are monic.l + O.. 3. but such that the reduced Grübner basis for (h. but keeping lex with z > y > x. and for each such solution. we will assume that 91. d E k[XI.) for h. . . in n variables whose total degree is small.J. .. . b..gj) = "L. we let X ij = Icm(Xi .g.) with respect to a certain lex order has polynomiaIs of very high degree. we have .j. . Show that S(~. Show that if S(f.J...xn ]· This is done in two steps. Use c to give a method for generating polynomials h...). j = l. . .t max (Ip(h ijv ) Ip(gv)) = Ip(S(gi.3.g). (The polynomials h ijv are obtained using the Division Algorithm. .)). note that there are 2 solutions for z. . f. ..g.t.t}.3...) from SYZ(g" . . g.~) = . How does it affect the computing time? g.} for (h.. MODULES AND GROBNER BASES c.) We now define for i.g) {U. For i E {l.. there are 5 solutions for y.2.. . a.··· . Experiment by changing the degree of the x. We first compute a Grübner basis G = {g" .Yt).J. z terms in the polynomials above. . etc.. Let J.4. . . ."S(f... E A = k[XI.. v=l for sorne hijv E A. We then show how to obtain SYZ(f" .J..134 CHAPTER 3.] e. .J.g..t}.9t} be a Gr6bner basis... .. In this section we show how ta compute SYZ(f" . How does it affect the computation and the computing time? 3. i i. S(gi."S(f. Computation of the Syzygy Module.··· .hijvgv . Let {YI. where we assume that the Yi '8 are monie. Give examples of polynomials which satisfy d.01+ 0. .xn ] such that d divides both J and g. y. . . Experiment by changing the term order and the order on the variables for the examples in this exercise. .. . then . Can you give a reason for the difference between the bases in a and b? d. X.). .6. we cau choose aij to be a . .9.'" . and sinee X:Xi = X.4.(lp( U. let It( Ui) = CiX:' where Ci E k and X. Il Si < j st). Now for each i E {l.ei iES E Syz(Xi 1 i E 8)..' . . .).'" . .2.1.) Ip(9i)) least.X.3 we have for sorne ai. Thus... Il S i < j S t} is a generatin9 set for SYZ(g1.'" .g.3. we see that LC.) such that (U1.. . sinee gt ] Sij 8(gi. for i E 8. .Ut) with X = max1'.Ut) E SYZ(91. . Since each eoordinate of the vector in the left-hand side of the equation above is homogeneous.u. PROOF.. . Suppose to the contrary that there exists (U1.. the collection {Si.'" .9t). by Proposition 3.. Let 8 = {i E {l. COMPUTATION OF THE SYZYGY MODULE 135 We note that Sij E SYZ(g". Since (U1.Xi iES = 0..9j) - Lhi.(Si. With the notation above. = O. Then we can choose such a (U1.i.9')· THEOREM 3..Ut) E SYZ(91. t} we deline u. is a power product. and 80 L ciX. E A.) . as follows: . {U Ui i = Ui ifij"8 -lt(UiJ ifi E 8.t} IIp(ui) Ip(9i) = X}.v9v v=l . ... Also. ..4. . .. h . j E S.(lp(uv)lp(9v)) is least.. .2.u~) + L i<j aij(hij11 ··.v. . i... '2 '2 812 = -X e...4. .. 1 X2 .uD J ij ij ' " aij (X X )+( ' ~ Tei . i < j :<.jES i.w)..92) Therefore = _wy2 + xwz = X X -W93..u..XZ. .jES We deline (V1.) = (u~. . We return to Example 3. . iES .Tej Ull'" ...jES lp(u~ + L i<j aijhijv)Xv i.hij')' We note that (V1. These form a reduced Grübner basis with respect ta the degrevlex ordering with x > y > z > w.136 CHAPTER 3. we have: Now S(91. t} we have Ip(vv)lp(9v) i. i < j:<.. X LCiX:ei+(U~) .Ut) fic (Sij Il:<. ..jES L i<j aijSij + (u~.2. v.2 2. AIso. max(lp(u~).. as mentioned above.)+ L '<j aij(hij1 . t). 92 = xy-wz.9. and 93 = y2 . .). .... . i < j. For each v E {l. ai] is a constant multiple of X .) .. . Let 91 = x 2 -wy. . sa hm = h '22 = 0 and h '23 -w. we have 'J X X Therefore lp(vv) Ip(9v) < X for each v E {l.--e2 . ' and hence for all i. h .jES < But. u. since (U1. .9. . .jES we have Ip(u~)Xv < X. .hijt).) and (U1.u~) i<j ~ i. .(hm.2.. t). Using the notation of the abave result. max(lp(aij) Ip(hijv)))Xv ' i<j i. . We will obtain the desired contradiction by proving that max'<:v".'" . by definition of u~. .23 ) = (y.... Sij E SYZ(91. . Then we have 'J .) E SYZ(91. . .(lP(vv) Ip(9v))< X. 0 EXAMPLE 3. t} violating the condition that = max1<:v<:.. MODULES AND GROBNER BASES constant multiple of X .(Sij 1 1 :<.-x. . that SYZ(g"g2.-x)). S(g" g3) = x 3z ... . By the definition of S)...xz.g. Finally.. .1. Set F = [ ft Js land 0 = [ g. X 23 X 23 S23 = X e2 . Sr} for Syz(O) (the Si'S are column vectors in A'). .3" h . " PROOF. .··· . SYZ(g"g2. g.OS = F . 0. h '32 = 0 and h 133 = -yw. we have S8 = L~~) hi8i for sorne hi E A.} of non-zero polynomials in A which may not form a Grübner basis.} for (f" . Let 8 = (a). and hence (Ts i 1 i = 1.y3w = xzg. g3) = ((y.J.1. r) ç Syz(F). . . 1 r 0= OSi = (FT)Si = F(Ts i ). i=l i=l i=l r s r . With the notation above we have . (-z....4.. J. 2 3 By Theorern 3..(h = -e.y. S(g2. w). .X e3 . We now tUTn our attention to computing SYZ(f" .Tl. . .h232. COMPUTATION OF THE SYZYGY MODULE 137 AIso..TS) = F . .. ..). Js)' We again assume that g" .y. We note that S13 = yS12 + XS231 80 we have. (Recall that S is obtained using the Division Aigorithm. .). Then 0 = Fs = OSs.. (y2 .-x. -x).g3) = x 2z .. there is a t x s matrix S and an s x t matrix T with entries in A such that F = OS and 0 = FT.. ." . . -x.) Now using Theorem 3.1. l· As we saw in Section 2.Js) = (Ts 8 = S - TSs + TS8 = (ls . 3 ).h233) = (-z. and hence the columns T" . . y..FTS = F . .TS)s + L hi(Tsi ) = L aiTi + L hi(Tsi ). . Finally. in fact. Moreover 1 if we let 15 be the s x s identity matrix. .. Syz(ft. We first compute a Grübner basis {g".4. and T is obtained by keei>ing track of the reductions during Buehberger's Algorithrn.(h23l..3.xz. THEOREM 3..g3) = ((y. ..-X . so that h l3l = xz.TSr. . (y 2 . as) E SYZ(f"". _x 2 + yw).ywg3.). g..w). we can compute a generating set {SI.. Therefore for each i = 1.F = 0.4. we have F(Is . Sr. are monie. h .4.(-z.T s ) ç AS. Thus we have TS8 = L~~) h i (T8.33 ) = X. g.32 . for a collection {J" .). J.T s of ls -TS are also in Syz(F). and hence Ss E SYZ(g"".. -x 2 + yw Z.3. 0.yzw = Zg" so that h 23l = Therefore h 232 = 0 and h233 = O. -x)). Therefore S13 X '3 X '3 e3 . y.0. we have Then 2We will often use the same letter for a finite set and a row vector corresponding to it.4.95 (Theorem 3. it is enough to reduce 8(91. -X. [ 91 92 93 94 95 1 1 l= 1 [h 12 hl [~ 0 0 0 -(x + y) -y x v -(x h) ] -y(y + 1) and 2 -x +xy+x T s To compute generators for SYZ(91.138 CHAPTER 3. 0. In view of Coroilary 3. 12. = (0..94. X.0.x 4 J and 5 4 g5 = x . hl· We find 2 G = [91 92 93 94 95 where gl = y4 ~ X3 ..is) ç (Ts l has already been noted.94. and 8(94.4.8(92. Moreover we have hl· = = = l.95) we need to reduce ail 8(9i.x 2y2.4.x -1. O.1). 92 = xy3 . Let F = [h 12 We first compute a Gr6bner basis G with respect to the lex term ordering with x < y for (h. Ey Theorem 3. g4 = x 4y . -y.92.x .x 3y. . . and h y4 . 93 = x 2y2 .TST)Tl)'" .92). when the context is cIear.4.8(93.95) which give the following respective syzygies: 81 = (X. 0 s). 0). -y + 1).4. -y.93.x 2.x 3y.4.92.T Therefore SYZ(fll'" . MODULES AND GROBNER BASES ) . 0).93. -1. The reverse inclusion EXAMPLE 3. S3 = (0.3 and Exercise 3. S4 82 = (0. Let h x 2y2 .9j)'S for 1 <:: i < j <:: 5 using 91. 12 xy3 . y).x 3.93).1. -x .x 3y.94).3. .. h) with respect to lex with x> y > z is G = {g"g2.t. . Verify that the reduced Gr6bner basis for! = (fI'!2.4.z].) = (TB" . .2. Consider the polynomials fI = xy + 1. . then the polynomials ft.9t. appear in the list g" .Z.x..t. = y .t.y..3._y3. b. g2 g3]=[fI h -z 0 x v -i z2 z ] ... show that [g. _xy+y3.TS did not give any syzygies that were not already in the submodule (TB" . . z]. T . If the Gr6bner basis for (fI.Y + x. .4.4 the rows of !.TB. h = xz + Y E Ql[x..4.18 do appear in the list gb'" ..) in the following examples: a. . 3. and g3 = Z2 + 1. fI = x 2 y .3.. However it is easy to see (Exercise 3.TS in the set of generators for Syz(f" . y]. Also. h) = (( -y. we show that these vectors are necessary.4. it is easy to verify that Therefore Syz(fI.1.. c. h. h = xy2 . x 2y_xy2) is in the submodule of A3 generated by (-y. if a reduced Grobner basis is computed. COMPUTATION OF THE SYZYGY MODULE 139 T [ -y ~2 = _X2 + y' ] _Xy+y3 .g3}.x. _y3. In this exercise. This is not always the case (see Exercise 3. as is done in all Computer Algebra Systems.18 may not appear in the list gb . Using the notation of Theorem 3.. . we note that in Example 3. and h = yz + 1 E Ql[x... .. where g.. Finally.xE Ql[x. fI = x 2 y + z.3.7.t. . . .)._x2y+xy'). _x 2y + xy2)). .4.4.. h =y2z+1 EQl[x... y.1) or by the improved Buchberger's Algorithm (Algorithm 3.gt· ExercÎses 3.3) that if the polynomials fI.). h =xz+y. a. Compute generators for Syz(fI.4. . [ x 2y _ xy' Y We note that (_x 2 +y2..However. 0). y.2)..g" then Syz(f" .3 we had to include the columns of the matrix l.) is computed using Buchberger's Algorithm (Algorithm 1.t.) of A'. (x 2 . .1). . . h = xz + 1. . then the polynomials h. Ts.O) and (x 2.. .z.. g2 = X . In Theorem 3.4. z]. fI =x2y+z. 4. .4 we defined t. + h.J. for Syz(gl...6.} ç {g" .4. .TB. g.l. Prave that {Si' 1 Tij E B} is a generating set for SYZ(g" . .). MODULES AND GROBNER BASES 3.5.r"r2)' " .4. Let il.5. b. ) xn]... Praye that S is not empty if and only if 9 E (fr.J. .4.).).. " Genera!ize Theorem 3.3. the columns of the matrix l.. with the notation of Theorem 3. + h 212 + . .4...J. ..5 wc defined the concept of reduction whieh in turn led to the Division Aigorithm.gt} be a Grübner basis and let Ti... This matrix has 2 non-zero colurons.} is a Grübner basis for l = (f" .4.... 3.1 as follows: Let {g" ...J. As before. Grübner Bases for Modules.)). . givillg the cquivalent conditions for Ct Gr6bller basis in Theorem 1. Praye that if S '" 0 then S = h + SYZ(f" . .9. Let us recall the ingredients for the methods wc used before for ideals.J. " Assume that {f" . Give a method for computing h. = g}. . 3.. g3 S. . in Section 1.) E A' 1 hd.3. As a result we will be able ta compute with submodules of Amin a way similar to the way we computed with ideals previously.. Namely. we continue to study such submodules.2 (and further in Theorcm 1.h.e.J.3 to the computations in Exercise 3.J2 + .. .) = {h + sis E Syz(f" . Let B ç {Tij 1 1 <:: i < j <:: t} be a generating set for Syz(lt(g.6 we dcfined the concept of a Grobuer basis.. Theil in Section 1. Let S be the set of solutions. c...3.).g... but now from the point of view adopted for ideals in the earlier parts of this book.It(g.6. TB are not necessary. a. Use the above ta find the solution set for the equation h . Compute I3 ..3. .4..4 and Corallary 3. Compute the matrix S such that [il 12 h = [g. with unknowns hl. 92..he concept of a power product and then defilled a term order on these power prodncts. g3). say 81. I.xnJ and consider the !inea!" equation 1 1 hd. . 3.J.) = (Ts . Syz(f" . S = {(h" .Tsr )..J.4.hs E k[Xl' .. 83.4. . d. First 1 in Section 1. The llext issu<" discuss('d in Section 1.. . we will genera!ize the theory of Grübner bases to submodules of Am. We have seen in the previous sections of this chapter that certain submodules of Am are important. 82.7. + h. Compute the 3 generators..g.J" 9 E klxl" . Prave that.. let A = k[Xl' .Ts3. (X 2y' _ X3 y ) + h 2 (Xy 3 _ x2y') + h3(y4 _ x3) = y7 _ y6 3. be as in Section 3. Verify that (Ts Ts"Ts 3) '" (Ts Ts 2.. b. that iS a total order with special properties with respect to divisibility (wc willneed ta define a concept of divisibility). In this section.140 CHAPTER 3. . = g.. c. .xnJ for a field k. 1 . that is.... say Tl. . .} and {g" .1). The idea is to mimic the constructions we used in Chapter 1 as much as possible.g. + h. Using these ideas. where h is a particular solution.)}. ... T2. e. Apply Exercise 3. 3.5. GROBNER BASES FOR MODULES 141 was how to compute Grübner bases, and for that we developed S-polynomials and Buchberger's Algorithm. We will follow this development very closely. Indeed many of the proofs in this and the next two sections are very similar to the corresponding ones for ideals and will be left to the exercises. We again need the standard basis el = {l,0, ... ,0),e2 = (0,1,0, ... ,0), ... ,em = (0,0, ... ,0,1) of Am. Then by a monomial 3 in Am we mean a vector of the type X ei (1 SiS m), where X is a power product in A. That is, a monomial is a column vector with ail coordinates equal ta zero except for one which is a power product of A. Monomials in Am will replace the notion of power products in the ring A. Sa, for example, (0,XîX3, 0) and (0,0,X2) are monomials in A3, but (O,XI +X2,0) and (0,X2,XI) are not. If X = Xei and Y = Yej are monomials in Am, we say that X divides Y provided that i = j and X divides Y Thus in A 3 we see that (0, XîX3, 0) divides (0, xix~, 0), but does not divide (0, XIX3, 0) or (xîx~, 0, 0). We note that in case X divides Y there is a power product Z in the ring A such that Y = Z X. In this case we define< Y=Y=Z X X . So, for example (0, xix~, 0) xîx~ 2 = - 2 - =X3· (0,X I X3,0) Xl x3 Similarly, bya term, we mean a vector of the type eX, where e E k-{O} and X is a monomial. Thus (0, 5xIx1, 0, 0) = 5X, where X = (0, xIx1, 0, 0) = xix1e2' is a term of A 4 but not a monomial. Also, if X = eX ei and Y = dYej are terms of Am, we say that X divides Y provided that i = j and X divides Y We write y X (0, 5xlx~, 0) (0, 2xIx3, 0) dY cX Sa, for exarnple, = 2xi x 3 = -:lx 3. 5xîx~ 5 We now can define a term order on the monomials of Am. 3.5.1. By a term order on the monomials of Am we mean a total arder, <, on these monomials satisfying the following two conditions: (i) X < ZX, for every monomial X of Am and power produet Z of 1 of A; DEFINITION 3In the case that m = 1 we have now called a monomial what we referred to before as a power product. From now on in the book we will use monomial and power product interchangeably for such elements in the ring A. 4Be careful about what we are doing here. We are "dividing" two vectors in Am to obtain an element in A; but we are only doing this in the very special case where each of the vectors has only one non-zero coordinate in the same spot which is a power product and one power product divides the other. The "quotient" is defined to be the quotient of those two power products. 142 CHAPTER 3. MODULES AND GROBNER BASES (ii) If X < Y, then ZX < ZY for aU monomials X, Y power product Z E A. E Am and every Looking at Definition 1.4.1 we see that Conditions (i) and (ii) there correspond to Conditions (i) and (ii) in Definition 3.5.1. Condition (i), specialized to the CaBe of m = l, simply says that Z > l, since the monomial X can be cancelled. This is Condition (i) of Definition 1.4.1. The corresponding second conditions are exactly parallel. If we are given a term order on A there are two natural ways of obtaining a term order on Am. These are given in the following two definitions. DEFINITION 3.5.2. For monomials X = Xei and Y = Yej of Am, we say that X<Y X<Y= or { X = Y andi <j. We caU this arder TOP for "term over position", sinee it places more importance on the term arder on A than on the position in the veetoT. So, for example, in the case of two variables and m power products of A with x < y, we see that = 2, using deglex on the (x,O) < (0, x) < (y,O) < (xy, 0). DEFINITION 3.5.3. For monomials X = Xe, and Y = Yej of Am, we say that X<Y={ ~;j i = j and X < Y. We call this arder POT for' "position over term", since it places more importance on the position in the veetoT than on the term arder on A. 80 in this case we have, again for the case of two variables and m deglex on the power products of A with x < y, = 2, using (x,O) < (y,O) < (xy,O) < (D,x). It is easily verified that these two orders satisfy the two conditions of Definition 3.5.1 (Exercise 3.5.1). Of course, each of these two orders could just as weil have been defined with a different ordering on the subscripts {l, ... , m}. In order to indicate which order we are using we will write, for example, el < e2 < ... < e m . There are many other examples of orders and we will use an order different frOID either one of the above in the next section. We note that we are using the symbol "<" in two different ways, bath for a term order on the power products of A and for a term order On the monomials of Am. The meaning will always be clear from the context. In analogy to Theorem 1.4.6 we have the following 3.5. GROBNER BASES FOR MODULES 143 LEMMA 3.5.4. Every term arder on the monomials of A= is a well-ardering. PROOF. The proof of this lemma is exactly the same as the proof of Theorem 1.4.6 except that the Hilbert Basis Theorem (Theorem 1.1.1) used there must be replaced by Theorem 3.1.1 (Exercise 3.5.2). 0 We now adopt sorne notation. We first fix a term order < on the monomials of A=. Then for ail f E A=, with f '" 0, we may write f = a,X, +a2X2 + ... + a·rXc, where, for 1 :::; i :S r , 0 #- ai E k and Xi is a monomial in ATn satisfying X, > X, > ... > Xr. We define • ImU) = X" the leading monomial of f; • leU) = a" the leading coefficient of f; • ItU) =a,X" the leading termof f. We define It(O) 0, Im(O) = 0, and le(O) = O. Note that, consistent with using monomials in Am instead of the power prod~ ucts in A, we now use leading "monomials" instead of leading "power products" , and use the symbol "lm" instead of "lp" . EXAMPLE 3.5.5. Let f = (2x 3y _ y3 +5x,3xy2 +4x+y') E A 2, with the lex ordering on A = lQI{x, yi with x < y. Then in the TOP ordering with e, < e2 of Definition 3.5.2 above, we have = f = _y3 e , + 3xy'e2 + y 2e2 + 2x 3ye, + 4xe2 + 5xe" and so ImU) = y3 e" le(!) = -1, and ItU) = _y3 e ,. On the other hand, in the POT ordering with e, < e, of Definition 3.5.3 above, we have f = 3xy2e2 + y'e, + 4xe2 - y3 e , + 2x 3ye, + 5xe" and sa ImU) = xy 2e2' leU) = 3 and ItU) = 3xy 2e2 . We note that lm, le and It are multiplicative, in the following sense: ImUg) = lp(f) lm(g), leUg) = le(f) le(g) , and lt(fg) = lt(f) It(g), for ail f E A and 9 E A= (Exercise 3.5.6). We now maye on ta the second ingredient in our construction of Grobner bases for modules, namely, reduction and the Division Algorithm. It should be emphasized that now that we have defined monomials (in place of power products), divisibility and quotients of monomials, and term orders, the definitions can he lifted word for word from Section 1.5. The basic idea behind the algorithm is the same as for polynomials: when dividing f hy fi"" ,f8' we want to cancel monomials of f using the leading terms of the fi'S, and continue this process until it cannot he done anymore. DEFINITION 3.5.6. Given f,g,h in A=, 9 '" 0, we say that f reduces ta h modulo g in one step, wntten f --"-+ h, 144 CHAPTER 3. MODULES AND GROBNER BASES if and only iflt(g) divides a term X that appears in f and h = f - I,-(;,)g· We can think of h in the definition as the remainder of a one step division of by 9 similar to the one seen in Section 1.5. Observe that the terms introduced by subtracting I,-(;,)g from f are smaller than the term X subtracted out of f. We can continue this process and subtract off ail possible terms in f divisible by lt(g). EXAMPLE 3.5.7. Let f = (-y3+2x 3 y,3xy2+y2+4x) andg = (x+l,y2+x) be in A 2 We use the lex ordering on A = Q[x, y], with x < y, and TOP with e, < e2 in A 2 Then, lt(g) = (0, y2) = y 2e2' and f f ~ ~ (-y3+2x3y-3x2-3x,y2-3x2+4x) (_y3 + 2x 3 y _ 3x 2 - 4x - 1, -3x 2 + 3x). Let f, h, and f" ... ,f, be veetors in Am, with f" ... , f, non-zero and let F = {fl>··· ,fs}. We say that f reduces to h modulo F, denoted DEFINITION 3.5.8. f --->+ h, F if and only if there exisls a sequence of indices i" i2, ... , it E {l, ... , s} and vectors hl,'" ) h t - 1 E Am such that EXAMPLE 3.5.9. Let f, = (xy - y,X 2),f2 = (X,y2 - x) E A 2 We use the lex ordering on A = Q[x, y], with x < y, and TOP with e, < e2 in A 2 Let F = {J"h} and f = (y2 + 2x2y,y2). Then f ~+ (y2+2y-x,-2x 3 -2x 2 +x), since (y2 + 2xy _ X, -2x3 Notice that thls last be reduced further by f, or f2' This is because lm(f,) = (xy,O) and no power product in the first coordinate of h is divisible by xy and lm(f 2) = (0, y2) and no power product in the second coordinate of h is divisible by y2. DEFINITION 3.5.10. A veetor r in Am is ealled reduced with respect ta a set F = {fl>'" , fs} of non-zero veetors in Am if T = 0 or no monomial that appears in T is divisible by any one of the lm(fi)' i = 1, ... , s. If f ~+ rand r is redueed with respect ta F, then we call r a rernainder for f with respect to F. + x) J.2., (y2 + 2y - x, _2x3 - 2X2 + x). vector h = (y2 + 2y - x, -2x 3 - 2X2 + x) cannot 3.5. GROBNER BASES FOR MODULES '45 The reduction process allows us ta give a Division Algorithm that mimies the Division Algorithm for polynomials. Given J, J" ... , J, E Am, with fI' ... ,fs oF 0, this algorithm returns quotients al) . - . ,as E A = k[Xl1 ... ,xnl, and a remainder r E A"t, which is reduced with respect to P, such that J=ad, +···+asJ,+r. This algorithm is given as Algorithm 3.5.1. INPUT: J,I" ... ,l, E Am with Ji of 0 (1 <:: i <:: s) OUTPUT: a" ... , a, E A, r E Am with J = ad, and r is redueed with respect to + ... + asJ, + r {J" ... , J s} and J max(lp(a,) lm(f ,), ... , !p(a s ) hn(f ,), lm(r)) = lm(f) INITIALIZATION: a, := 0, a2 := 0, ... , as := 0, r := 0, 9 := WHILE 9 of 0 DO IF there exists i sueh that lm(fi) divides lm(g) THEN Choose i least sueh that lm(fi) divides lm(g) ai 9 ELSE := := ai + lt(fi) lt(g) lt(g) 9 - lt(fi) Ji r r := + lt(g) g:=g-lt(g) ALGORlTHM 3.5.1. Division Aigorithm in Am Note that the step r:= r+lt(g) in the ELSE part of the algorithm is used to put in the remainder the terms that are not divisible by any lt(fi) and the step 9 := 9 -lt(g) is used to continue in the algorithm to attempt to divide into the next lower term of g. EXAMPLE 3.5.11. We will redo Example 3.5.9 going step by step through Algorithm 3.5.1. INITIALIZATION: a, := 0, a2 := 0, r := 0, 9 := (y2 + 2x2y, y2) First pass through the WHILE loop: (xy,O) = lm(f ,) does not divides lm(g) = (0, y2) (0,y2) = hn(f2) divides lm(g) = (0,y2) 146 CHAPTER 3. MODULES AND CROBNER BASES a a 2.·- 2 + It(f2)It(g) - 1 + 2x2y y2) _ l '= g _ It(g) f = (y2 g . It(f2) 2 (O,y') (O,y2) (x y2 - x) 1 = (y2+2x2y-x,X) Second pass through the WHILE loop: Neithèr Im(f,) nor Im(f2) divides Im(g) = (y2,0) r := r + It(g) = (y2, 0) g := g -It(g) = (2x2y - X, x) Third pass through the WHILE loop: (xy,O) = Im(fl) divides Im(g) = (x 2y, 0) a .- a + It(g) - 2x 1·1 It(fl)- g := (2x2y - x, x) - (~~::o~) (xy - y, x 2) = (2xy - x, -2x 3 + x) Fourth pass through the WHILE loop: (xy,O) = Im(fl) divides Im(g) = (xy,O) al := al + I~~~:) = 2x + 2 g := (2xy - X, -2x 3 + x) - ?::,~) (xy - y, x 2 ) = (2y - x, -2x 3 - 2x2 + x) Fifth pass through the WHILE loop: Neither Im(fl) nor Im(f2) divide Im(g) = (y,O) r := r + It(g) = (y2 + 2y, 0) g := g -It(g) = (-x, -2x 3 - 2x2 + x). The remaining four passes through the WHILE loop, one for each of the four remaining terms in g, will be similar ta the last one, since neither Im(!l) nor Im(f2) divides any of the remaining terms of g. Sa we finally get, as we did in Example 3.5.9, that f~+(y +2y-x,-2x' -2x 2 +:z:)=r F 2 '3 and, mQreovcr f=(2x+2)fl +f 2 +r. THEO REM 3.5.12. Given a set F = {fl, ... ,fJ of non-zero vectors and 3.5.1) will pradnce polynomials al, ... ,a,5,E A and a vectorr E Am Buch that f in A=, the Division Algorithm (Algorithm f=adl + .. ·+a,f.,+r, with r reduced with respect to F, and Im(f) = max(lp(aJJlm(f 1)'''' ,Ip(a.,) Im(fJ, Im( r)J. 5. The following statements are equivalent for a submodule M ç Am and G = {g" . . The proof is basically the same as the one for the ideal case and is left to the exercises (Exercise 3. it generates..4 instead of Theorem 1. then M = (g" . (0. If G = {g" .. We nQW give the characterizations of a Grobner basis analogous to those in Theorems 1. r2 are reduced with respect to G. we note the Corollaries of this result which correspond to the Corollaries of Theorem 1..6. (ii) f E M if and only if f -S+ o. • Lei where L = lcm(X.0.gt} contained in the submodule M is called a Grübner basis for M if and only if for all f E M.5.5.. DEFINITION 3.6. 0 We now have what we need to define Grübner bases in modules. f -S+ T2. 0)) = (x 2 y3 z .gt} ç M with gi oJ 0 (1 :'Ô i:'Ô t). xy3)) = (0. . for a subset W of Am. (iv) Lt(G) = Lt(M). 0). we mean . . We continue to emphasize that we need only copy what was done in the polynomial case. and Tl. We will now introduce the analogue of S-polynomials.g. 80 let X = X ei and Y = Yej be two monomials in Am. . Y). G is a Grobner basis for M).4.5.gt).16.6. COROLLARY 3.gt} is a Grobner basis for the submodule M of Am.5. .t(lp(htllm(gi)). GROBNER BASES FOR MODULES 147 PROOF. +htg t and Im(j) = max'<. there exists i such that lm(gi) divides Im(j) (that is.. (G).9). Then by the least common multiple of X and Y (denoted lcm( X. Lt(W) = (lt(w) 1 W E W) ç Am.15... . lcm((x 2 yz. if i = j..ht E A such that f = h. (xy3.7.3. So let M be a submodule of Am. then Tl = T2..11). For example. THEOREM 3.14. the leading term module of W to be the submodule of Am.5. (iii) For aU f E M.t} such that lm(gi) divides lm(j). if f -S+ Tl..5. + . except that it is not yet clear what the least common multiple of two monomials should be.5. 0).0). A set of non-zero vectors G = {g" .. there exists i E {l.6. . .13. (v) For aU f E Am.2 (Exercises 3. We say that the set G is a Grübner basis provided G is a Grobner basis for the submodule. We first define.5.i<. (i) For all f E M.5. Y)).9 except that we use Lemma 3.10 and 3. 0) and lcm((x 2 y.2 and Theorem 1. there exists h" . The proof is exactly the same as the proof of Theorem 1. For completeness sake... Every non-zero submodule M of Am has a Grobner basis.ifioJj. COROLLARY 3. On the other hand.5. then S(f.} be a set of non-zero vectors in Am. Although this algorithm is exactly the same as Algorithm 1. 0) < (0.lm(g)).g.18. MODULES AND GROBNER BASES DEFINITION 3. (x 2 y. x < y and A2 with the TOP ordering with e. x 3 y 3) wlllle the leading monomial of the S-polynomial is (x 2 y 3.5.3x Y + Y)=(5 X +5 X -3 XY 1 2 3 1 3 '5 x -3 Y 1 3) · Note that the leading monomial of each of the summands is (0.1.14). Let 0 vector io 1. Algorithm 1. (x +1. This last Theorem allows us to give the analog of Buchberger's Algorithm. We consider A = <Ql[x. S((x 2 EXAMPLE 3. for computing Gr6bner bases. Let G = {g"". The L S(f.g) is set up to cancel out these leading monomials in the most efficient way.2. Namely.5.1.) of Am if and only if for aU i io j.148 CHAPTER 3. we define their least cornmon mult~ple ta be the zero vector making their S-polynomial the zero vector.g) is caUed the S-polynomials of 1 and g. we are interested in linear combinations of 1 and 9 in which the leading monomials cancel out. Then G is a Griibner basis for the submodule M = (g"".7.5.7. .5 xy3) (0. it follows the proof of Theorem 1.l.4 exactly. x 3 y 3).3x y) x2 2 3 y2 2 3 1 4 1 2 5(x +1. Then. 5xy3 + x). The proof of the correctness of the algorithm is left to the exercises (Exercise 3. Let L = lcm(lrn(f).7.5xy3+ x ). We willleave the proof of tills Theorem to the exercises (Exercise 3. < e2. 5We use the term "S-polynomial" even though the result is clearly a vector with polynomial coordinates.5xy +x)-3(x y .g.g = E Am. . y] with the deglex ordering with + 1.17.7.5. It is given as Algorithm 3.lt(g)g The motivation for tills definition is the same as the one used for the definition of S-polynomials in Definition 1. if lm(f) and lm(g) have their respective non-zero entries in the same coordinate. 3x3y + y)) = (0 x 3y 3) (0 x3y3) 2 .13). L lt(f/ .' 3 (x 2 y. in this case.19. we will restate it here for the convenience of the reader.3x 3 y+y) = (0. THEOREM 3. This cannot happen if lm(f) and lm(g) have their respective non-zero entries in different coordinates and 80. .5. We compute a Grübner basis for M = (f"!2'!3'!4) using Algorithm 3.y.fO (1 Si S s) OUTPUT: G= {g" ...5. The only Spolynomials with non~zero L in Definition 3.f0DO Choose any {J. One readily sees that S(f2'!4) -S+ o. .y.fOTHEN g := g U {{ u. Let G = {J"!2.3.x).J. 0).5.!4}. yi with x > y and the TOP order on (<Ql[x. We use the deglex term order on <Ql[x.g}} S(f.y. and we add it to G.y. and we add it to G. 0) -S+ (0. 0).. = (O. This vector is reduced with respect to G. -!4 = (-y.5.0).5.y.x).2. .2.fs) INITIALIZATION: G := F. . > e2 > e3. y])3 with e. -x . where h is reduced with respect to G IFh.s. S(f. -x . !4 = (y.y2 . This new vector generates only one S-polynomial that needs to be considered.{{J. Next.x) -S+ (-x. The new vectors!5 and ! 6 generate only one S-polynomial that needs to be considered.x . !6) = yfs . Am with!i. WHILEg.x!6 = (0. S(f5. a Grübnerbasis for (f" . h} G:=GU{h} 1 for ail u E G} ALGORITHM EXAMPLE 3. GROBNER BASES FOR MODULES 149 INPUT: F= {J" . h = (X.O). -2xy .g) -S+ h. !2}' {J. Again. this vector is reduced with respect to G.y2.y. .y. -2xy . {J2. Consider the following vectors of (<Ql[x. -2xy .!J ç. so we set! 7 = (0. .. y])3: !2 !. !2) = y!. = (O.. and we add it to G.y2.20.O).xy-x).'!4) =!. so we set !s = (-x..17 are the Olles corresponding to g = {{lI.g} E g g := {{Ji' !j} 1 !i .x .O. Buchberger's Algorithm for Modules 3. 0).x.O).h'!4}. = (0..x. This vector is reduced with respect to G.gt}.f !j E G} g:= g . so we set!6 = (-y. !4}}· We compute S(lI. (0.5. Prove the analog of Exercise l. Finally. but we will state the main Definition and Theorem here for completeness.5: Let < be a term order on Am. In that example we had that G = {fI' f2' h. This new vector generates two S.5. We will not prove the results here as..yfs = (0. Thus for all i.5.3.5. x).0 --->+ 0 and S(f7. (0. 3.4x . Am is a reduced Grobner basis if. We go back to Example 3.4. This Grobner basis is effectively computable once M has been given as generated by a finite set of veetors in Am. f5' fs. 0). x .21. THEOREM 3.5. Prove that the POT and TOP orders of Definitions 3.5. Then every non-zero submodule M of Am has a unique reduced Grobner basis with respect ta this term arder. f3 --->+ (0. fs} is a Grübner basis for M. Prove that.!7.X -y.2. 1 1 2 fs = (3 S(f3. The proof will be left for the exercises (Exercise 3.5..!S ) = 2x 2 f3 +y 2x '2 xy2 + 2 Y3)G . 0).22. .5. gi is reduced with respect ta G .150 CHAPTER 3. f 7 . . } (0. they exactly parallel the ones in Chapter 1.0) 3 1 2 --->+ O. Fix a term arder. 0) and add it to G. For X.y2 .!S. _2X2 + ~x + ~y.{gJ and lc(g. y .4.gi} ç:.X < ZX for every monomial X of Am and power product Z '" 1 of A. EXAMPLE 3.23.O).5. f 4 . y monomials in Am.5. DEFINITION 3. 0) .2 are term orders on Am.t. MODULES AND GROBNER BASES We set fs = (0. Complete the proof of Lemma 3. and assume that < is also a well-ordering. again. x + y.4. 1 (y. 3.6: Let < be a total order on the monomials of Am satisfying Condition (ii) of Definition 3. Y. y. 211) { (0.. 3. for ail i. Moreover.5. no non-zero term in Yi is divisible by any lm(gj) for any j '" i. G Therefore G = {f1. Prove the analog of Proposition 1. We first observe that since lm(fl) divides lm(f2) and lm(f4) we may eliminate f2 and f4 from G and still have a Grübner basis.polynomials we need to consider.1. A Grobner basis G = {gl' . -y. -x . Exercises 3.!8} is a Grübner basis for M.y. xy +1 2x + 2 Y' 0).2 Y . f 3.3 and 3. .5. (x. Therefore the reduced Grübner basis for M is 2. f4' J5.2 xy . we conclude this section by noting that the results in Chapter 1 concerning reduced Gr6bner bases hold in this context as weil.20.1.5. then X:::.4.) = 1 for aU i = 1.4Y' 0 . .f s ) 2 = Xf7 .x ..!2. if X divides Y.17). GROBNER BASES FOR MODULES 151 3.5.f4}' where f.h.9. z2 + xz).7. '!sl. with the deglex term order on A = lQl[x. b. Use lex on A = lQl[x.O.y.5.5. < e2 < e3. prove that for ail f E A and 9 E A= we have lt(fg) = lt(f) lt(g). Prove Theorem 3. Prove Corollary 3. 0). y2 + yz. Then change the order to deglex with y > x and the POT ordering with e..xy2+x 2 ). z] with x > y > z. > e2 > e3. y] with the TOP ordering on A3 with el> e2 > e3· b.y2).x). and !c(fg) = !c(f) !c(g).} be non-zero vectors in A= and 0 of h E A.!2.xy. b. Prove that lt. and f4 = (y.xy. f2 = (y. 3.5. > e2 > e3. lm. > e2 > e3. hg.5.5.0).y). Complete the proof of Theorem 3.5.y.x).5.O).!3.y] with x> y and the TOP ordering on A3 with e. and the TOP ordering on A3 with e. h = (O.6. 1.5. by F = {t.12.0).xy2+X 2 ).} is a Grübner ba3is. 3.5. a.O).5. 3. Divide f = (x 2y+y.x. = (xy.1 to find the quotients and remainders of the following divisions: a. 3.O.5.12. x. 3. f = (x 2 + xy.3.10. where f. f 2 = (y. Prove that Algorithm 3.8.y).xy+y2.x). < e2 < e3. Use deglex on A = lQl[x.x.11. You should do the computations without a Computer Algebra System.!4) ç A3. and f4 = (y.5. change the order to lex with z > y > x and the TOP ordering with e. Then change the order to degrevlex with z > y > x and the POT ordering with e.!2. As in Example 3.19.14. and le are multiplicative. f3 = (O.. f3 = (x+y. y] with the POT ordering on A 3 with el > €2 > €3.13.O. Prove Corollary 3. and the TOP ordering on A3 with e. . Use lex with x > y on A = lQl[x.5.h. Namely.5. y3_1). 1.x.15..y). .} is a Grübner ba3is if and only if {hg .g. .5. y] with y> x and the TOP ordering on A3 with e.x. Write the following vectors as the sum of terms in decreasing order according to the indicated term orders: a. 3.6. > e2 > e3. with the degrevlex term order on A = lQl[x.11.5.15... and f4 = (y.14. lm(fg) = lp(f) lm(g).g. 3. x.5. where f. Divide f = (x 2y+y. change the order to lex with y > x and the TOP ordering with e.!2. Prove the analog of Exercise 1.f3.O).5. f3 = (y.y2). Prove Theorem 3. < e2 < e3. y.13: Let {g"". x). y] with x > y. and f4 = (x. Prove that {g"".5. M = (f . Use deglex with x > y on A = lQl[x. = (x-y. Finally.!2. follow Algorithm 3. 3. f 2 = (yx.16. = (x 2 . M = (f.xy+y2. . . 3. f2 = (y. Compute the Grübner bases for the following modules with respect to the indicated term orders.2 produces a Grübner basis for (f"". x 3+1. < e2 < e3. = (x 2. f = (X 2y_xy2. " 3. by F = {t. f4}' where f.O.5.x). Finally.f4 ) ç A3. . we show in the present section how to perform effectively the following tasks: (i) Given f E Am. c. Prove Theorem 3.3 is carried over ta the module setting.5. variables are larger than the e variables we have the TOP ordering in Amand if the e variables are larger than the x variables we have the POT ordering in Am. Show that the Division Algorithm for modules (Algorithm 3. 3.5.. h. (iv) Find a basis for the k-vector space Am lM.5. e m ].8. ) In]. critl of Section 3.5.... find hl.2. el. < em .5. . . if the . e2. we will take a different view of the module Am which allows us to implement Grübner basis theory in Computer Algebra Systems that don't have a built-in module facility.5.16.··· .5. .el" . X n .5.6.1 is false in the modnle case. 1 Xn' el. Find the reduced Grübner bases for the examples in Exercise 3. .em ].. . Let el..19. Fix a term order on k[Xl' . e m simply by sending (fI.. J.Xn .· . Then.1) corresponds directly to the polynomial Division Algorithm (Algorithm 1.f.X n .3..152 CHAPTER 3. and 3... As in the case of ideals in A (see Section 2. 1 em sueh that el < e2 < . Give an example that shows that the analog of Lemma 3. another submodule of Am. . we must set lcm(X ei. we show how the theory of elimination we introduced in Section 2. and if M' ç M.1. Moreover.5. d. in the above correspondence.. Let M = (f 1'" .em be new variables and consider the polynomial ring k[Xl 1 ' " . f .5. . Consider any order on the variables eb e2.) be a submodule of Am.) to fIel + . Show that the usual Buchberger's Algorithm (Algorithm 1.. Elementary Applications of Grobner Bases for Modules.e m ] generated by el. 3...6. Note that division and quotient of monomials in Am just mean the lisuai division and quotient in k[Xl1'" ... . and if so.10 become the usual ones in k[xj.el.··· .1) performed in k[Xl1'" 1 In) el. that is. 3. b. . . (ii) Given M'.15 using the above method. .Xn.. Prave the following: a..··· . Redo the computations in Exercise 3.22. . .. 3.7.. 3..5.18.. whether M' = M. + Jmem E k[Xl.5. E A such that f = hd 1 +···+h..15. Consider an elimination order between the x and e variables. (iii) Find coset representatives for the elements of Am lM. lem]' Then note that Definitions 3.e'TrJ We identify Am with the A-submodule of k[XI"" . determine whether M' is contained in M. . 1 e m ] can be used to compute Grobner bases in Am with the following modification: In the definition of the least common multiple.5. determine whether f is in M (this is the module membership problem).1). e2.3 cannot be used in Algorithm 3. MODULES AND GROBNER BASES 3.. Yej) = 0 for power products X and Y in the x variables when i # j.1).17. In this exercise. el. '. J.). x 3.x.=(xy. -x).-l.. we apply the Division Algorithm for modules presented in Section 3. -x' + 2x..y). e. yi.O).14 that J E Am.6. and hence J is not in M.)t). 96 = (x 2 2x.O.=(x 2 +x. where 9.96}.y+x 2. the vector (-x. .-x2+x-1.5 (Algorithm 3.O) is reduced with respect to G.g.x. we have G = FT. Moreover. ..1) J = h'9. x . (T is obtained.!. J3 = (-y.5.93. 6We remind the reader that when we write f ~ h we mean that h = f .x). We have already noted in Theorem JE M =J --->+ O. by keeping track of the reductions during Buchberger's Algorithm for modules. J. if we view F and G as matrices whose columns are the f /8 and the 9/S respectively.J 4}.} be a Grübner basis for M with respect to sorne term order. 94 = (x 2. Let 3. where J.5. = (x 3 + x.6.. _x 3)..x.)t+x'.x 2 93 X.1. To determine whether J is in M. X. - = ()t+x'.x. 1 f 8' EXAMPLE 3.1.)t). xy + x).} be a set of non-zero vectors in Am and let M = (f" . + h.X g. Let A = <Qllx. We consider the submodule M of A3 generated by F = {f"J2.. as in the ideal case.y). 92 = (x.94. We start with Task (i). x 5 x4 - 3x 3 + x 2 + 2x).x.y. + .9t· As in the ideal case.x). ELEMENTARY APPLICATIONS OF GROBNER BASES FOR MODULES 153 Let F = {J" .5. J4 = (x'. J2 J3 = (x 2 +x.1) can be transformed to express the vector J as a linear combination of the vectors fI' . = (-y. ...y).5 (Algorithm 3. = (xy.g.1) to get (3.x) -'.y. - 95 = (x 2. > e2 A 3 . We use the lex term ordering on A with x < y and TOP with e. we perform the Division Algorithm presented in Section 3.) Therefore Equation (3.!3.)t + x' .92.. (-X.6.y)..2x. . The reduced Grübner basis for M is G = {9. Consider the vector J = (. > e3 on J. if J E M.3. Let G = {9" .1 1 0) • ~ Since xe" and e2 cannot be divided by any Im(9i). y).. To indicate the leading term of a vector we will underline it.95.6.1)6: J ~(-x3-2x. we can find an s x t matrix T with polynomial entrÎes such that.O). G So we can determine algorithmically whether J E Mor not.. J4 = (x 2.9. . Let M' be another submodule of Am. This can be verified algorithmically using the method described above. J.x. x 5 + yx).2 xy2 + x 3y . .y_x6+x5+x2_x)x5) - ~ (_x 7 _ x5 .3x + 2.3) h. Again x~ (-yx _ x 7 +x. l where (3. Therefore 9 is in M.2y . +J2+ X J3-J 4· Now we turn our attention to Task (ii). Sa we _x 2 _y x+y x(l .x)(-J3 + J 4) +XJ4 4 X J.0. can be verified algorithmically using the above method. Then M' c:: M if and only if J. _x 6 -+x 5 . again.3xy .X 4 g.x)g3 + xg 4 4 4 -X (-J.x 3y + 2x2y + 2xy . .x 5 ) -x 4 .0). + (x 4 + 1)g2 + (x 5 .. .).g1 --> (0.J. if M' c:: M. we can compute reduced . . xy + x 5 ) xy ~(_x7_X5+X. Moreover 1 we get 9 X5 g3 + X4 g2 . J3 J4 [" 1 -x x-l T .x 3 .xg 3 + xg 4 + g2 . h2 h3 h4 = = = = y2 + 2x2y _ xy .6. Y _ x 6 + x 5 + 2x2 7 _ x 5 X. MODULES AND GROBNER BASES _ Now consider the vector 9 = (yx 5 we perform the Division Algorithm: 9 yx + x.x 2y . + (x 4 + 1)g2 + (x 5 . -x4g.2) 0 1 0 -1 0 0 -1 1 0 0 0 1 f /8.yx4 +y + 2X2 .2y + x 4 . Alternatively.2X 3 + x 2 + 4x _xy2 + y2 .y) xy-y-x h. and this. .yx+x 5 ) - x~ (-yx _ x -~3 (_x7 _ .xfs + (x -1)J4) + (x + 1)(f2 . . Moreover.6.x 6 + x 5 + 2x2 x5 + x3 + X. + x 5) X..154 CHAPTER 3.J 4 ) +(x 5 . yx 4 + y + 2X2 .R.. say M' = (f. + J2 . + X. Y .2x2 . Therefore we have 9 -x4g. E M for i = l. h2 h3 h4 = [ J.x.x)g3 + xg4· We nQW want ta express 9 as a linear combination of the original consider the matrix T that transforms G into F. We have (3. then M' = M if and only if M c:: M'.x 2y + 2y + 2x2 + X .3x 2 + X + 2 _y2 . such that e f ---->+ r. PROPOSITION 3. .4). We next turn our attention to Task (Hi).y).4) and is left to the reader as an exercise (Exercise 3.g~. As in the ideal case (Proposition 2. We go back to Example 3. and therefore Mn = M.) divides X.f3. Let 1 and g be vectors in Am. _X ) .6.6.6. PROPOSITION 3.5.5.1.3. where g'3 = (x 6 " x 7 _x 7 ) g'4 = (x 7 + x 5 .6. and we denote it by N e (!).x.1. Am N c(!) The proof of this result is similar to the one for the ideal case (PropoSition 2. Then 1 with respect to 1+M = g + M in Am /M <==? Nc(!) = Ne(g).2. x 6 _ 1 S S X . that is. the modules M and M'are not equal.5). _x 6 + 2x S . we have the following Proposition whose proof we also leave to the exercises (Exercise 3.22) to determine whether M = M'. we find coset representatives for the quotient module Am / M. ELEMENTARY APPLICATIONS OF GROBNER BASES FOR MODULES 155 Grübner bases for M and M'and use the fact that reduced Grübner baBes for subrnodules of Am are unique (Theorem 3.6.6. = (x'.6).g2. we can compute that the reduced Grobner basis for Mil is the same as the reduced Grôbner basis for M. Let M' be the submodule of A3 generated by {f 2. Therefore {Ne(f) IlE Am} is a set of coset representatives for the quotient module Am/M.6.1.3.12' 13.1. we call this vector r the normal form of G.14 that there exists a unique vector r E A ~. EXAMPLE 3. A ba. Now ifwe consider the submodule Mn of A3 generated by {I 1. reduced with respect to G.g~. then again MI! is a submodule of M.1 that g E M. since we showed in Example 3. _ g.sis for the k-vector space Am /M consists of aU the cosets of monomials X E Am such that no lm(g.g}.g~}.g~. As in the ideal case. We know from Theorems 3. the reduced Grobner basis for M' is given by {g~. Since the reduced Grübner basis for M' is not equal to the one for M.12 and 3.4. M' is a submodule of M. using the same order we did in Example 3. Moreover the map Ne: Am ----> >---+ 1 is k-linear.g}.6. g~ = (x 6 - 2x S . Moreover. Let G be a Grübner basis for M and let 1 E Am. It solves Task (iv).6.5.x9 _ x8 3x 7 + x 6 + 2x S ).f4. However. For example..6. .6.First.. we see that Im(g) can only involve the x variables as weil. .y. We can use this result to compute intersection of submodules of A= and ideal quotients of two submodules of Am. The leading terms of the reduced Grobner basis are: Therefore a basis for the Q. So let 0 of f E MnAm.5..go}.3. except for the proof of Theorem 3.y.g51 and g6' Therefore by Theorem 3. as in the ideal case (Proposition 2. .4 is the special case m = 1 in Theorem 3. . since the order is TOP on Am we see that the polynomials in ail of the coordinates of 9 must contain only x variables. Clearly (GnAm) ç: MnAm. ..6.y.6... THEOREM 3..6. .1.6. First. let G be a Grobner basis for M with respect ta the TOP monomial ordering on (A[YI. and consider a non-zero module M ç: (A[YI' .. We again go back to Example 3.])m = (k[XI' . Again the proofs are very similar to the ones for the ideal case and.6. Then sinee we are using an elimination order with the y variables larger than the x variables we see that the polynomial in the coordinate of 9 giving rise to Im(g) cau contain only x variables. .4 in the module context (in fact Theorem 2. we consider the theory of elimination presented in Section 2. EXAMPLE 3. With the notation as above. Finally. Since the coordinates of f involve only the x variables. are left to the reader.6.156 CHAPTER 3.6).3). namely gl.6. we wish to eliminate the variables YI...'" . Then there is agE G such that Im(g) divides Im(f).3. The next result is the analog of Theorem 2. we wish to "eliminate" sorne of the variables.y.] with the Y variables larger than the x variables (see Section 2.6. we choose an elimination order on the power products of A[YI.7. As in the ideal case (Section 2.3.6.5) we have ... ) YR. .1 we saw that the reduced Grübner basis for M has three vectors in x alone.X n . Let YI.])m.8).])m.6.. D We will give an example in the exercises where the above result is false if the TOP ordering is replaced by the POT ordering on Am (see Exercise 3. .3. PROOF. YI. be new variables. we wish to compute generators (and a Grübner basis) for the module MnATn.YR. Mn (Ql[x])3 is generated by {g"g5. In Example 3. MODULES AND GROBNER BASES EXAMPLE 3.3). Then G n Am is a Grobner basis for M n Am. but now in the module setting.vector space A 3 !M is To conclude this section. that is. . Since we have a method for computing intersection of ideals (Proposition 2.6.. EXAMPLE 3. ELEMENTARY APPLICATIONS OF CROBNER BASES FOR MODULES 157 PROPOSITION 3.6. . we only need to show how to compute N: (f) for a single vector f E Am.3.9yx 2 . and let N be the submodule of A3 generated by the vector 91 = (y. Let M = (J" .8. .) be submod· ules of A= and let w be a new variable. . .9. neither {f" .. 9yx _7x 7 + 2x 6 + 25x 5 + 7x 4 .3x6 + x 5 + 2x4 .9x 2.(1 . 9y 2x .6. The reduced Grübner basis for (Wf" wJz.9x 3 .6... Let M be the submodule of A3 of Example 3.).1.9yx.3.) ç (A[W])3 with respect to the TOP term ordering with e.9.yx 6 .w)9.9yx 3 .6.} nor {9l'··· . N : (f.) and N = (91'··· ..6.x8 ..7yx 7 + 2yx 6 + 25yx S + 7yx 4 .10. Note that N: M is an ideal in A.6. . We note that in Proposition 3. a Grübner basis for Mn N is then given by G n A=. two of which are in A3 : h. Then . yx S _ yx 7 _ 3yx 6 + yx S + 2yx4 ). > e2 > e3 using the lex ordering in A[w] with w > y > x has 8 vectors.W)9t) ç (A[w])m. Let M and N be two submodules of Am. h2 + 2yx 5 + 25yx4 + 7yx 3 .8. As a consequence of this result we obtain a method for computing generators for the module Mn N ç Am: we first compute a Grübner basis G for L with respect to the TOP ordering on monomials of (A[w])m. (9y2 . As in the ideal case (Lemmas 2.5 or equivalently Proposition 3. wh. x.w)9l'··· .J. ule of Am.11.f.x 7 .7yx 6 Therefore Mn N = (h" h 2 ) ç A"- DEFINITION 3.11) we have LEMMA 3. The ideal quotient N: M is defined to be N: M = {f E A 1 f M ç N} ç A.9t} need be a Grübner basis.3.10 and 2. Let M = (f" . Consider the module L = (wfr.3..Js ) ç Am and let N be any other submod· N: M = n i=l . ThenMnN=LnA=.8 with m = 1). xy).6. wJ 4' (1 .9yx 2) (yx 7 . using an elimination order on the power products in A[w] with w larger than the x variables.3yx 5 + yx 4 + 2yx3 .wf" (1 . by Lemma 3.12. M: (YI) = (9y-7x6+2x5+25x4+7x3_9x2-9x.6. In Exarnple 3. . M: (Y2) = (x 7 _x6_3x5+x4+2x3.13.6. Thus we may compute N: (f) by first computing a set of generators for N n (f) and dividing these generators by f using the Division Algorithm.7x 6 + 2x 5 + 25x4 + 7x 3 . by Lemma 3.11.7x 6 + 2x 5 + 25x 4 + 7x 3 . - 3x 5 Therefore M: (Y"Y2) = (UI.3x 5 + x 4 + 2x 3)YI· h.X7 -x 6 -3x 5 +x4 +2x 3) çA. x 2) E Ao.9X)Y2· Therefore. We proceed as in Example 3.158 CHAPTER 3.11. three of which do not contain the variable w.9x2 .12.9x). We wish ta compute M: (Y" Y2). Note that (9y .6.9 using the same term arder. we first need ta compute M: (Y2). Ta compute this intersection.) = (h" h2). Then N: (f) = {a E A 1 y = af E N n (f)}.5x 3 .6. MODULES AND GROBNER BASES LEMMA 3.x 6 _ 3x 5 + x 4 + 2x3).2x 3 .6.9y+2x6_7x5_2x4+16x3+9x2_9x) ÇA.9 we saw that Mn (Y. (1-w)(x 7 _x6 _3x 5 +x4 +2x 3).6. and we find where h3 h4 (x 7 . (1-w)(9y+2x 6 -7x 5 -2x 4 + 16x3 +9x 2 -9x)) of A[w] with respect ta the lex term ordering with w > y > x.9X)YI (x 7 . Let N be a submodule of Am and let f be a vector in Am. Now.6. The Grübner basis has five polynomials .9x 2 x 6 + 2X5 + x 4 .x 6 . Now let Y2 = (y + x 2.3x 2.6.9x 2 . w(x 7 . EXAMPLE 3. By Lemma 3. we find the Grübner basis for the ideal (w(9y .2x4 + 16x 3 + 9x2 . h2 Therfore. Ul x7 - x6 U2 U3 9yx 3y2 - + x 4 + 2x 3 5x 6 + 4x 5 + 14x4 .U3) ÇA.U2. The quotients obtained are then a set of generators for N: (f).x 6 . by Lemma 3.12.3x 5 + x 4 + 2x 3)Y2 (9y+ 2X6 -7x 5 . y. J3).x. 9 = (_y2 +x _y.y. .1.x + yz + z)). Compute a Grübner basis for M = (J" J2. = (O. Consider the following 3 vectors in (<QJ[x. Prove Proposition 3. 9 = (x. Consider the following vectors in (<QJ[x. J ' where fi E Am. J4) with respect to the TOP ordering on (<QJ[x.zx . yx' _ y' +2xy _ y. J2 = (O. Y + x). ExercÎses 3.y. Determine whether J + M = 9 + M for the following examples: a. Prove that this inclusion is strict.y).x. x)) and M' = ((-y. J 3. y]) 3 : J. You should do the following without the use of a Computer Algebra Sysa.y2 X + y2 +yx+2y . y3 + 2y2 - yx .4. Show that J E (J"J 2. J2 = (x. J.0). a. (y2 +y. (x.6. xy + y. tem. M=M'.6. y.y2 + y). we introduce these syzygies. y. using deglex on Q[x.3.1. y). . Let 9 = (-x. yx + y. Z])3: M = ((x 2 . 3.6. 3. yz + z).y).3. 0. and f 3' b.6. y]) 3 with e. M'ÇM.3xy + x) is in M and express it as a linear combination of J" J2.3x. Use a to show that the vector (X 2y_y2+xy2. J4 = (y2.5. 0).xy2 _y2 +x 2+2xy-xy.y2_X).y. y] with x > y. J = (y3 + y' +x - y. J 3 = (y.x).y2 + y). x 2 y + xy2 .x. J3)' c. b.xy2). we give these applications.6. Show that 9 ct (J" fz. 0. 3.6.x.6.y.3. (0. ELEMENTARY APPLICATIONS OF GROBNER BASES FOR MODULES 159 Sorne of the computations we have performed in thîs section can also be done more efliciently using the syzygy module of the matrix [ J. Consider the following two submodules of (Q[x. J = (2x. = (xy . y])' : J.y3 + 2y2.6. Prove Proposition 3.x 2 _ y). Let J = (_X. In the next section. You should do the following without the use of a Computer Algebra System. (xz + x.2.6. b. Determine whether any of the following holds: MÇM'.xy-x). Moreover 1 express f as a linear combination of fI' f 2. Clearly we have (J 2' g) ç (J" J 2' g). -xz . y2 + y. xz . Use this to compute the matrix T which gives the Grübner basis vectors in tenus of the original vectors._x 3y2 + 2xy ..6.-y+x). and then. > e2 > e3. J 3'/4' 3. x 3y2 + 2xy .6.O). 3. x).= (y.8.4. Consider the module M of Example 3. (x. yx 2 _x 3 +y2 +y). in Section 3. y.12. (0. z2). (x . (x. that is.m.6. O. h = (0. MODULES AND GROBNER BASES 3. i.p(ei). 3. a. o.6. a homomorphism .p.p = 'Y.6.x.6.y.y.x 2.x).8. X + y).7. (x. (y.9. x.x. 3. Mn (Q[x. y.1l. where M = ((0. [Hint: Let {YI"" . Prove that Mn (Q[yJ)" contains a non-zero vectar. Compute generators for the intersection of the following two submodules of (Q[x. (x. Prove that G = {fI' f 2} is a Gr6bner basis and note G n (Q[y])2 = 0. Prove Lemma 3. 0. b. A theorem of Module Theory states that there exists a "lifting".6.Yt} be a Grobner basis for im(q. In this exercise we show that this is necessary.12. 3.10. Find a Q-basis for the vector spoce A 3 lM for eoch module M in Exercise 3. Consider module homomorphisms q. yJ)" with el > e2 and lex on Q[x. x. x. Mn (Q[x. X + y).11.14. That is. z)). Compute generators for M: N.6.. x)) and N = ((x 2. 0.6. b. (x. show how to compute . M 2 = ((x.y)) ç (Q[X. (y.y])3. y])2. 3.p: Am ----+ AS such that q.13. 3. (y. the following diagram commutes: a.x). such that imb) ç im(q. x)).5. Show how to compute . Compute generators for the following modules: a.y). .8 for the computation of the intersection of more than two submodules of Am. Prove Lemma 3.y. x).. x.x.6. Prove Proposition 3. Consider the vectors fI = (y.160 CHAPTER 3.8. xV-x).6. x+1) E (Q[x.15.16. z])3: M = ((x. Use this to compute generatars for the intersection of the following three submodules of (Q[x.0)) andN= ((y. i = 1.15. xz. In Theorem 3. Consider the submodule M of (Q[x. y2).-x)).e.6. 3. We use the POT ordering on (Q[x. y])3 : M = ((x.x)). (x. y. xv. State and prove the analog of Exercise 2. yi with x > y. M3 = ((y. c. z])3. 3. yI)". z])'3.3.xy).6 we required that the order be TOP. 3. y])3 : MI = ((x. x)). Mn (Q[y.6.y.6. 0).6.y).6. (y2.).y.-y). (y.) and .: A' ----+ A' and T Am ----> A'. x 2. . ... if f.f. g. + fm..x). fs 1is a vector (h" J" . .f. We first compute a Grobner basis {g" . . . we !'Ssume that !c(gi) = 1..(e2) = (x 2 y_x.fm.. then Syz(J. of column vectors in Am... . f.) is the set of all simultaneous polynomial solutions Xl..4)... . f s land is a submodule of AS.(ed = (x 2 +xy.(h" . and q. . . x 2 .} for (f" . + h... As in the case of the syzygy module of a 1 x s matrix [ fI fs of polynomials in A. 1 fm'X.f. .) ç Am and compute Syz(g" . + .. So let us first start with G = [ g.) is done in two steps. .t}.: A3 ---->A3 byq. . .. xy .) ç AS from Syz(g.2).fs E Am.hs) = I::~... . . ....xs + .4).g. .) can be viewed as the set of all polynomial solutions h E AS of the system of homogeneous linear equations Fh = 0 with polynomial coefficients.Xs of the system f11X' f21X. . Let i of j E {l. and wedefine q... 3.: AS ----> Am by q. Syzygies for Modules. We now turn our attention to computing the syzygy module of a matrix [ f. SYZYGIES FOR MODULES 161 apply the method used to solve Task (i) at the beginning of the section.7...). Let Im(gi) = Xi and 1 .).y). That is. In other words Syz(F) = Syz(f l' . q. X .. = (J11"" '!m'). i=l The set of aU such syzygies is caUed the syzygy module of F and is denoted by Syz(f" .) ÇA'. Let A = lQ[x..) orbySyz(F).y2..s>".. l. We define 'Y' A ----> A 3 bYi'(l) = (x2y2+x2.g. fs = (J.] b. This computation is very similar to the computation of the syzygy module of the 1 x s matrix [ fI fs of polynomials in A (see Section 3. hifi · As in the case where m = 1 (see Section 3.. + fIsXs + .(e3) = (y2 + X + y.1 we defined the map q. .). More formally we have 1 1 DEFINITION 3.y2+ x.) E AS such that s Lhifi=O..h. A syzygy of the m x s matrix F = . y].xy2_1). Compute 'IjJ in the following example. the computation of Syz(f. . Recall that in Section 3.Xs o o o. As we did in the polynomial case.. We follow c10sely the construction we used for the ideal case (see Section 3. .. .7.f..xy_1).f. We then obtain Syz(f" ..g.7.3. the kernel of this map is called the syzygy module of [ f. . ...f..1. You should first verify that im(!') ç im(q.y3+xy+y2. Let f. . . For example. 0).g5. such that (3.7. .2. To obtain the generators for Syz(g"g2. We now give the analog of Theorem 3.g6)' 8(g5. The proof is identical except that instead of Proposition 3. g5..162 CHAPTER 3.. J By Theorems 3. X 1 1 =lx . g3..7. With the notation above. We first state the analog of Proposition 3. v=l 8(gi.7. Recall that the set {g"g2. for sorne hijv E A. g2 = (0. g4. PROPOSITION 3.) is generated by " ij X . yJ)" with e..7.g6)' we follow Theorem 3. sinee .X.2.g5)' 8(g2. where g.g.g. Xj).gj) = Lhijvg v .3 and 80 we compute and reduce aH S-polynomials.y2 -x-y. g6 = (0.g6) and 8(g3.g3. . We go back to Example 3. xy + 2x + 2Y' 0).O).y.. -y. the collection {Sij Il :'Ô i < j :'Ô t} is a generating set for Syz(G) = Syz(g" . The S-polynomials wc need to compute are 8(g2.5).J E 1.1.5.7. y] with x > y and the TOP order on (lQJ[x. g4 g3 = (x.).4.=lY' 0).x).2. > e2 > e3.4. g6} is a Gr6bner basis with respect to the deglex term order on lQJ[x.1) We easily see that Sij E Syz(g" .23. MODULES AND GROBNER BASES X ij = lcm(X i . we have .7.5. .8 ' j . We leave the proofs of both the Proposition and the Theorem as exercises (Exercises 3.).7. { }} X ej E A ~..g4). .2. 1 . . 1 = (y... = (O.5.19. 2 - 1 g5 = (0. Syz(X Xij { Xi €i - . Then the S-polynomial of gi and gj is given by Xij X ij ( .3.12 and 3.3 we use Proposition 3.g4. THEOREM 3. 0).7.4 and 3. EXAMPLE 3. X + y.3.gj ) = X· gi ..X·gj· 8gi . yx2 (yx . We consider the submodule M of A3 = (<Q[x. 93 = (1{+x .0). x 2 3 - 2x.7). 1) 846 (x 4 5 4x + 2x + 4. -1.3.x 5 .94.93) Y91 .x 5 (_x 5 (_x 5 _ - - yx.X 93 3 (yx . EXAMPLE 3.95.96} forms a Grübner basis with respect to the TOP term ordering in A3 with e.0. x 3 .7. > e2 > e3 and with lex in A with y > x. -2.7.x. SYZYGIES FOR MODULES 163 S(93' 94) = Y93 -X94 = 92 + 295' we have 834 = (0..95. _x 4 + 2X 3 - -yx) (_X S - + x 2 . . -x. -yx .0). -yx) 2x 3 .x. (0. x + 2. ~x3 . We saw that {9.825.1.7. _x + 2x.x. _x 2 + X. -x).7.1) 1 3 1 3 1 (0'-4'0.96) can be computed using Theorem 3.1{).92.0). x + 2).3. 1{ + x 2 2 - x. -yx) X2. _x4 + 2x3 .7. 813 = (y + x 2 . yI)' of Example 3. -yx) x 3 . -1.~) - 2x .0. The other syzygies are computed in a similar way (Exercise 3.4'-Y.0) X3 _x 3 ) (X2 .6.92.-. _x 2 + 2x 1- x5 - x4 - 3x 3 + x 2 + 2x). The other generators of the syzygy module are obtained similarly (Exercise 3. 2 92 = (x.94.2) to obtain (0. 94 = (x . y 2 2 2 + x2 X - X - 2. The syzygy module SYZ(91. 2.7.0.856) (See Exercise 3.0). x 2 . since the remaining S-polynomials are aU zero. For example S(91.x 4 2x 3 x3 + x 3 .x. y.3): 825 ( _x 2 + X + 2. 0). + x 2 . y 2 _x + x + 3x x - 2x. 2 x 3 _x 3 ) ' 6 9· = (x 2 9 5 = (x .93. where 91 = (x 3 + x.93. These are the only generators needed.x.0.2)· Note that we have not included 826 because it is in (834.5.0.x 4 + x 3 . _x3 x 3 4 l _x 2 . -y + 2. -yx . 7. .2) and (3. Ta compute Syz(J" f2' fa.x.6. fa = (-y. . We go back to Example 3. .!2. Also.!4) we use Theorem 3..7.0.y+X 2.T s be the columns of the matrix Is -TB.3).. .7..x). for a non-zero matrix = [ f.} for (F) and set G = [ y.. is the s x s identity matrix. f. .0.y. .!4) given in Example 3. = (xy. f4 = (x .1 that _the original vectors are f.. say SI)'" . With the notation above we have Syz(J..) = (Ts" .5.Sr) and we let Tl.164 CHAPTER 3. .!3.·.6.6 and compute TS'3 TS 25 TS 46 (0..6.7. + y 2 x 3 _ y 2 x 2 _ y 2 x _ yx 3 _ x 5 + x 4 + x 3 . there is a t x s matrix S and an s x t matrix T with entries in A sueh that F = GS and G = FT (S is obtained using the Division Algorithm and T is obtained by keeping track of the reductions during Buchberger's Algorithm for modules.7. f2 = (x 2 +X.4.). _y3 _ y 2x 2 + yx 2 + x4. The proof of the following result is simitar to the one for the ideal case (Theorem 3. of column vectors in Am..fJ. in that example we saw that a basis for SYZ(Y"Y2'Y3'Y4'Y5'Y6) is S13.r.3) and we leave it to the reader (Exercise 3.y). THEOREM 3.y. f2 f3 f 4 1= [ y. EXAMPLE 3.f.0) (y3 y3 x + 2y2x2 _ y2x + yx4 _ yx3. . Now we need the matrix S that expresses F in terms of G. Y2 Y3 Y4 ys Y6 1 0 0 -1 1 0 0 0 0 0 1 0 0 . and S46· Recall also that the matrix T which gives G in tenns of F is given in Example 3.0.y).7.0. F 1 1. 2 The Grübner basis G for (J'.9).'" . Recall from Example 3.) ÇA'.. We first compute a Grübner basis {y" .7. _y3 x + y3 _ y2x3 + 2y 2x 2 _ yx3 _ x 4 _ x3). MODULES AND GROBNER BASES We now consider the computation of SYZ(fl.y)..6.5 consists of the six vectors gi' i = 1.1 in Equations (3. We have -1 1 X 0 0 0 0 1 0 1 0 0 S [ f.0) (0. . As in the ideal case. where l..6.x. .6.) As in the ideal case we compute generators of Syz(y" .y.Ts" T" . y. S25. .. ) with respect to a certain order which we deline next (see 8chreyer [Schre]).7.0) (x 2 + x.7.1) form a Grübner basis for 8yz(9" .9. . : 0 Therefore In the last example. .92) = ((0.3. EXAMPLE 3.··· .8. We conclude this section by showing that the generators for 8yz(9. we have [91 92 1= [J. y). SYZYGIES FOR MODULES 165 Then 14 -TB = [~ 0 0 0 0 0 0 0 0 o 1. and lm(92) = ye2.) computed in Theorem 3. This result is technical and will only be used in 8ection 3. 9. .. -2)) ç A 3.) = ye. and iJ = (x 2 + X. Also. J 3) with respect to the TOP term ordering on A 2 with e. J3) = ((l. fz We have Jol = [91 92 1[~ ~1 s ~ n· Therefore 8yz(f" fz. where 91 92 (y+x 2. the rows of the matrix 1. Let A =Q[x.6 give a complete set of generators for 8yz(F) as the next example shows. Since Im(9.y] and let J.7.12' J 3.7..TB did not contribute to the syzygy module 8YZ(f.y). = (y+2x2+x..0)).4. J 2 = (-y+ x.3 (or in the polynomial case Theorem 3. J2 J3 1[ ~1 ~ T l '---v----' and [J. y).. J 4). l. y) be vectors in A2 Then the reduced Grübner basis G for (J" J2. we see that 8yz(91.10. > e2 and the lex ordering on A with y > X is {9" 92}. It is not true in general that the TS i in Theorem 3. and hence ZXei < ZYeJ .5. then Im(ZXg i ) = Zlm(Yg j ) = Im(ZYg j ) and j < i. DEFINITION 3.j E {l.. . and let i. y] with x < y. and 93 = y4_X 3 .t}. 0 EXAMPLE 3. First. for otherwise.9.9. Z be power products in A such that Z of 1.9..1. Iflm(Xg i ) = Zlm(Xg i ) = Im(Yg j ) and j <'i.t}.g. Z be power products in A. Finally. i Then < is a term oredering on A rn . if Im(Xg i ) = Im(Ygi)' then X lm(gi) = Im(Xg i ) = Im(Ygi) = Y lm(gi)' and hence X Yei = Y. .Yt be non-zero vectors in Am and let < be a term arder in Am. Let X.. We first show that < is a totaJ arder..7. . we have one of X ei < Yei or < Xei· We now verify that < is a term arder as defined in Definition 3.. g. and X of Y. Y.166 CHAPTER 3. The term order defined in Lemma 3. If i = j E {l. then either Im(Xg i ) = hn(Ygj) and one of i < j or j < i holds. We first consider the case where m = 1 and polynomials 91 = x 2y2 _x 3y. Let 911 . by the term order < on Am).. We use the lex term ordering in lQI[x.9 is called the arder on At induced by [ g.. if i of j E {l. let X. . that is. MODULES AND GROBNER BASES LEMMA 3. Note that the hypotheses of the lemma do not require that {g"".7. We define an order < on the monomials of At as follows: X e· < Y e· . then one of hn(Xgi ) < Im(Ygi) or Im(Ygi) < Im(Xg i ) holds. since Yi i- O. 1 PROOF OF LEMMA 3. . one of Xei < Yej or Yej < Xei holds.7. sa ZXei < ZYej.5. Let i E {l. . .t}. t}.7. In the notation of Lemma 3.) = lp(yg. Then Im(Xg i ) < Zlm(Xg i ) = Im(ZXg i ). J {::=? Im(Xgi) < Im(YgJ) or { Im(Xg ) = Im(Ygj) and j < i.10.) = x'y3 < xy4 = lp(xg3)' Note that < is neither TOP nor POT as defined in Section 3. The reader should note that when Im(Xg i ) = hn(Ygj) in the lemma. In any case..} be a Grübner basis. Assume that Xei < YeJ' If Im(Xg i ) < Im(Ygj)' then Im(ZXg i ) = Zlm(Xg i ) < Zlm(Yg j ) = Im(ZYg j ).. . we have since lp(xg. Therefore. 92 = xy3 _x 2y2. or one of Im(Xg i ) < Im(Ygj) or Im(Yg j ) < Im(Xg i ) holds. the j and i are reversed. implicitly. and hence Xei < ZXej.7. . Let X.11. . (and of course. we have X ei < Yej when j < i. Y be power products in A.. .7. 9.. then R :2: i. Define a new vector Since S is a syzygy of [ g. Note that we have lm(s) = Y.hij ... Let us use the six vectors gi' i = l.gt). lm(sij) = . We first prove that for 1 S i < j S t.j such that 1 S i < j Stand lm(sij) divides lm(s). be a monomial that appears in (h ij " .gi)}· We observe that if RES.) and Cp = lc(a. PROOF.12.Ye lt(g.7.7..5. Therefore since i < j. . . . Let G = {g" . lm(xg4) = xye33..t}. gt ] . . These vectors deterrnine a term ordering on A6 as described in Lemma 3.7.5. we will use e3i and e6i for the ith basis vector in A 3 and A 6 respectively.3. by Equation (3. . For this i define < ----. With the notation of Theorem 3. .) = lm(Y.. . Then lm(Xg.g. In order ta distinguish the basis vectors ei in A3 and A6.!. the collection {Sij Il Si < j S t} is a Griibner basis for Syz(g" .1).) = x~: gi lm x~ Yj Xij..g. ale"~ where a.7. In A 3 we are using lex with y > x and TOP with e33 < e32 < e3'· Using the order induced by [ g..7.. we have 1m( Sij) Note that we have lm (X) = (X.. But 1m(S(gi. X': X Xi We now show that {Sij Il Si < j S t} is a Grübner basis for Syz(g" .13 we need ta show that there exist i.3. . .7.13..we have L c. Moreover.ei. Let S E Syz(g" .) with respect ta < . since lm(xg6) = x6e33.).ei for sorne i E {l. E A. SYZYGIES FOR MODULES 167 EXAMPLE 3.) = yx3e3" and THEOREM 3. By Defintion 3. . lm(g2) = ye32.).} be a Griibner basis.7. . Now let Xe.) S lm(S(gi' gj)). S = {R E {l..1.t} Ilm(Yeg. Let Ye = lp(a. g6 ] we have xe66 < e62 < xe64 < ye61. First we write S = I::~.gj)) < lm therefore X e~ ( X gi ) .. by the definition of < .. and lm(yg. : ei for each 1 Si < j S t.gt) with respect ta the term arder < on monomials of At induced by X [ gl gt ].. ...6 of Example 3.) 'ES = o. 3. of Syz(f.Js E k[XI. where the sum is over all j E S.2. . C < m } .~I Udi' 3.2. J. in general.7. Xe X m Thus we have ..4. .J.7. 3...2.15). Now Im(s') = Im(s) = liei and thus.7.7.5.fI.4 in the module case (generalizing Theorem 3..).7. In this case we obtain J = -~ I:..4.3.7. So.7.7. J.us) E G with u oF 0 and u E k.5. we have. mES.3.2.} is nat necessarily a Gr6bner basis.5 (Algorithm 3. desired. 1 where {JI"" .7.1.5 in the module case. .J. Prove Proposition 3. S = L atm (Xem X-ei. Use Exercise 3.7. fI. that s' is in the em Xt submodule of A t generated by B = {Xem --ee .6. Ul.)..7.7. 3. Note that we have already seen in Exercise 3. since j > i for all j oF i E S. and use it to describe how crit2 can be implemented in Aigorithm 3.3.16 that critl cannot be used in the module case. '\' Xem) .168 CHAPTER 3.e m E Ale.7. .) if and only if in the Grëbner basis.Xem l. where the corresponding order on monomials of AS would be the one induced by [ J.. that is. there is a vector (u. We saw in Theorem 3.7.) It(gt) 1. a Grëbner basis with respect ta the arder induced by [ JI J./s)' These generators do not farm. J..5.7. Complete the computations in Example 3. Let J. . give a definition of Gr6bner bases for modules in terms of the syzygy module of the leading terms.6 how to compute generators for SYZ(f1l'" ..7.3). j immediately that for sorne j E S.7 to state and prove the analog of Exercise 3. 3.5.5. MODULES AND GROBNER BASES S' Therefore is a syzygy of [ It(g. Prove Theorem 3. 1 Exercises 3.. It follows ei = Im( Si)) divides Im( S') = hn( s) as i such that = li We might hope that a result similar to Theorem 3. (see Exercise 3.mES l m e<m for sorne aem E A.4. . by Proposition 3. . 0 -?:' oF Ip(aij)-i7. Complete the computations in Example 3. .7. 3.) with respect to the POT ordering with the first coordinate largest.2.3.J.6.··· .'" .xn].) given in Theorem 3. we would use the algorithm presented in Section 3. . G. Show that J E (fI.'" . f.2) starting with the generators of Syz(f" . State and prove the analog of Corollary 3.8. 3.13 holds for finding a Grëbner basis for Syz(f" . Noting that the indices of the non-zero coordinates of s'are in S.7. .7. State and prove the analog of Theorem 3.7. ta obtaln a Grëbner basis for Syz(fl. we see that Ciliei = It(s') = Llt(aijriJ ei. . y.14.x). Prove that if S # 0 then S = h+Syz(f"". f2 = (x 2 .x 2y + yx 2.y). a.y). f 3 =(x+y. f 4) in thefollowingexamples. . where h is a particular solution. f4 = (-x. X.4. xy). Prove Theorem 3. Namely. > e2.) with respect to the order induced by [ f"". 91 = (0.7...y). .y])2 with e. hs E A.1.13.y])2 3. f3 = (x 3 . in general.y])2 b.3. f3. 94 = (x .15. 92 = (y. = (x+y.7..x.7. f. 93 = (0. (y2 . Compute generators for Syz(f" f2. 94 = (0.4.f.f. f 2 =(x-y. x). Prove that Sis not empty if and only if 9 E (f"". 3. f. y])2 We use the lex order on Q[x.y.3 in the module case. x 3 ).. = 9. 3.y.fs)}. .y. Repeat the exercise using deglex with x > y on Q[x. . f. Write f as the sum of terms in descending order according to the order on (Q[x. Use the deglex order on Q[x. We consider the linear equation hd. x 2). f2 = (-y+x.9 E Am.6 do not. y] with x > y and the POT ordering on (Q[X. y2).l.7.11. 0. f3 = (x. 3. the generators for SYZ(f"". .2y+x). x) +h 2(x. and 93 = (x 2 . compute generators for Syz(gl' . 92 = (xy + y. State and prove the analog of Exercise 3.y])3. . Consider the following analog of Exercise 3.f.7. a. y] with x > y and the POT ordering on (Q[x.7. y])2 with e. Verify that the following vectors form a Grübner basis with respect to the indicated term arder. y.).7.7.x.. y])3 induced by [91 92 93 1 ..y3 + x 2y) E (Q[x. We mentioned at the end of the section that a result similar to Theorem 3. You should do this without the use of a Computer Algebra System. 93 = (2x.7.y).x).y).) obtained in Theorem 3.y) E (Q[X.7..x). (x+y. form a Grübner basis for Syz(f " . b..) = {h+s 1 s E Syz(f"". Use the above to find the solution set for the equation h . y] with y > x and the POT ordering on (Q[x. = 1.6.gt). Let f"".12. x) +h3( -x. Consider the vectors b. . x) +h4(X.x 2).7. = (x 2. 3. 91 92 = (0. SYZYGIES FOR MODULES 169 3. c.7. f.9.10. y])2 with el > e2.y+x).hs). x). y) = (y.y + x 2). -x+y. Consider the vector f = (xy2 + y3.6. . a. f4 (_X+y2. .x).x.y3).13 does not hold for the generators of Syz(f "" . > e2. y]. . with unknowns h" .6. . Consider 9..y) E (Q[X. = (xy . 3. and verify Theorem 3. Use the deglex order on Q[x.y). Give a method for computing h. +h2!2 + .1sl obtained in Theorem 3.x.y. Let S ç A' be the set of all solutions (h" . . 95 = (0.._x+y3) E (Qlx.13. +h. x .f.7. = (xy+y. _y3 _y2x+yx2 +x3.22 to give a method to decide if M = (91"" .. y.y).7. Let M = (f1"" .x). for i = 1. 0.~1 aijJ j . J 2.x. .y)). 0).s.O. a generating set {S1'''' . It is easy ta verify that 8yz(f 1.96)' 3.t. for some aij E A.17.7. d.6 to find a generating set for 8YZ(91 .J4' 93 = (x. _X 2.x).16. say Syz(f1"" . Let esi. . 92 = (O.X +y)J4.) ç AS.} ç A' for 8yz(f1'''' .O). y3 x _ yx2.7. You should first verify that M= (f1.16. 0). _y3 x + y3 _yx 2 _x 3)). we give a more efficient way ta solve the prablem raised in Exercise 3.y) E (lQIlx. -y.9.J'} for a submodule M of A= (not necessarily a Grübner basis).J. x 2.J2.9.93. 0)). 93 = (O.J 2 + XJ3 + (-x + I)J 4' 95 = (0.6 gives Syz(fp J 2. .. give a method ta find bij E A such that Ji = L~~1 bij 9j' Use the praof of Theorem 3. Now consider the polynomials 91 = (0.). 0.) = (S1.6 has nothing to do with the theory of Grübner bases. J 5 = (X. (Note that the proof of Theorem 3.7.16: we are given a generating set {J 1"" . yJ)3. Now consider vectors gi = 2.9t).y. x) = J1 . If so.94..95.J. Use Theorem 3.!3.94. a. -x. 0) = J2 . . 94 = (y.J 4..!4) = (9u92.J.9. . 94 = (_x 2.xy2 +xy)fa + (_x 2y + xy2 + 2X2 .) and if 80 to find generators for SYZ(gl' .s.170 CHAPTER 3..t.!3. Verify that the reduced Grübner basis for (fI' J 2. (Müller IMa90]) In this exercise. Verify that Theorem 3.x. Apply the methods given in a ta the following example. .!.y+x. We have the same hypotheses as in Exercise 3.7..). x.-x.y.y2 .)..X). 95 = (x.16. As in Exercise 3. and a collection of vectors {91"" .) b.x.'" . and 96 = (_x 2.O. Verifythat SYZ(91. and J4 = (x.(O..).y])3.95) = ((-x. J3' J4' h) with respect ta the TOP ordering with e1 > e2 > e3 and lex on IQIlx.5.x.} such that 9i = L.y .x).X) E (lQIlx.7. .9. -x.=1 aijfjl for sorne aij E A. Assume that we have a generating set for Syz(f1"" . 0) = fa .-x.) be a submodule of A=. we wish to determine whether M = (91"" .. c. 2y+2x). y) = J 4.y).. Verify that the two vectors given in c do not form a Grübner basis with respect to the order induced by [ J 1 J 2 J 3 J 4 3.X.y.! 3' J 4) = ((y3 + y2x. J 2 = (x. . 0. a.O. fa = (-y.x). for i = hl· "1. -x. J4' h) = (( -x.:.O.y. b.7.92 = (-y-x. -x.X3) = (x 2 _y2+y)J1 +(-2x 2+ y2 _ y)J2 + (yx 2 . 0. y] with y > x is given by the vectors 91 = (O.7. i = 1. Consider J 1 = (xy..S and . (y.92' 93. _x 2 +x) = (-x-y+ I)J 1 + (x+y-l)J 2 + (-:yx+x)fa + (yx-y-x+l)J 4. MODULES AND GROBNER BASES J4 = (-X+y. . . ..~:~.. . (s" 0). E At such that (esi. . Applications of Syzygies. Prove that {(a" etl). 3.s. again. Consider a new order < on As+t defined as follows: for Ul. In !inear aigebra over fields. As we have noted before.Vt) E At.· . . we use the Grobner Basis Aigorithm ta transform the matrix iuto the matrix [~ Syz(g.. . a.7.8.. The purpose of this section is to reconsider and solve more efficiently various problems. The matrix B is then the inverse of A. for i = l.xnJ for a field k). one reduces the matrix [ Ails J ta the row reduced echelon form [ Is 1 B ] .gt) if and only the reduced Grobner basis for W with respect ta < is the union of the two sets {( esi.(at. . Prove that M = (g" .. . Redo the computation in Exercise 3.. .3. .. Finally.s} and {(O.. These problems were solved previously using elimination... Prove that M = (g" .. . . where ls is the s x s identity matrix.us) E A'.gt). O)} generates W b. bd 1 i = 1.C}.~. . Prove that the set {(O. bi ) E W.) 1 i = 1. where ls is the sxs identity matrix)..~ . .C} i8 a generating set for Syz(g" . A = k[x" .g. d.t....v = (V" . . ... . . Fix an order <s and an order <t in AS and At respectively.. This exercise answers the question of whether T is anto and whether there is a linear transformation T': M ---)... . .... V2 E At.gt) if and only if there are b" ..j = 1. .. . . M defined by a matrix B sueh that (T' 0 T) (M) = M (B is an "inverse" of Ain the sense that BA-ls ç Syz(J" . Let W = {(u.. v. ud i = ~. . The method presented here is a generalization of a linear algebra method ta compute the inverse of the matrix of a linear transformation...ais) E AS. One can think of the vectors a.ti) 1 i = 1. .) ] . let ai = (ail. such as computing the intersection of submodules of Am (where. as the rows of a matrix A which defines a linear transformation T: M ----} M. c. .(se...t be the standard basis for AS and At respectively. In the case of a module M. ett). lt.v) E AS+' 1 u = (u" . for i = 1..b.. APPLICATIONS OF SYZYGIES 171 etj.). U2 E AS and VI. ta compute the inverse of an s x s matrix A... .16 b using this method.f..8.g). + a2h + .IS. . .. f s = [ ~ ] .Yt} is a Grübner basis). Then a polynomial h is in I n J if and only if h= ad.. where the latter may be computed using any term order whatsoever.as.) is a syzygy of the matrix [1 91 Yt ].bt E A such that nJ (381) is a syzygy of [i f. Thus we have essentially shown gt ] = [ ~ o fI J. . . In J = {h 1 there exist polynomials al..... 1 as. .Yt) (we will not assume that either {J" . if and only if there are polynomials Then it is easily seen that h is in I al. the intersection of two ideals in A. and h = b'Yl + b2Y2 + . .1...g!1'" .. .. . o 0 YI Yt o ].. This is the same as the two conditions that (-h. . ..8. . We begin by considering the simplest case..) is a syzygy of the matrix [1 fI J. ] and (-h. . J. . then {h" .. This case.. Let land J be ideals in A. any elimination ordering) is very inefficient from a computational point of view.s can aIl be done using syzygies. PROPOSITION 3... ] ...gt = [ ~ ].J...gt) and the first coordinate oJ hi is hi Jor 1 :0: i:O: r. b . J. b.hr is a generatingsetforSyz(i. .. f. cantains aIl of the essential ingredients of the rnost general situation.. . = [ ~ ] . ideal quotients and kernels of homomorphism.172 CHAPTER 3. ifh l1 ··· . h r } is a generatiny set for I n J.. . Moreover. we will show how the computation of intersections.) and J = (YI.as. .. mueh effort in the theory of Grobner bases has been expended in order ta avoid the use of elimination in computations involving Grobner bases.. bl ... Indeed. bl . = [ .I I . .. ..bt E A... a" . . .} or {YI.. . Usiny the notation above.. . .g.· . + a. nevertheless. . a. . MODULES AND GROBNER BASES the lex term ordering (or more generally. Here. bl.. It is then easy to put"these two conditions into a single condition for syzygies of vectors in A 2 Let i = [ ~ ] . Assume that I = (fI..ht such that fs YI gt ]}. . Even in A (when m = 1) these computations using syzygies tUTU out to be more efficient than using eliroination orders in A. + bty" for sorne polynomials al. 1). We wish to compute l let 't n J.8.lOy).. 2xy _ 2y 2 Y . (x 3 + 2y3 + 2x2 .xy_ y 2 -2x+2y-2. APPLICATIONS OF SYZYGIES 173 PROOF.8.f" YI' . _2 y 3 + 2y 2 + 1. 2xy 2 + y2 + 2y + l. .x . 0.x 2 + xy + y2 + X + 3y.l.l.xy2 -2xy+x-y+2. Assume that the first coordinate of hi is hi for 1 SiS r. x 2y2 _x 2y_5xy2 +3xy-4 y 2 . then the first coordinate of the vector i8 a linear combination of the first coordinates of the other vectors. Then {h" .). -x+4). These polynomials do not form a Griibner basis for InJ with respect to the given term order.x 2y + y. . We use the same notation as above. we get the much simpleT generating set We note that in Proposition 3. Computing a Grobner basi8 for this ideal from the given generators. y3 + x). as we just saw. since if a vector is a linear combination of other vectors.2xy+y2 +x -1. Assume we have a fixed term order < on A and consider the corresponding POT order on Al+s+t with el targest. and (-4xy2 -lOy3 -4xlOy. 2xy 2 . (xy3 + xy2 + 3y3 . 0.. . 0). xy2 . For the first statement we simply note that h E l nJ if and only if -h E In J.. > e2 is generated by (x 2y + y. we only obtained a generating set for In J.lOy 3 - 4x . he} be a Grobner basis for Syz( i.1. We will consider the following ideals in <QI [x.y.8. f"" . We use the deglex ordering with x > y in A.2y2 +x + 2y. -1).2y2 + X + 2y. that i8.-1.8. .x3y-l.=[1] f =[Xy-x-y-l] 1'1 0 12 f =[X2+l] 0 ' YI = [ -x20+XY] 'Y2 = [ X 2yO+y] 'Y3 = [y3~X]' Then we can compute that the module of syzygies with respect to TOP and e.. x 3 + 2y3 + 2x 2 .2. Let {h" .3. So we .2x-4y-2.3. x 2y_xy2 _x 2 -5xy+6y2+x+ 10. Of course. -4xy2 . yl: 1= (xy .y .8. x 2 + 1) and J = (_x 2 + xy. xy3 + xy2 + 3y3 . and this accounts for the difference between the statement of the proposition and Expression (3. _y3 +3y2 . So the first coordinates of these vectors farm a generating set for In J.. . Alternatively we have THEOREM 3. -y. _ xy3 +2xy2 +6y3 -xy-8y2 +4y2. we could obtain a Griibner basis for In J by applying Buchberger's Aigorithm to this set of generators. -1).3. The second statement follows from the first statement. In J = (x 2y + y.x 2 + xy + y2 + X + 3y. . 0 EXAMPLE 3.y2 .he} is a Grobner basis for l n J with respect ta < . x 2y2 -x 2y-6xy2 +xy+2 y 2 +9y-l.3y . We define [ H. Reading off the first coordinates of these vectors. D We note that in Theorem 3. We redo the computation of the previous example.IH21·· ·IHr J as the S x (t .91'··· . . For simplicity we use the Grübner bases for the ideals I and J with respect ta deglex with x > y../2. EIl h]EIl[::] 14 93 [h h 14 h o 0 0 0 0 0 92 93 J.··· .. Since hl. each Hi is an s x t i matrix for 1 <:: i <:: T. the TOP ordering would not work. FUrther.. y2 + 1) and J = (y3 + y.. We define Hl H 2 EIJ Hr ta be the (SI + . -1. sa let h E InJ.0) = It(h) = I:lt(G) lt(h i ). Thus. 0)..9. + q 1 . we obtain the same Grübner basis we did in the previous example.8. . y2+1.3 shows. Sa using i = [ ~ ] . _y3_ y). . let H" . + . -1..0.4.y).. (0. and the fust coordinate is largest. 1 ]. + tr) matrix with the matrices H l1 . . Let H.y). we can find polynomials C. Hr be Si x t i matrices for <:: i <:: T..a" b . x-y.. One containment is clear.0. let Hl. .1 we cau choose h = (h. we made essential use of the POT ordering...] EIl Similarly.hr is a Gr6bner basis for this module. . whose lower right"':hand corner matrix is the 82 x t 2 matrix H2 and the Test of whose entries consist entirely of zeros. ).. a" . + .. We compute that I = (x + y. as should be evident from the proof of the theorem. and H 2 ) ta be the matrix whose upper left-hand corner matrix is the 81 x tl matrix Hl.Hr clown the diagonal and zeros elsewhere. From Proposition 3.. since the arder is POT. . It is then clear that we have lt(h) = I: It(G) lt(h i ) where i ranges ovef the same i's as before. (xy2 +x.. -x+y.J. -x-y. Cr such that h = I:~~1 Cihi and Im(h) = max'<:i<:r(IP(Ci) lm(h i )).. EXAMPLE 3. where the sum is over al! i such that Im(h) = Ip(ci) Im(h i ). as an example in Exercise 3. (y3 +y. Then. In particular. . _yZ -1).J" ..9. EIl [fI h . 0. > e2 > e3 > e4 > es. for example. we compute a Grübner basis for SYZ(i. This gives the desired result...... be an SI x t . -x.. 0. . We need ta show that (lt(h . -y.8. matrix and H 2 be an S2 x t 2 matrix. (x 2 _y2.. . this time.. where we may assume that h '" o.. 0.3. MODULES AND GROBNER BASES PROOF. 1. . and obtain (0.91 = [ y3 ~y ] .) " E Syz(i. + Sr) X (t . 0. We define the (SI + 82) X (t..). y ] .1. 0.92) with respect ta POT with e.8. + t2) matrix H. At this point it is convenient ta compactify the notation. . h '" 0.8.174 CHAPTER 3. 0. -x .92 = [ X ~y ] . we see that (lt(h).b. H 2 (called the direct sum of H. X .lt(hr )) = Lt(I n J). 0). = [ x.12 = [ y2.'" 1 Hr be matrices where. ..9" ' for sorne polynomials al. x . We then compute. in fact. . Ij = (fj1 .I=denotesthe m x m identity matrix. that Syz(H) = ((0. .z. and if the POT order is used on A l+tl +t2+··+tr ta compute a Grübner basis for Syz(H). let F be the 1 x s matrix [ fI J..0. For example. Then. 0. . This is the same as the two conditions that (-h. J and " (-h. 0. (xz+y2-2yz+y-z. y.. using the notation above in the discussion of l n J. -x + y)).. + ad. 9. al. and h = b 9..1. and (-h..2. b1 ..3. 0. y . Let A = <QJ[x. -z. H= 1 'x-y 1 0 [ 1 0 y-z 0 0 z-x 0 0 0 x-l 0 0 ° y 0 0 y+l 0 0 x+l z-l ° ]. y-l. -x . .ij")' and define the 1 x t j matrix Fj = [fJ. APPLICATIONS OF SYZYGIES 175 matrix whose first t l columns are the columns of Hl. x + z . . in exactly the same way as above. "r. -y. So let h. 0.fs) and N = (91"" . y + z .as) is a syzygy of the matrix [ I=lf . We next consider the case of the intersection of tWQ submodules M. a Grübner basis for h n h n· .. N of Am. -z.y).Z-X).x 2 _y2 _x+y. To compute h n h n h we set J. z] and consider the term order deglex with x> y > z.Ir be ideals in A.. J.as. + a2f 2 + . then the set of first coordinates of Syz(H) is l n J. h.1. J (here.)· Then a vector h E A= is in M n N if and only if h = ad . Assume that M = (f" ..f.9... etc. . . (x 2 y2 _ X + y. -y.. x . Leth = (x-y. -y+z-l. As a second illustration of this notation we consider the intersection of more than two ideals.8..x+l. using the TOP ordering on A 3 with el > e2 > €3.z + 3. EIl F 2 EIl .b 1 .. -X. . 0. We note that this is not a Grübner basis for h n h n h. -y. then the set of first coordinates is.b. z..1. y + 1).a" .1.0.y2z_yz2 _y2 +yz. n Ir. g. (0.1.z-l). from which we read off that h nh nh = (xy_y2. -yz + y.0. .. -y+2z-l. . .1. 0. .y) andh = (y+l... fJ2 fJ" for 1 " j Let i be the r x 1 matrix (colunm vector) ail of whose entries are 1.3. x-l.y2. yz-y-2). x .0. J and let G be the 1 x t matrix [ g.z.. 0). .2. -yz + z2 + Y . y. Y . y. + b. y. If we let H = [ ilF EIl G J= [~ ~ Js o 0 g. n Ir.)isasyzygyofthematrix [1=191'''' .ij2.z + 2.bt E A.0. we see that the set of first coordinates of a generating set for Syz (H) is a generating set for ft n h n . EIl Fr J .z... .. (xy .xz+y2 -2yz+y-z)..as) is the vector in Am+s whose first . Y + z .I2 = (x-l.5. -yz + y + 2z .. 0. (y2 Z .y2 + yz. + b2 92 + .. -y + z . .0).yz2 .8.y-z. whose next t 2 columns are thecolumns of H2. EXAMPLE 3. -y+z+l. Set H = [ ilF. . MODULES AND GROBNER BASES m coordinates are those of -hl.6. 1 .xz . we obtain Syz(H) = ((xy-2x+y. -xz +x +y .xz .8. (x.yZ Mi + yz + 2y + 2z)). We again let A = lQI[x.) and J = (91.+s+t.. In general. Recall that I: J = {h E 7For a matrix S. y. (x 2z-2x 2+xz+4x-2y. we denote its transpose by tS.2x2 + xz + 4x - 2y.yZ -y. then the set of vectors h which are the first m coordinates of vectors in Syz(H) is the intersection of the Mi's. and N = ((x+1. (x 2z .. y2_y_z.2x +y. z] and consider the terrn order deglex with x> y > z.z .z). . -xz +x+y . This can be computed as a special case of the computation of intersection of ideals as we did in Section 2.) be ideals in A.f. y). -y+1. the set of vectors which consists of the first m coordinates of each of the vectors of a set of generators for Syz( H) is a generating set for the intersection of the r modules MiWe now turn to the ideal quotient. as in Proposition 3.xy .3. 80 we set 0 1 0 x-y x H= 0 1 [ 1 0 o 1 z 0 0 y 0 0 0 x+1 y Then. -y+1. Then. if we have r submodules matrices Fi (1 ~ i ~ T) and generated by the columns of the m x t i H = [ . EXAMPLE 3. [ Imllml·· ·IIm r copies llFl EIl Fz EIl··· EIl F.y2 +yz + 2y+ 2z.2.2. On the other hand it is easy to recognize it directly as a syzygy computation. x . using the TOP ordering on A 4 with el > e2 > e3 > €4. where again lm denotes the m x m identity matrix.176 CHAPTER 3.2. Let (sothat7 J= t [Imllm ]). Moreover. l.andletFbethemxsrnatrixF= [J.. we see that the set of vectors h which are the first m coordinates of vectors in Syz(H) is Mn N. J. the set of vectors which consist of the first m coordinates of each of the vectors of a set of generators of Syz (H) is a generating set for M n N.2)).xy . (x-l. Moreover 1 as in the case of the intersection of two modules. Let M = ((x-y.y . z)..8. . . xyz .1. xyz . 1). l. Set H = [ JIF EIl G 1. z)). So let l = (f" . y)). from which we read off that MnN = ((xy . It is then easy to put these two conditions into a single condition for syzygies of vectors in Am.l and G be the m x t matrix G = [ 91 9.9. x . 6..J. we obtain I: J = (y. xy + y2)). Let J"".8. We begin with the latter as it is a special case of what we have already done.l" ·l'em llF EB F 8l . .J. x 2 . X + y). We wish to show how to compute the intersection of two (or more) submodules of M.10) to the exercises (Exercise 3.3. Let 9 be the column vector 9 = (g"". . Then using the TOP ordering on A2 with e.. Let l = (x(x + y)2. We now consider modules which are given by a presentation as defined in Section 3. x 2 ). y. ~ 1. be in Am. J. Since M = Am IN we see that ann(M) = N: Am. i = 1.7. (X2+xy~y2. .8. and we will do so now. find the ideal quotient of two submodules of M and find the annihilator of M. > e2. [ 'e. APPLICATIONS OF SYZYGIES 177 AI hJ ç.··· . . 8l F 1 l.. Now consider H = [x 2 x(x + y)2 x+y ° ° Y 0 X(X+y)2 Y °1 . let N = (f"". y] and consider the term order lex with x> y. from which we read off that Computing a Grübner basis for this ideal (or by simply looking at it).8. x+y). ~xy2.3. Although we have relegated the computation of the ideal quotient of two modules to the exercises. 8l F t 1. Thus letting F = [ J..t. . and H = [ . x 2 +3xy+2y2). 0. . 0. I} = {h E A 1 hgi . . m copies .. y) and J = (x 2. copies Then 1: J is the set of ail first coordinates of Syz(H).. 1.) and consider M = A m IN.) and let F be the row vector F = [h J. 0.em and so h E ann(M) if and only if hei EN for all i = 1.}. .g.. We simply observe that A= is generated by the usual standard basis e" . we compute Syz(H) = (( ~y. (an m x s matrix). ~x+y. We willleave the computation of the ideal quotient of two submodules of A= (Definition 3.12.. is a linear combination of h.5).J. we set H = [ glF 8l F 8l . EXAMPLE 3.m. We let A = Ql[x. We define the annihilator of M to be the ideal ann(M) = {h E A 1 hM = {O}}. in this special case it is easy to write clown the correct matrix whose syzygies allow us to computeann(M). (~xy~2y2.. in agreement with the computation of the same ideal in Example 2. ~ xy2.. In this case.1. 1. We now go on to consider the following Question. Let N C M be submodules of Am. SO in this case 1 o H= [ 0 1 x2+Y xz . xz . + N) then K... . = (g. .g. j=1 i=l j=1 for sorne al. We note that. we see that Syz(H) = ((x 2yz . MODULES AND GROBNER BASES we see that ann(M) is the set of ail fust coordinates of Syz(H). Then using TOP with el > e2 > e3 > e4 and degrevlex with x > y > Z.xy2 .yz. . from which we conc1ude that ann(M) = (x 2yz . n M2. setting F = [ f. We wish to compute M..br. (We note that the matrix t [ tell-" 1 te m ] is the m 2 x 1 matrix of the vectors ei stacked up on top of each other. f. Alternatively. The same sort idea will allow one to construct M. + N. There are submodules K tl K 2 of Am. ] . .x)) C A 2 and let M = A2 IN. n K 2 )IN.8). (xy .. where 5 = [ ' [ Iml1m ]1 [ GIF ] El) [ HIF ] ]. = (g"". .178 CHAPTER 3. = {h E Am 1 h + N E Md.b~. b~ E A. G = [ g. we see that a coset p+N E M.8.. ..bl. we see that the set of aIl sueh p's is just the set of first m coordinates of Syz( 5).fN.t. continuing to use the notation above. g.4).x 0 o o x 2 +y xz-y 0 0 xy yz z-x ~ J.. br E A " such that t r p+N= Lai(gi+N) andp+N= Lbi(hi+N).: M 2 (Exercise 3.H = [ h.yz). .yz. We consider N = ((x 2 + y. .a.a~.8. .y).y 0 xy . K. nM2 if and only ifthere are a" . .) + N = (g"". + N. .xyz2 + x 3 . or alternatively if M. E A and b .g" f"". .g.. sueh that Ml = K.8. -xz + y.. Clearly M. -xv + yz.x 2z + y2z + xy . ..xyz2 + x 3 . . z .yz z. .. -x + z.xy2 . and setting M 2 = (h.. We now consider two submodules M M 2 of M = AmiN and we wish to " compute their intersection. . h r + N).). n M 2 = (K..at)a~.x 2z + y2z + xy . We would like to determine a presentation of MIN (see Definition 3. and lm the m x m identity matrix. Since we know how ta compute KI n K 2 we can compute Ml n M 2 ..fN and M 2 = K. . Thus.. . ] .) EXAMPLE 3. x 2 + y)). i=l i=l This last statement is eQuivalent to t s r $ p = Laigi + Lajf j and p = Lbihd Lbjf j i=l . With the notation abave.J. 0.PT E A. Then K = (h" . is in N. .10.y)+x(-y. Y + x 2 )).at E A Buch that Let H = [ J. g. SA we need ta compute an explicit set of generators for K.x). . Therefore MIN ~ A3IK. >----+ MIN Ji + N (for 1 ::. 1.y].). (1.. (-x. y). . . y + x 2 ) we see that N = = (xy. 3. we set x2 X _x 3 y y:x 2 ] and compute Syz(H) = ((0.x)) C A 2 and (_x 3 .h.. sinee EXAMPLE 3 = ((xy.x)and (xy. M N = (( _x ...: A' €i ---. .0. J..+t be a generating set for Syz(H). + h.0. 0)) (this last computation was done using TOP with e.hT ).x)-x(x 2 . where we do not assume that either generating set is a Grübner basis.g. x). 0. (-x. We note that h = (h" . (x 2 . . Then the desired presentation is A'1 K ~ MIN.) is in K if and on1y if hrf 1 + . x. s). J.. Thus > e2 > and lex with y > x). K = ((0..y). -1)...3. maps A' onto MIN sinee M = (f" .xn ]' denote the vector whose coordinates are the first s coordinates of Pi' 1 :S i ::. APPLICATIONS OF SYZYGIES 179 Sa assume that M = (f" . x..J. . y)). i ::. .). y.. We define an·A-module homomorphism q.. We have proved THEOREM 3. g.y) (0. 0. . Let K = ker(q. Sa ta determine K C A 3 such that MIN _y X A3IK. in turn. (0.. 0. 0). and let hi E klxl. (-y..) and N = (g" . x. 0. (1.J.8.8. ~ c M.9. Let A = Qlx. It is clear that q. x).8.. Then.). . is true if and only if there are polynomials al. which. y) + x( -y. -1). ... r.. let Plo . . de) c A.. Then the desired presentation is A' 1 K '" MIN.10).).. We note that li = (h. Xn]' denote the veetor whose coordinates are the first s coordinates of Pi (1 <.9t E k[x" . + (ad I)Yt· + .. = (a.... ..11 (see Exercise 3. is an element of 1.8. . .180 CHAPTER 3. . + I)I. +1. .(-y. MODULES AND GROBNER BASES We close this section by generalizing this last result to the case of an affine algebra.. h. G H = [ FIGID EIl DEll· . >---> MIN ei li + N (for 1 <.... xnlf1. We have that Ne M as before. (x2 . + I)I. +1..y). . It is clear that </> maps A' onto MIN sinee M = (1" ... is in N. Let K = ker(</». let Pl"" .y +X2)). i <.xn ]'" and M = (1" . With the notation above. exeept that we will now assumle that the ring is A = Q[x. for a vector b = (b" . + . r J. which is true if and only if there are polynomials a" ... We again consider Ne M ta be submodules of ATn.ylf1 where 1 = (x 3 +y2. 80. Then We note that ail the computations that we have done in this section could be formulated using Theorem 3. at E k[x" .!..8.lsl for !" . !. 80 to determine K C A3 such that MIN". + (h. x n ] such that (h. + h.y).x 2 )... we set b= (b. ... and let hi E k[x" . 80 we may assume that N = (y" .X n ]'+t+m' be a generating set of 8yz(H). of course.12.... + 1)?h + . We will redo Example 3.. xn]m. the coordinates of the elements of Am are cosets of 1.) for 9. ...··· ..+1) is in K if and only if (h. + (h.y. In this case we let A = k[x" .xn]m. E k[x" .bm+I) E Am. 80 now M = ((xy.·.x)) C A 2 and N = ((_x3 . but now.8.. bm ) E k[x" .8. .. + . - (a'9.. de and set 1' 1 l. EIl Dl· m copies We have proved THEOREM 3.. We deline an A-module homomorphism </>: A' --.x). . s).. .. ..8.. + at9t) [!. + I)1.11. This last statement is readily seen to be equivalent to the statement that every coordinate of hrf. . (Ü.. + I)1. we set _y X x2 X . Thus in this case. + .· . EXAMPLE 3. A 3 IK. 80 we need ta compute an explicit set of generators of K.. 9t and D = [ d. . for an ideal 1 = (d" ...Pr E k[ Xl. .!. we let F = [ 9. ..10.t. . i <. 6.1. Compute generators for ann((Q[x.12 using the technique of syzygies presented in this section. 0. In Theorem 3. This is not the case as is easily seen by repeating the exercise with the ideals 1= (x + y. in the module case.1.x.3 we required that the order be POT.3. y])2 with e. x 2 . -x). (0. 0. (xy . (1. 0. 0. y.3.3. -1. 0. 0. 3.8. 3. (-x. 1).11 using the technique of syzygies presented in this section.6 and 2. In tbis exercise.12 and Exercise 2.8. 0. (-y.0. x. 0. 0).8. x + 1). y]. One might think that the problem comes from the fact that the generators for l and J above did not form a Grübner basis with respect to deglex with y > x to start with. 0.8. x 2 + 1) and J = (y2.11 and 3. y])3.8. Consider the following two submodules of (Q[x.y).0. Show that these generators do not form a Grübner basis for InJ with respect to deglex with y> x. (0. -1. yx + x.3. where M and N are modules. we show that the TOP ordering would not give rise to a Grübner basis for the intersection of two ideals. 0. 1. > e2 and lex with y> x). APPLICATIONS OF SYZYGIES 181 and compute Syz(H) to be generated by (0.7. 3.9.0.xy.1.y) using the same orders as above. 3. 0.6. x. Consider the submodule N of (Q[x. -x . Redo the computation of the intersection of the ideaJs given in Example 2. 3. x). y. and (0.3. 0). 0. -1. where land J are ideals. (x 2.0. (-x. x. K. Exercises 2. y. x. 0. y .4. = ((xy.8. 0. y[ with y > x. 0. 0. 0). 1. Compute generators for In J by computing generators for the syzygy module of H=[l 2 x xy+x 10 ° xy2+y ° 2° 1 2 ° y° x-xyx -y using the TOP ordering on (Q[x.15 using this technique. 0.0.6.3. (0. 0.5. 0.x. > e2 and deglex on Q[x.8.0. (-x. y])3. -1.y) of Q[x. 1.3. 0).13 and Exercise 3. 0.2.x. 0. ExercÎses 3. 0. 0.Y.0. y. Redo the computation of the ideal quotient of the modules given in Example 3. 0).x 2. y. N = ((xy . )). Thus K = ((0. Redo the computation of the ideal quotient given in Example 2. x 2 . 0. (1. y]) 3 / N).6.xy + y)).6.8.xy +x).x. x.y)) and .0). x 2 +x) (tbis last computation using TOP with e. 0.8. X . 0.y. xy2 +y) and J = (y2.6.Y +x). Give the analog of the matrix H UBed to compute generators for 1: J. Consider the ideals l = (x 2. (x + y. find H such that M: N can be obtained from Syz( H). Exercises 3.x. (-x.7 using the technique of syzygies presented in this section. 3.6. That is.x.0. Redo the computation of the intersection of the modules given in Example 3.x).0. and the ideal quotient of two submodules of Am/N.8.8. all the computations done in this section can be viewed as the computations of kernels of certain A-module homomorphisms. y)).jN) n (K 2 /N) ç (iQl[x.10. a.x 2-2xY-X+y2+ y ). xnl be an idea!.'" . y2+y. where K" K 2 and N are the modules of Exercise 3.) ç k[XI.11 to generalize all the computations performed in this section to the case where A i8 an affine ring.8. 3..h) E A' œA m 1 cj.6 using this technique. MODULES AND GROBNER BASES K 2 = ((0. Give the analog of the matrix H used to compute generators for (K. Reda the computations in Example 2. the annihilator of M = Am/N. We define the pullback of this diagram to be the submodule PB of A'œAm defincd as follows PB = {(g.7. h"". x n ]/ J if and only if there is an element in the reduced Gr6bner basis for Syz(1. Ali the computations done in this section can be done using only Theorem 3. O.1.x n ].2xy+y2 -x+y). Let l = (fI.8. Namely. Find the presentation of Kz/ N. where l = (d" . Prove that N = ((y2 +y.8. We illustrate this in the present exercise. (y..xn]/I.1.jN) n (Kz/N) which can be used to compute generators for (K. of the form (1. We get a commutative diagram .9. J" f).jN): (K2/N). for A = k[XI"" .12.8. (0. in this case. . y]) 3 /N.8.··· . 3. x-y)...11.y2 -xy+ x 2 + y)) is a submodule of K. Shaw that.x 2 . Consider the following diagram of free modules. Compute generators for the ideal quotient (K. That i8. .8. and K 2 . with respect to any POT order with the first coordinate largest.O.jN): (K2/N). J. fI. where A = k[Xl 1 '" . h" hl.8. Compute generators for the module (K. (0.(g) = -y(h)}.182 CHAPTER 3. 3. the intersection of two submodules of Am. b.9.de). x-y. where K 2 and N are as in Exercise 3. give the analog of the matrix H that i8 used ta compute generators for the intersection of two ideals in A. y. the ideal quotient of two submodules of Am. (0.7. .10 and Exercise 2. . the intersection of submodules of Am/N. y2+y).8. 3. Show that J + l has an inverse in k[XI"" . h + J is the inverse of J + J. 3.xY+X. Follow the construction of Theorem 3. Show how to compute generators for M: N. N) is an Amodule... b.1. we wish to "describe" Hom( M. N) by elements of A: for a E A and </J E Hom(M. it is easily verilied that. .9.p in Hom( M. .p)(m) = </J(rn) + . c...8. . by giving generatars (elements of Am lU) for V. Show how to compute generators for the annihilator of an element of AS IN using the pullback. where N is a submodule of AS. Again.s be the standard basis for A'. . We will then obtain a presentation for Hom(M. Given two finitely generated A-modules M and N. 3.9. i = 1. with this multiplication. Then show how to compute generators for ann(A' IN). N). N) as a submodule of sorne quotient Am lU.. We deline the usual addition of two elements </J and . As mentioned earlier. We are interested in the study of the set of ail A-module homomorphisms between M and N. or. Hom(M. s. d. COMPUTATION OF HOM 183 A' ~~</J.3): by computing a presentation of V. with the projection onto Am and AS respectively. we deline a</J by (a</J)(m) = a(</J(m)) = </J(am) for all m E M.9. It is easily verilied that Hom(M. . Also.----_ A' where 'Trm and 1rs are the compositions of the inclusion PB ~ AS E9 Am. there are two ways one can "describe" an A-module V (see the discussion following Lemma 3.i = 1. We can also deline multiplication of elements in Hom(M.3. Show how to compute generators for the intersection of two submodules of a free module using the pullback. Since we will need it later on. we first consider the easy case of computing Hom(A'. We deline Hom(M. if V is a submodule of sorne quotient Am lU. N): (</J + . Let €i. An element </J of Hom(A'. a. where M and N are two submodules of Am. N) using Theorem 3.. Show how to compute generators for PB.p(m) for ail m E M. At) is uniquely determined by the images </J( €i) E At. N). we consider two A-modules M and N. Computation of Hom.N) = {</J: M ----+ N 1 </J is an A-module homomorphism}. N) is an abelian group under this addition. . As before we let A = k[x" . x n ].At). We will lirst give generators for Hom(M.. 184 CHAPTER 3. At) ~ ASt. suppose that we have a presentation of the A-module M. we need to assume that we are given M and N explicitly. we assume that we are given presentations of M and N.x). Now. We will denote this matrix by cp also.. cp(l.9. J6). A') with the column vector cp' in A" formed by concatenating the columns of cp in order.9. Sequence (3. 0) = (2x+1. Then the idea is to compute Hom(A' 1 L. cp is given by matrix multiplication by the matrix whose columns are the cp(ei)'s. That is. x. in order to describe Hom(M. To do this. x 2 . 13.x2 . Note that cp is a t x s matrix. We further identify the matrix cp in Hom(A'. J5. we identify the matrix [j: j: j: 1 with the column vectors (h. 1) = (_x 2 . As an example of the above identification. This identification gives an explicit isomorphism Hom(AS. suppose we have given bycp(f"h) = ((2x+1)h -x 2 j"xh +j. as an- other example. say M"'A'IL and N"'A'IK. 0. M "" A'1 L. J4. 0) and cp(O. Then we have a sequence of A-module homomorphisms (3.x). as in linear algebra.1.1) is called a short exact sequence. _x 2 . MODULES AND GROBNER BASES Then. 1.1) where the map i: L ~ AS is the inclusion map. 12.(x 2 -x)h). 1. N) explicitly. Or. and 7r: AS --+ M is the map which sends the standard basis of A' onto the generating set of M corresponding to the standard basis in the isomorphism M "" AS IL. Then it is readily verified that cp is alQ[x]-module homomorphism. Hence the matrix associated with cp is and so the vector in lQ[x]6 associated with cp is (2x + 1. In general a sequence of A-modules and A-module homomorphisms BRecall our space saving notEttion for column vectors! . x.9.N) --> AM EXAMPLE 3. Moreover. At) above. At1K) by adapting the ideas in the computation of Hom(A'. sinee it is readily verified that it gives a one to one and onto A-module homomorphism Hom(M. p >---> ~ Hom(M"N) and Hom(N.p c""'.9.3) and (3.6) Hom(P. M 2 )/ker(.2) Let P be anothec A-module. (3.9.9.N) . We will only be interested in the exact sequence ASl -S AS ~ M -------7 O. We use this last sequence.. and 1r is anto.!2. and N and an A-module homomorphism <p: M. N).9.5) and (3. if Ml = ASl) M 2 = A S 2. COMPUTATION OF HOM 185 is called exact if im(Œi+l) = ker(Œi) for each i. .M. Then the two sequences (3.. in this case.po) = Hom(P. Hom(P.2..1) is exact.9. M 2 ) .) . M 3 ) ~ Hom(P.. It is easy to verify that 0<P and <Po are A-module homomorphisms.9.p<p. Assume that we have an exact sequence of A-modules and Amodule homomorphisms (3.. For example.9. In order to do this we review the elementary homological properties of Hom.M2 ) <P a p. M2. M. It is easy to check that Sequence (3. LEMMA 3. M 3 ) ----> 0 are exact. In parlicular.!2. means that i is one to one. say L~AS. and a similar one for N to compute Hom(M.4) Hom(P. Now we lind a presentation of L./L" and thus we have another short exact sequence 0----4 Ll ~ ASl ~ L -------7 O. This leads ta an exact sequence where r = i a 1f1 (the exactness is easily checked).9. .2.) . .g.p 0 <P p Hom(N. which. M 2 )/ im(<po). We refer the reader to [Hun]. then 0<P and <Po are given by matrix multiplication. ker(1f) = im(i).9. and N = At l and the maps are given by matrices. e. What we need is summarized in Lemma 3.3. Hom(P. We assume we are given A-modules Ml. -----> M 2 • We deline two maps Hom(M2 . 13) Hom(A S" At.. N) and Hom(A".9.A t )jim(8) Hom(A". a similar ..3) we let P = N and replace Sequence (3.9. N) ---+ 0.9.10) kerh). and '1 =or.N) Hom(A". o and p. for P ---+ = AS and J. Ll) are easily seen to be the vectors which generate L (resp.9.) So..9) where 0.9. Ll) is given by a matrix which we will also denote r (resp.8) to obtain the exact sequences (3. we will identify M and N with their respective presentations.12) where 8' (3. For this we first compute presentations of Hom(A'.' Ast. and then we can compute ker('f*) using Theorem 3.N) ~ ~ Hom(A'.6) Hom(A'. = 7r~ where D = D.9. Hom(M. since the image of r (resp.7) to obtain the exact sequence (3. At)j im(8').!-' ---+ (3.A") ~Hom(A'.N).2) with Sequence (3.9. The colmnns of r (resp. is a t x tl matrix.186 CHAPTER 3. the map r (resp. AtjK). These give. Now in Sequence (3. if M and N are not given initially by presentations. then compute the map ')'* corresponding ta 'Y between the two presentations.11) Hom(AS.9..8. Hom(A".9. Now in Sequence (3.8) As before.7) (3. MODULES AND GROBNER BASES We now return to the computation of Hom(M. Ll).9.9.9. K).At) --"-+Hom(A'. it now suffices to describe im(8) as a submodule of Ast in arder ta obtain a presentation of Hom(A'.. N) . K). Ll) is L (resp. N). = Llo and p. N).. Thus from Equation (3.' = 1f~ for P = A".) 8' Hom(A S" At) Hom(A". that is.9.2) with Sequence (3. N) and A'jL (resp. Thus to get a presentation of Hom(M.9.2. ta ease notation. as noted above. From now on. we have the two exact sequences (3.N) ---+0. we will assume that M = A' j Land N = At jK.--. At) 3. using (3.. (However. replacing Sequence (3.9. N) 3. N) for two A-modules M and N. we will have to take into account the isomorphisms between M (resp.9.4) we let P = AS and P = A" respectively. and Since. N) Œ =01f --"-+ Hom(A'. We have that r is an s x 81 matrix and D.' (3.5) we see that Hom(M. N) it suffices ta obtain a presentation of kerh). Of course. Hom(AS. where the right-hand side expression is a matrix multiplication. Thus we have (3. LEMMA 3. columns of ~ EIl··· EIl ~.) = ~.. x s matrices . '-----v------' This analysis can be scopies continued and we see that 8 = ~ EIl .. we will adopt the notation for making new matrices out of old Olles presented fight after Example 3. '-----v------' scopies PROOF. aiter concatenation of columns. by concatenating the columIlB of the matrix .pij) = ~ 0 . ~ Hom(AS. In arder ta facilitate the statements of the following results. and Wij with the corresponding matrices ~ and tPij respectively.6.2.pi.pi.p12.) --> Ham(AS. These matrices correspond to the colunm vectors 'l/Jij E Astt obtained.6. That is --> e---.14) Ham(AS. EIl ~ ..p22. Now the t x s matrix . EIl ~).pij whose entries are aU zero except for the ij entry which is equal to one.pij. in effect ordered the . Because of the identification of the homomorphisms .3. ~.pij.At) corresponds to a map 8* which is given by an st x sh matrix S. EIl ~. we denote the module generated by the COIUIDIlB of a matrix T by (T)). the submodule of Ast generated by the COIUIDIlB of 8 (in general.'lj.plj form the standard basis of As t .. as before. ) f1WtIl correspond to the ftrst t.). 8(.. At. ~..3.pij). The matrix 8 above is the matrix given by ~ EIl ..pI/s (or equivalently the . Similarly the matrices ~.8.'S) as follows: Now for each i and j.N).ij is the matrix whose j-th column is the i-th calumn of ~ with all other entries equal to zero.ij under 8*. Therefore. + 1 through 2t. The map 6: Ham(AS.2 '-----v------' scopies correspond to calumns t. the columns of S are the vectors obtained from the matrices 8(. N) ~ A st /(~ EIl .9. . At) by concatenating the columns of 8( .~. 8' Ast 8<//.. COMPUTATION OF HOM 187 statement halds far Ham(A". The columIlB of the matrix 8 are then the images of the vectors 1f.9. A basis for Hom(AS. the matrices f1Wu) f1W2b . scopies . '-----v------' 8 0 Now the image of 8 is the column space of 8. At. we have 8(. We note that by concatenating the colurrllls the way we do. we have.9.. .pt.At.pij. That is.. EIl ~. copies of ~ EIl . that is. The vectors .) is the set oft.pij) E Hom(A'. To do this. because of . 1t corresponds ta a map '"'1* ~ f--+ Hom(A".3. At)1 im(6) and Hom(A". PROOF. it suffices to show that "I(im(6)) ç im(6').ij and 7.:.188 CHAPTER 3.4. At) </. N) ~ Hom(A'. At)1 im(6'i. SI copies LEMMA 'Y' (</.' + (~ EIl . then 7(6(1/. :. At) </. MODULES AND GROBNER BASES Similarly (3. EIl Do) scopies A"t I(Do EIl':' EIl Do). Then 7* is given by a matrix T whose columns are the vectors "j*('ljJij ) E AStt. As mentioned above. 0 r. -2.9. We next show that the map T Hom(A'. Again.. EIl Do). the columns ofT are the vectors 7'(1/. N). we see that '1 corresponds to a map 1": As t --.:. A. N) and ofHom(A". 'Y(</./Jij be defÎned as in Lemma 3.7.N) ---->Hom(A". The matrix r 0 It is defined ta be the st x Slt matrix obtained by replacing each entry 'Yij of r by the square matrix 'YijI" where It is the t x t identity matrix.' + (Do EIl .N) </. Using the isomorphisms A.t ~ Hom(A'.13) we have Hom(A'. These vectors are obtained by concatenating the columns of "I(-I/Jij ).. T is the transpose of SI the tensor praduct r ® It.15) Hom(A".) = (Do 0 1/. We now wish ta describe T. that is. + im(6)) = 7(</.* scopies S1 copies S1 copies . E Hom(A'. 0 r. N) ~ A"t 1 (~EIl .N) ~ Hom(A".) 0 r = 6'(1/.A"). So let 1/.9. Moreover. 0 r) E im(6').N) is induced by the map 7.N) </.) + im(6'). There is an S1 t x st matrix T which defines 'Y' in the sense that T</.ij) = 1/. Thus 'Y' is induced by "l'and Hence is given by the matrix T.9. AStt. 81 copies Therefore we have a presentation of Hom(A'.ij 0 r by concatenating columns.)) = "I(Do 0 1/. Now from Equation (3.9.At) and A"t ~ Hom(A".) E A"t obtained from the matrices "I( 1/. We now return ta the map Hom(A'.A t ). Let 'lj.t1(Do EIl .: . as desired. EIl Do)) = s copies 3. EIl Do). f--+ Hom(A". We lirst deline the map copies Hom(A'. 0 Recal! that Hom(M.b22f.pt1r have their respective first.16) Hom(M.pijr is the t x SI matrix whose i~th row is the j~th row of r.Wt2f correspond ta vectors which form the columns of [ 1'2IIth22Itl" '11'2s I t 1' where [')'21 122 l'2s ] is the second row of r. N) is isomorphic to ker{')') and hence to ker{')"). Therefore (3.pi/S as before. E9 ~ (see Theorem 3.p2Ir.9). row equal ta the first row of r. .9. We first compute the kernel of the hOlilomorphism AS' -->A"t/(~E9':'E9~) 81 copies given by 1>' >----> T1>' + (~E9 . scopies A presentation of Hom(M. . We compute Hom(M. The corresponding vectors in ASlt obtained by concatenation of columns are then the columns of the transpose of the matrix [ 1'l1IthI2Itl" 'l1'b l t 1' where [1'11 1'12 1'15 1is the first row of rand It is the t x t identity matrix.. where the right~hand side expression is a matrix multi~ plication.5.pijr. . N) can then be computed again using the method given in Theorem 3.. Similarly. N) "" (U)/(~ E9':' E9 ~). etc. .8. Let U be the matrix whose colunms generate this kernel... (iii) Compute the matrix U defined by the kernel of the composite map A st ~ ASIt ------+ ASl t /(~ EB':' EB il). We can continue this analysis and we see that T is obtained as indicated in the statement of the lemma.pij and r with the matrices . E9 ~). '1. SI copies . the matrices 't/J12r.9. (i) Use the generators of Land K as columns to define the matrices rand ~ respectively..3. 80 we now compute the kernel of 1'*.9.. To summarize we now state THEOREM 3.8. wehave 'Y(. Thns the columns of U are given by the first st coordinates of the generators of the syzygy module of the columns of T and those of ~ E9 .pij and r respectively. For each i and j.. second.pij) = . 80 if we order the . COMPUTATION OF HOM 189 the identification of the maps .p11r.9. (ii) Let T = '(r@It). the matrices . ... N) as follows. Let M and N be A~modules given by explicit presentations M"" A' / Land N "" At / K. . Let L = SYZ(fI' f" f3' f 4 ) and compute L to be L= ((-l..190 CHAPTER 3. To do this we use Theorem 3.-l).9. We let A = <Ql[x. using TOP with el > e..x-z)) ÇA 4 . y.8. MODULES AND GROBNER BASES using Theorem 3. g. we now give an example. > .-l. Let M = (fI' f" fs.g2.8. This gives Equation (3. 0 0 0 0 0 x. = [ ~: ]. We also let K = Syz(gl. Z g3 = [ x / : yz' ] .9. where gl = [ ~~ ].-yx+yz-z'.9 . We first need to compute presentations of M and N.g3) ÇA'. (iv) Compute a presentation of Hom(M.z'. Ta illustrate the Theorem. z] with the lex order with x > y > z.16).g"g3) and compute K to be K= ((yx. f 4 ) ç A3. we compute the syzygy module of the .6.O.8.9. where and let N = (gl.(-z.z -1 z' -1 x-z "1 0 and Ll = [ -yx :~z -y+z z' ] . r= [ -. EXAMPLE 3. Thus the matrix T is -1 0 0 x+z 0 0 -1 x+z 0 0 0 0 -1 0 0 -z x+z 0 0 0 0 0 -z 0 0 z' 0 0 -1 0 0 0 0 0 0 -z 0 z' 0 -1 0 0 0 0 0 0 -1 0 0 0 -1 0 0 0 0 -1 0 0 x-z 0 0 x-z 0 z' -1 x-z To compute the kernel of the map Al' ----> AB j(Ll EIl Ll) given by T composed with the projection A6 ----> AB j (Ll EIl Ll).9. on al! the modules considered..x+z.-y+z)) ÇA"So we have the presentations M'l'!A 4 jL We let and N'l'!A 3 jK. N) using Theorem 3. A 3 /K) '-" (U)/(il.9.N) '-" Hom(A 4 /L. EIl il.EIlil.16) we have Hom(M. EIl il.).z2 ~y+z ~y yz 0 y 0 From Equation (3.9. EIl il. EIl il. COMPUTATION OF HOM 191 columns of T and of ~yx yx +yz ~ z2 ~y+z o o o o o o yx ~yx+yz ~ z2 ~y+z The first 12 entries of the generators of that syzygy module are the columns of the matrix U.EIlil. EIl il. = yx ~yx + yz ~ Z2 0 0 0 0 0 0 0 0 0 ~y+z 0 0 0 yx ~yx + yz ~ ~y+z z2 0 0 0 0 0 0 ~yx 0 0 0 0 0 0 yx + yz 0 0 0 ~ Z2 ~y+z 0 0 0 0 0 0 0 0 0 yx ~yx+yz ~ z2 ~y+z .3. Finally to compute a presentation of (U) / (il. where ~1 0 ~1 u= 0 0 0 0 0 x 0 0 1 0 0 0 0 ~1 0 0 0 0 0 x 0 0 1 0 0 0 0 0 0 X 0 0 1 x 0 0 1 0 0 0 0 0 z 0 0 0 x 0 0 1 0 0 0 0 0 z 0 0 0 x 0 0 1 0 0 0 0 0 z 0 yz Z2 ~y+z ~y y ~y Y 0 0 0 0 ~yz 0 0 0 0 0 yz .EIlil.) we compute the syzygy module of the eight columns of U and the four columns of il. EIl il. . ~1 ~ 0 0 o 0 0 x 0 0 <P6 = [ 0 0 1 0 n n <P7 = [ yz Z2 -y+z ~ -y 0 0 0 -yz ] [ y yz z2 0 0 Y y.. Let ei. 0 <P2 = [ ~1 ~ 0 0 0 o 0] xl. t4 = -y+z -x 0 0 0 0 0 -x Therefore Hom(M. A3 / K) is generated by the cosets of the eight matrices <Pl = [ ~1 0 0 0 0 0 Xl] o . We recall that we identify the homomorphism <P with the matrix that defines it.t4)' We now show how to generate A-module homomorphisms from A 4/ L to A3 / K. . o 0 0 0 <P3 = 0 [ <P4 = [~ 1 0 0 0 0 0 ~ ].192 CHAPTER 3. . K. MODULES AND GROBNER BASES There are 4 generators for this syzygy module and we need the first 8 coordinates of each of them. by the matrices obtained from the columns of U.t3.8. .8. i=l 8 where ai E A.z2 -y+z . The fact that this homomorphism is well-defined follows from the construction of U.. Therefore Hom(A4 / L.as) E AB.'" . Using this vector we construct a 3 x 4 matrix whose colmnns are the 4 cOIlBecutive 3-tuples of the vector Ua..A3/K) "" A8/(tl. This matrix defines a homomorphism from A 4/ L to A3 / K.<P8 = ~y 0 0 0 -y+z -y] Y . the product Ua is a vector in A 12 . which give 0 0 0 y -y tl = Y -y 0 1 0 . i = 1.8 be the standard basis of A8 Then Hom(A4/L. that is. Thus any <P E Hom(A4/L. .A3/K) is generated by the matrices obtained froID U ei.i '" 1.. and hence from M to N.t2. .z2 0 yz . 0 It can easily be verified that. . For a = (al. we have <PiL C. t2 = 0 0 0 0 0 1 .. o 1 <P5 = [ o n. t3 = 0 0 0 0 yz . . for i = 1.A3/L) can be expressed in terms of the <Pi'S <P = I: ai <Pi ..N) "" Hom(A4/L. . 3. In particular. e2. and f4 can be computed in a similar way. correspond- . Let tP be any homomorphism in Hom(A 4/ L. N) as follows.4. Then we deline a homomorphism 1>' in Hom(M. +K) +( -a2 + xas + (yz . g2 g3 e.3. + xa4 + yaS)g. In particular Hom(M.z2)a7 . + as [ ~y ] +K [ -a2 + -a3 :a:':(:~"-+):î~7 . But 1>'(fi) is just the image of 1>(ei + L) under the isomorphism A 3 / K '" N. e3.ya8)(e~ + K) +(-a3 +Xa6 + (-y + z)a7)(e~ + K).+z. First. + (-a2 + xas + (yz +( -a3 + xa6 + (-y + z)a7)g3· z2)a7 . N) is generated by the homomorphisms 1>. A3 / K). [ ~1 ] + a2 + a6 [ ~1 ] + a3 + a7 [ ~J + ~ a4 [ ] ] +a5 [ ~ ] [ ~ ] [ ~.2. COMPUTATION OF HOM 193 To obtain a description of Hom(M. it is enough to determine 1>'(fi) for i = 1. That is (-a. and e4 are the standard basis vectors for A 4 . A3 / K) we make use of the isomorphisms ~ M f .+K where the vectors el. N) from Hom(A4/L. +L e2 +L e3 +L e4+ L A3/K e~ +K e~+K ~ ~ N >---+ >---+ c--> g.yaS)g2 The images of f2' Is.9. f2 f3 f4 ~ >---+ >---+ >---+ c--> A 4/L e.yas ] + xa6 + (-y + z)a7 +K (-al +xa4+yas)(e. 1>' (f 1) is the image under the isollorphism A 3 / K '" N of 1>(el + L) i=l a. and the vectors el' €21 and e~ are the standard basis vectors for A 3 . say 1> = I:~~l ai1>i. M fI f2 f3 f4 1>'3 .y. M f.-y. g. Exercises 3. g. z]. yz2)). 0 -yzgj + yzg. f. --+ --+ --+ --+ --+ h f4 --+ 1>'7'.y).(x.1 that M has a presentation. N xg. M = A 31L. --+ --+ --+ --+ --+ h f4 N -g3 0 xg3 g3 N xg 3 g3 0 zg3 1>'4'.. f. Let M be a finitely generated A-module. . Free Resolutions. and a presentation of Hom(M.x)) ç (Qlx.x.10. (x'.yg..x+y. In this section we first show how to compute an explicit free resolution of a finitely generated A-module. 0 xg. 0 xg . f. N xg. the free resolution together with the results from the last section will be used to give an outline of the method for the computation of the Ext functor.(O.x-z). We will then use this free resolution to prove a theorem of commutative algebra concerning A.(-xz.x.xy). Let A = Qlx.y])3 and N = ((xy. y-z.(x.O..(y.x'. and a presentation of Hom(M.z2)g. 3.1) . g.z2)g2 +(-y + z)g3 -yg. f.(y.y.y). + yg. N). xz 2).9. Finally. 1>'. y-z). . 3. f3 f4 --+ --+ --+ --+ --+ N -g..y-z)) and K = ((x+y. f3 f4 --+ --+ --+ --+ --+ N (yz .x. and N = A 41K. + yg. 1>'8'. Consider the following modules M= ((O.x).x)) ç (Qlx.x-z.y])4 Compute generators for. M f. MODULES AND GROBNER BASES ing to the homomorphisms 1>i which generate Hom(A 4IL. 0 zg. M ==: A SO / Mo for sorne So and sorne submodule Mo of A SO • We also have the following short exact sequence (3. g.y. x-z. --+ --+ --+ --+ --+ h f4 N -g. --+ --+ --+ --+ N yg. where A = klxl.194 CHAPTER 3. . N). --+ 1>'6'. where L= ((x. --+ --+ --+ --+ --+ --+ --+ --+ M f. f3 f4 --+ 1>'5'. . xy+yz. M f.10.1. Compute generators for. y-z. f2 f3 f4 M f.y. Xn].x). x-z. 0 (yz .z).9. 0 zg. that is. A3IK): 1>'l '. We saw in Section 3. M f.. y. +( -y + z)g3 -yg. f.2. M f. (xy. f. (xy.x'. 0 ---7 0. FREE RESOLUTIONS 195 ta indicate that A" IMo is a presentation of M. Thus we have obtained an exact sequence DEFINITION 3. Mo. The exact sequence obtained in (3. and <P.pl ASO ---7 !/Jo M ---7. and M. A" /\ A" • ASQ 'M . We continue this process recursively: at the jth step.l . 0 0 \/ /\ M2 \/ Mo 0 0 /\ 0 FIGURE 3. Now we find a presentation of Mo.10.1.3. -----+ ASl 1f1 -----+ M. = io 0 "1 (the exactness is easily checked).3. we find a presentation A'j IMj of M j . 0 where <Po = "0. We then obtain a sequence of module homomorphisms as in Figure 3. -----+ ASl -----+ . This leads ta an exact sequence O -----+ M 1 i.10.2) is called a free resolution of the module M.10. o 0 \/ M.3. and KO: A 050 ---7 M is the map which sends the standard basis of A" onto the generating set of M corresponding to the standard basis in the isomorphism M ~ A 'olMo. The map io: Mo --+ A" is the inclusion map. Presentations of M. . say Mo~AS'IM" and thus we have another short exact sequence O ---7 M 1 i. wesay that the free resolution has finite length and that its length is <::: C. ... and hence svj.... Sa. = 0.10. We say that A has global dimension less than or equal ta n. at all... We will show below that finitely generated A-modules have free resolutions of finite length.Xi do notappear in lm(gv)... . Otherwise. then X vp. .10. If lm(gv) and lm(g".e~ the standard basis for Am.. . ..) for 1 <::: v < p <::: t. . and the other is defined in At with respect ta whlch the set {Sij Il <::: i < j <::: t} is a Grübner basis·for SYZ(gl' . ..t}. In particular.. if Xl. by Theorem 3. .Xi. Ifthevariablesxl. We will assume that they are ordered as follows: the gi'S are arranged in such a way that if v < Ji.196 CHAPTER 3. then the variables Xl.J. we need ta be careful which arder we put on At Notice that the arder on At changes if we reorder the vectors Y1' ... We will use the same notation for bath orders. <::: t. Then. . Leti E {l...Xi do not appear in any Im(9v) for v = 1. Xi+l do not appear in lm(sv". First we have a technicallemma. Let v. . . we will nse < and we will write lm(gi) and !m( Sij). A free resolution of a module M is used ta "measure" how far away from a free module M is and ta give sorne useful numerical invariants for the module M and the ring A. Im(g. In fact.7.m} (that is.. free modules are the analog of vector spaces. lm(gv) and lm(g".Xi.ej for sorne j E {l. that is.t = 0.) have the same non-zero coordinate) then Xv > X". Recall that it is a Grübner basis for Syz(gl'··· .gt' AIso. and lm(gJ = Xvej.13. there exists a finite resolution of length at most n.. . let Xv = Im(gvJ = Xvej and X"..7. . Let {gl' .: Am be a Grübner basis for M = (JI'··· .2) it turns out that there exists an Csuch that A'j = 0 for all j 2: C. E A and sorne j E {l.). . . .gt). Ji. we will denote by el. E {l. for each A-module M.2. . We will now nse the theory we have developed sa far ta prove this result (Theorem 3. Let {Sij Il <::: i < j <::: t} ç At be as in Theorem 3. for some v E {l. Xl does not appear in any lm(sj". .n-l}...gt) with respect ta the term arder < on monomials of At induced by [ gl gt by Theorem 3. v < Ji.). .ej for sorne power products Xv.4). .t.. so no variables appear in sv".3. If in Sequence (3. for every Ji. <::: t....) do not involve the same coordinate... . We note that we are using two different term orders: one in Am which is used ta compute the Grübner basis {gl' . In fact. Xi+l do not appear in any lm(sv". MODULES AND GROBNER BASES In module theory.. The context should indicate which module we are working in and which order we are using. > Xn (regardless of the arder that was used ta compute the Grübner basis for M). for 1 <::: j < Ji. .gt} c.. = !m(g) = X". . . Binee we will be using the standard bases for bath At and Am. the number of variables in A. ..t}.) with respect ta sorne term arder on the monomials of Am. PROOF.10. with respect ta the lex ordering with Xl > X2 > . and by e~... then Xll'" . X".J = X". such that v < Ji. . 1' LEMMA 3.et the standard basis for At. .gt} for M.13.m}.7. ..Xi.3x 3 +x2 + 2x).x 4 . . y. y > e2 > e3 + x2 - X. let us assume that Xl.x .. -y .0).Xi+l do not appear in any lm(sv~). x 2 . x 5 . Every finitely generated A -module has a free resolution of length less than or equal ta n. _x 3)..0.x 2 -x. (X .. -1) (x 2 . y])3 of Example 3.. FREE RESOLUTIONS 197 where XV{J. .3. = (x 3 + X.x 5 _x 4 .. 1.x 2 + X Using the order induced by the y's we have Im(s'2) Im(s34) Im(s56) THEO REM 3. We re-order the y's as required in Lenmm 3. Y6 = (x 2 - 2x. . lcm(Xp.10. ..Xi do nat appear in anylm(gv)' v= 1.3x 3 + x 2 + 2x). thenxl. Xi+l do nat appear in ---..2). -y .x. _x 4 + x 3 + 4x 2 .. -y 2 X - x2 . _x 2 + 2x. First. Y2 = (x 3 +x.7. .X. -x).2x .X 2.4.x 4 . Y6 form a EXAMPLE X Grôbner basis with respect to the TOP term ordering in A3 with e. _x 2 + X + 2.l Xv).6. . Y5 = (x 2. Y4' Y5..Xi do not appear in Xp. and we get = (y+x 2. and 80 do not appear in Im(svjt). Since Xv > X~ with respect to the lex term ordering with Xl > X2 > .-x 2 +2x. and the power of xH1 in Xv i8 at least as large as the power of Xi+l in X w Therefore the power of Xi+l in XVIl. . Y6 = (x 2 . .!::..· .2x.i8. as usual.10. The syzygies are now 812 (x 3 2 + X.3x3 + x 2 + 2x. x 3. We saw that the vectors y" Y2' Y3. 0).1.2 with y > x.4.10. 0) 834 856 + 2.!!:.t. Y5 = (x 2. x2 3 - X. As in Example 3.-x). Thus Xv ifx1. for 1 :S v < '" :S t. . . x 5 ..5 we let M be the submodule of A 3 = (<Ql[x.3. where n is the number of variables in A. then Xl.2x. > Xn. -X. 0 3. . where y.y).. Y2 = (x. . The last statement is proved in exactly the same way.i8 the same as the power of Xi+l in XV) and hence the variables Xl.. and with lex in A and y > x.10. _x 3 .. _x 2 + 2x.Xi.Xi do not appear in Xv. then. we see that the module Lt(g" .. Xll'" ) Xi) Xi+l do not appear in the leading monomials of the elements of the Grübner basis for ker(1).6. If Mo = {O}. t}..2.14. Also. otherwise we may assume that G = {gI. .198 CHAPTER 3. We then choose A'H' to be a free module whose basis maps onto Gand we let 1>j+1 be the projection map.Xi do not appear in the leading monomials of the g~. we see that for ail f E AS" f + Mo = Nc(f) + Mo.).2. If X" . If Xl appears in lm(gj) for some j. generated by the ej's that appear in sorne lm(gi). lm(f) is divisible by some lm(gi) for some i E {l. Moreover Nc(f) E M'. by Proposition 3.. Then M has a presentation M ~ AS..2...i > O.Therefore 7f is one ta one. . by Theorem 3.. so}. where Nc(f) is the remainder of f under division by G.). .) <:: AS..'S. Therefore 1f is onto.2 recursively. IMo is isomorphic to a free module.g.) = 8yz(g" . by Lemma 3.. Then none of the variables Xl.). we choose a monomial order on the free module A'j and find a Grübner basis G for ker(1)j). we see that no variables appear in the leading monomials of the elements of the Grübner basis for ker(1)n_i).) is the free submodule of AS. At the jth step.. then. We will prove that M has a free resolution of length less than or equal to n .g..i = O~ we see that .10..g. Assume also that the g/8 are arranged as in Lemma 3.2.) and ker(1). Consider the map 1f: M' f ---> e--. . and j E {l.10.i. CASE 2. Finally. We construct a free resolution recursively as follows. AsoIMo~M f+Mo· It is easy to see that 1f is an A-module homomorphism. We need ta show that M ~ AS.gt} fOrIns a Gr6bner basis for Mo with respect to sorne term order in A 80.10. . . .X n appear in any of the lm(gj)'s. Note that ker( 1>0) = Mo = (g" . But this is impossible unless f = 0..g. then we are done. Let Xl. Let M' be the free submodule of AS.. We will show that M ~ M'.. generated by the other ej's. . since M' is generated by those ej's which do not appear as leading monomials of any gi. . if f E M'and f E Mo..10. and hence 'if iB an isomorphism.. sinee the monomials of Nc(f) are exactly those monomials which are not divisible by any e J which appear in some lm(gi).Xi be the variables that do not appear in any of the lm(gv)'s (i = 0 is possible). .10. 80 ifwe apply Lemma 3.'" . . where Mo = (g" . CASE 1. Let M be any finitely generated A-module. . We arrange the elements of G according to Lemma 3.5. n . n = i. By the case n .3.. Since the leading monomials of the gi '8 are of the farm aej for sorne a E k. by Lemma 3. Xl does not appear in the leading monomials of the elements of the Grübner basis for ker(1). .. IMo. then. MODULES AND GROBNER BASES PROOF. 10..5.94. 856) = (0.0).2 aod Lemma 3.92.. z. z.4 we cao compute the maps between thern yielding another complex which at the ith spot looks like We can use Theorem 3. z).. again using Theorem 3. Therefore if we replace A'n~' by AS"~'/ker(4)n_i). N) .z)) ç. To conclude this section.93. . ri ASi-l r ---+ .0. For the A-module N we form the usual complex which at the ith position looks like .. y.0.9s. A3: 812.z.3.94.. (x. Hom(AS. x . D EXAMPLE 3. we see that SYZ(S121 834.96)' Also.. 4>. 834. We continue Exemple 3.9.. (y. im(1i+1) is obtained using the columns of the matrix that determines T i + 1. ---+ 2 ASl fI ---+ AS' ----Jo M' ---+ ° 1 which wecompute as above. S34.. N) "" ker(Ti )/ im(Ti+1).3 we cao compute presentations of these Hom modules aod using Lemma 3. Therefore A3 "" (S12.8.: (C " A3 C2 .9.92... y). S56) and so wc have the following free resolution for M = (91. z. As in Lemma 3.+" N) ~ Hom(A". z])3. x. z). Compute a free resolution for the module M = ((x... Thus we can compute Exti(M.~" N) ---> . FREE RESOLUTIONS 199 is free.93.. we use the techniques developed 80 far ta outline the computation of Ext n (M. .10. y.S12+f2S34+C3SS6' ---> As we saw above. ExercÏses 3...1.8. (lQI[x. Since the leading term of and 856 involve different basis vectors. is (0. ---> Hom(As. (y. Also.9s...9..C3 ) ---> A6 C.96) = (f"!2'!3'!4) ç...--t ASi+l fi+l ---------+ ASi . (y.10. x . we get the desired resolution. .. (y.:!'-. x). where The kernel ofthis map is SYZ(91. We begin with a free resolution for ao A-module M • • • fi+2 -----.9 to compute ker(Ti ). We will assume that the reader is familiar with this concept and we will only give an indication on how ta go about the computation.10. z). N). the kernel of 4>.9.3.0). . The theory of Grobner bases over rings has complications that did not appear in the theory over fields. In Section 4. y] = (k[x])[y]). In Section 4.e. in Section 4. and a result of Zacharias [Za] which gives a method for computing a complete set of coset representatives of A modulo an ideal. We are now going to generalize the theory to the case where A = R[Xl1'" . for example. to discuss questions related to rings of quotients and use this material to give an algorithm to determine whether an ideal in A is a prime ideal.xn ] for a Noetherian commutative ring R.1 we give the basic definitions and lay the foundations for the theory of Gr6bner bases over rings. for a field k. where k is a field. a unique factorization domain (UPD). in Section 4. the theory of Grobner bases over rings will allow us ta use inductive techniques on the number of variables.'" .3 we give the usual applications including elimination.X n ]. Next. Moreover. Indeed. Z20 and Z[ Al.2. Moreover.4 where it lS used ta test whether Ideals are prime. Grobner Bases over Rings In the previous three chapters we considered the theory of Grobner bases in the ring A = k[Xl. An algorithm for constructing Grobner bases will be presented in Section 4. or a principal ideal domain (PID). y] can be viewed as a polynomial ring in one variable y over the ring k[x] (i.4. We then go on. many of the results will not hold in this generality. many of the basic techniques will become more complicated because we now have ta deal with Ideals of coefficients in the ring R. We will give many of the same type of applications we gave in the previous chapters in this more general context. We will have to assume certain computability conditions on R in order for this algorithm to be effective. We use a method presented by Moller [MoSS] to compute the appropriate syzygies needed for this algorithm. k[x. In the last section.5 we specialize the ring R to be a PID and show that in this case we may again use the notion of S-polynomials to compute Gr6bner bases. k[x.Chapter 4. we use Lazard's result to compute the primary 201 . computing syzygy modules. We then give examples of computing Grobner bases over the rings Z. We conclude that section by giving a structure theorem of Lazard [LazS5] for Grobner bases in polynomial rings in one variable over a PID. We give an example of this technique in Section 4. Sometimes we will need to be more specifie and require R ta be an Integral domain. 4. we will mimic the constructions of Chapter 1 as much as we cano We will assume that we have a term order < on the power products in the variables Xl) . leading coefficient and leading term of J. given three polynomials J.1. We assume that we have a term order < on the power products.. .5 and 3. we need R to be a PID.. The correct way to do this is to work with syzygies in the ring R.4. we write J = alXI + .X n ]....Xn . this becomes a very important issue.. for a field k.. 1 X s are power products in Xl. •. . that A is also a Noetherian ring.X n . For J E A. . in the variables Xl..Xn with Xl > X 2 > . We then have from the Hilbert Basis Theorem 1.9.5. . After we define this new concept of reduction. g. So to reiterate the setup. h in k[XI' . So we need to change our very definition of Grübner basis. we must modify our previous concept of reduction. In the case of ideals of polynomials with coefficients in a field. if and only if lp(g) divides a term X that appears .202 CHAPTER 4. IpU) = Xl' leU) = al and ItU) = alXI (called the leading power product. . As we did for modules in Section 3. w€ need the concept of reduction. we know that < is a well-ordering on 1rn (the point here is that this is a property of the power products 1r n . It turns out that in order to have a reasonable theory of reduction and Grobner bases using the same ideas of reduction as in the field case..5. . Basic Definitions. There is one major exception and that is in the definition of Grübner basis itself. and leading coefficient of a polynomial in A = R[xl. . see Section 1. even though the proof of Theorem 1. not of the polynomial ring klxl. Since we want to define reduction and Gr6bner bases in the context of rings more general than PID's. of the equivalent conditions for a Gr6bner basis over a field given in Theorem 1. respectively). the definition again involved the concept of dividing one leading term by another.. We now turn our attention to the concept of reduction. 1rn . Then frOID Theorem 1. . we have the lisuai concepts of leading power product. .3.xn ]). In those sections we required the notion of divisibility of leading terms. GROBNER BASES OVER RINGS decomposition of ideals in Bueh rings. that in the case where R = k is a field.. In this section we develop the theory of Grübner bases for polynomials with coefficients in a Noetherian Ting R. '. we assume that we are given a Noetherian commutative ring R and we let A = Rlxl. we say that J ~ h.1 will work for us in our new situation. + a. leading term. .. When the coefficients are not elements of a field. ... and also in the case of modules. with g i 0.6.. we will be able to pattern our results again on what we did in Chapter 1. > X.xn ].. We will explore this in Section 4.6 used the Hilbert Basis Theorem in klxl. .xn ].xn ]. Actually..as E R.5.. . Ji 0. .1. but not al!. 4. We define as before.With respect ta this term order. ai # 0 and Xl. Recall...X" where al. . . It turns out that many. there we were nat concerned about whether we were dividing leading terms or dividing leading power products since one was a non-zero element of the field k times the other and this had no effect on the divisibility. Next... .) ... we want to caueel the leading term of 1 using the leading terms of fJ. The key idea in resolving this diffieulty is to use a linear eombination of the the leading terms of the divisors. /J = 7x + 3.Is· We should be allowed ta use any li which has the property that lp(j.Xn ] with fJ. the theory requires. 12. providedthat lp(g) divides lp(j) and h = 1 . however) the results involving reduced polynomials are no longer valid and 80 reducing tenus that are not leading terms is unnecessary (see Exercise 4. of course. it is convenient ta ouly reduce the leading term of 1.··· . .} in A. We thus arrive at 1 ••• 1 DEFINITION 4.2. Sa with F = {/J. h = llx3 2y 2 + 1 and h = 3y . .6).. and we want ta divide 1 by fJ. We could build this into our definition just as we must require that lp(g) divide lp(j). EXAMPLE 4. we have cauceled out the leading term of 1 using the polynomials in the set F.1.. we could fephrase our definition of reduction over a field k by saying that 1 ~ h. We will use the deglex ordering with x < y. X. we say that 1 reduces ta h modulo F in one step..4. That is..(y/J . . . The first difference is that in the case of rings.).. .jg) and sa tbe leading term of 1 has been canceled).1. Also.hh}.::gjg (note that lt(j) = ltn:i.. sinee lp(f) = xy is not divisible by lp(h) = x8) So we have done what we said we wanted ta do. In the case of rings. are valid with this restricted concept of reduction. + csXsIs ) lor CI.) divides lp(j) aud sa what we desire is that lt(j) be a linear combination of these lt(j. . as... Sa we nQW assume that we are considering polynomials f and fI fs in A = R[XI.1. We see that this notion requires that we divide by lt(g) = lc(g) lp(g)... E R and power products XI. Given two polynomials 1 and h and a set of non-zero polynomials F = {/J.(CIXt!1 + . as it must..1. where lp(f) = Xi lp(/t) lor aU i Buch that ct # 0 and lt(f) = CjXIlt(/J) + . .). . lt(f.. excluding the oues making explicit use of reduced polynomials. # 0.1.5..2xh) and xy = lt(f) = Y lt(f. .. namely. With this in mind. The problem with this is that over a ring R we may not be able ta divide by the ring element lc(g). Let R = Z and let 1 = xy -1.1. whose leading power products divide lp(j).ltt) g. but this turns out ta be tao restrictive when we are attempting ta divide by more than one polynomial. denoted F 1 ----> h. + esX. The case where R is not a field differs in two ways.j..1 since h = 1 . we see that 1 ~ h. l. . 1 could not have been reduced using only one of the polynomials /J. h· . It is readily seen that all the results on Grübner bases in Chapter 1.2x lt(h)· (Here C2 = 0. il and only il h= 1 . BASIC DEFINITIONS 203 in 1 and h = 1 . ta eliminate lt(j). where h = -3y -lOx . . c. 1.) in Definition 4. . ta compute bl .) with Ci # 0 is actually used to help cancellt(f).1.am E R. We are réstricted to such J by the requirement that Ip(f) = Xi Ip(f. This equation can be solved if and only if le(f) E (Ie(!.3.. there is an algarithm ta determine whether a E (al.(clXdl + .2xh) = -3y .. .) 1 j E J) R. . . + ambm .1. Its purpose is to ensure that each CiXi It(fi) in the difference h = f .(blyft + b3x/3) = f .1.(yft .4.) divides Ip(f). We have Ip(f) = xy and so J = {l. This condition is one of two conditions R must satisfy in order ta compute the objects of interest ta us in this chapter.f. .l <:: j <:: s}. we could end up with a term c Ip(f) remaining in h.f. We Hst these two conditions in the following DEFINITION 4.(2ft + 312) = -2y.3}.1.2 in this context. !c(!. we must be careful about this. + c.) JEJ for b/s in R. if we only required Ip(f) = Ip(CiXdi) for all i such that Ci # 0. LEMMA 4. where for a subset C ç: R we denote by (C) R the ideal in the ring R generated by the elements of C. Once we have the b/s then we can reduce f: EXAMPLE 4. Thus we need ta solve !c(f) = 1 = bl 1e(fJ) + b3 Ie(h) = 7b l + 3b3· We choose the solution bl = 1.1. We first find the set J = {j IIp(!.. and 12 = y. We put Exanlple 4. With the notation of Definition 4. ft = 5x 2 + y.·· .204 CHAPTER 4. To see this consider the case where R = ZIO with deglex and x > y and let f = 3y.Then we must solve the equation (4. a=) R and if it is.lOx .13). Having been careful with the definition we have the following cruciallemma whose easy proof we leave ta the exercises (Exercise 4. For example.X..1.1.1.1 we have Ip(h) < Ip(f).} be a set of non-zero polynomials in A. which is readily done. al. . GROBNER BASES OVER RlNGS We draw attention ta the condition Ip(f) = Xi Ip(fi) for aIl i such that Ci # O. Let f E A and let F = {ft.5. . b3 = -2 and thus we reduce f as h = f . ...1) Ie(f) = I). bm E R such that a = al bl + . . We now examine how we would determine whether f is reducible modulo F.1.. We will say that linear equations are solvable in R provided that (i) Given a.. Because of the possibility of zero divisoIs in the ring R. It is thus clear that we must be able ta solve linear equations in the ring R in order ta be able ta reduce in A.. Then It(f) = 2lt(fd + 3It(h) and Ip(f) = Ip(2ft) = Ip(3h) = y whereas h = f . then f ... where R = Z and f = xyl. We continue Example 4.1. without the assumption that linear equatiolls are solvable in R.1.5 and F = {f" fz.} if r cannat be reduced. be polynomials in A.). we will otherwise be informaI about our assumptiolls on the ring. . Zm..fs}. l In the case of ideals and modules where the coefficients were in a field we required that every term in r could not be reduced. We say that f reduces ta h modulo F.. .1 ~ -lOx . Qlxl.. . we need to iterate our reduction process. DEFINITION 4.6.8.ht-l E F A such that We note that if f ~+ h.am E R.6. In the literature this latter concept is sometimes refered to as "reduced". The first condition in Definition 4.f. In particular. if and only if there exist polynomials h" . what we present in this chapter is valid.1. denoted f ----->+ h.4. However.5 is the one discussed above which we found necessary in order to make the reduction process computable. since f ~ -3y -lOx . h = llx3 _2 y 2 + 1. We will always assume that !inear equations are solvable in R when an algorithm is presented. there is an algorithm that computes a set of generators for the R-module Examples of such rings include Z. . We note that this reduction could not have been done in one step.. h. BASIC DEFINITIONS 205 (ii) Given a" .6. EXAMPLE 4.2.2. with JI. The first reduction is the one noted in the previous example and the second is obtained by simply adding h to -3y . of 0.lOx . .xn ]. ... as has been the case throughout this book. and let F = {JI..J.. A polynomial r is caUed minimal l with respect to a set of non-zero polynomial.. .J.1.. . . Let f..7. Zli] where i 2 = -l.1. . F = {f" . We see that f ~+ -lOx .1) to be presented in the next section for computing Grübner bases in A is actually implementable. and ZIR]. Here we are only requiring that lt(r) cannot be reduced modulo F.h E (f" . The second condition is needed to guarantee that the algorithm (Algorithm 4. and should be viewed as mathematical existence statements. and JI.1.f. .. As in the case of ideals and modules where the coefficients lie in a field.. . h}. DEFINITION 4.JI = 7x+3...1. .13 = 3y. but we adopt the word "minimal" so that we are consistent with the similar concepts in the rest of this book. lt(j. is minimal with respect ta a set oJ non-zero polynomials F = {h.) + ...X. r. GROBNER BASES OVER RINGS Wc recall the notation that for a subset W is denoted ç A. A polynomial r E A. 0.. Jd since only h has the property that lp(ji) divides lp(r) = x and kIr) = -10 't (k(h))z = (7)z...J. such that E A such that F f ---------++ r.) > Ip(r2) > . l:S. + c. if lt(r) E Lt(F).(c.10.. This process must end since the order on the power products is a well-ordering. lp(j) > lp(r. and set F = A. the leading term ideal of W E Lt(W) = ({lt(w) 1 w We have the following eaBy W}). either rI is minimal with respect ta F or rI ~ T2' Continuing in this way we get where we have.j. with r # {J" .) for sorne polynomiaIs hi E A..1 we have that lt(r) = c.. Ifr is not minimal.hS1 rare computable. -+c.!.!.hs h.1. lt(j. PROOF..J..s If linear equations are solvable in R. If we expand this equation out into its individual terms. . . We have THEOREM 4. Either J is minimal with respect to F or J .1. + . + r lp(j) =max((max lp(hi)lp(ji)). 0 We note that in Example 4. It immediately follows that lt(r) E Lt(F)..}.) for Ci E R and power products Xi.6 is minimal with respect ta F = {J" h. . that is. . 7.) is a reduction ofr. Moreover. by Lemma 4.tS..1.. . we see that the only power product that cau DCCUI with a non-zero coefficient is Ip(r).lp(r)).} ÇA iJ and only iflt(r) 't Lt(F).1..· ...1. PROOF.!:. h.. . We now have By the definition of reduction. . the polynomial r = -lOx . minimal with respect ta F. Let J.. LEMMA 4.. Similarly..X..206 CHAPTER 4. + .. Conversely.J... E A with h. + h. .X. then hlJ . and 80 we obtain the desired polynomial r. lt(j..lt(h) + . It is then clear that r .. . Then there is an r J with = E # 0.J.j. . + h. we may assume that each hi is a term CiXi. there are hI. then lt(r) = h.9.X. then r can be reduced and so from Definition 4.3. X2s It(f.1.) = C2lX2llt(fI) + . The first pass through the WHlLE loop was done in Example 4. ....) and Ip(f) = X. + c. + c. we are assuming that Ci = 0 for ail i sueh that Ip(fi) does not divide Ip(r).. + C"X It(f.....(c.1.) for ail i sueh that r := r .1..2 and gaveusr= f-(yfI + (-2x)h) = -3y-lOx-1.Xs sueh that It(r) = C. 7 and gave us .4. + C2. + . + C2sX2sfs for sorne C21.i Ip(!. for ail i ·such that Cli # o. . BASIC DEFINITIONS 207 for sorne CIl.... ..} INITIALIZATION: h. f. We reconsider Examples 4.. for all i sueh that C2i # O. + hsf.1. D f ..1.X.X1S1 where It(f) = CllXlllt(fI) + .. Ip(r)) and r is minimal with respect to {fI. X 2" where It(r. Assuming that !inear equations are solvable in R we can determine whether the Ci '8 exist -and compute them if they do. h 2 =0.. we get a representation of the desired type for f ..1.. := 0. We note that in obtaining It(r) = C.··· .T2 = C21X21h + .Xd. . r := f WHILE there is an i sueh that Ip(fi) divides Ip(r) and there are Cll .··....).1.1. .Xs It(fs) and Ci Ip(r) = Xi Ip(!. where f = hd.1. INPUT: f. . . .lt(fI) + . namely . Tl . ..Xs in Aigorithm 4..2 and 4. # 0 DO + .) = X 2i Ip(fi).1. E A with!.. + esXsf.C2s E R and sorne power produets X 2l .r.X.f. . =y.Cl s ER and sorne power products X I1 . This gives a representation of the desired type for f .r of the The method for eomputing r is given as Aigorithm 4. Division Algorithm over Rings EXAMPLE 4.fI.1. Since Ip(f) > Ip(r. andh3 = -2x.7 using Aigorithm 4..1. .) > Ip(rz).Cs E R and power products Xl.1.. .11.1. . := 0. h. h" r.r2.. . . + r with Ip(f) = max(maxl<:i<:s (Ip(h i ) Ip(fi)).) FOR i = 1 to s DO ALGORITHM 4.s Continuing in this way we eventually obtain a representation for desired type. # 0 (1 SiS s) OUTPUT: h". The last statement is clear. + .. . h.) and Ip(r. . The second pass through the WHlLE was done in Example 4. Similarly. xnl. we need not have lp(h i ) Ip(9i) = Ip(h i9. On the other hand. ht E A such that lp(f) = maxl<:i<:t(lp(hi) Ip(9i)). COROLLARY 4.. then f E I.. Thus. andh3 = -2x-1. PROOF. GROBNER BASES OVER RINGS r= (-3y-lOx-1)-(-h) = -lOx-6.10. We use lp(f) = maxl <i<t(lp(h Ip(h i9i).1. + ht9t such that lp(f) = maxl<i<t(lp(hi) Ip(9i))' It is then easily seen that lt(f) = l:lt(h i ) lt(fi) where the sum is over ail i satisfying lp(f) = lp(hi ) Ip(9i)' Thus lt(f) E Lt(G). Then 3yft -2xh = 3y-2x E I while It(3y . The WHILE loop stops. Then we know from Theorem 4.} be a set of non-zero polynomials in I. .10. IfG is a Grobner basis for the ideal I in A.12.1.9.92).lt(92)). Then the followin9 are equivalent. Let R = Z and A = Z[x.1. then I = (G).14.6y). 0 It is important ta notice the form of Statement (iii) in ..1. 92} is a Grübner basis for!'. it is easily seen that Lt(!') = (2x. We know that if f -S+ 0. as another example.12. (i)==? (ii). .6). whieh is a contradiction. Converselyassume that f E I.xy) = (2x. . (ii) For any polynomial f E Awe have f E I if and only if f -S+ O. that f -S+ r with r minimal. since over i ) Ip(9i)) instead of lp(f) = maxl <i<t -a ring R with zero divisors.9. But f Eland f . For f E I we need to show that lt(f) E Lt(G).rEl imply that rEl and so lt(r) E Lt(!) = Lt(G).. as desired. .). Let ft = 4x+1. We have that f = h l 91 + . and thus {ft. (iii) For all f E l. Then by simply looking at alllinear combinations of 91 and 92.. h = 6y+1 and I = (ft.. let 91 = 2x+ l. A set G of non-zero polynomials contained in A is called a Grübner basis provided that G is a Grobner basis for (G).1. f = h l 91 + . h 2 =0. and sa {91. lt(r) cl Lt(G).2x) = 3y cl (lt(ft). + ht9t for some polynomialB hl. DEFINITION 4. (ii)==? (iii).the Theorem. If r oF 0 then.3y.3y) = (lt(91). sinee only lp(ft) divides lp(r) but there is no Cl sueh that -lOx = lt(r) = ellt(ft) = cl(7x). . . We may nQW give the first characteristic properties of Grobner bases over A = R[Xl. f = yft + (-2x -l)h + (-lOx . A set G of non-zero polynomials contained in an ideal I i8 called a Grübner basis for I provided that G satisfies any one of the three equivalent conditions of Theorem 4. Let I be an ideal of A and let G = {91.1... by Lemma 4. THEOREM 4.1. This is the special case of r = 0 of Theorem 4.13. 92 = 3y+ 1 and set!' = (91. EXAMPLE 4. yi with the aeglex ordering with x < y. hl =y. (iii)==?(i). . (i) Lt(G) = Lt(!).15..1.. for a ring R.h). lt(h)) = (4x.208 CHAPTER 4. h} is not a Grobner basis for I. . This is immediate from part (iii) of Theorem 4. wheTe PROOF.1.. even in the case where F is a Grobner basis (see Exercise 4.' .X n ].1. Let 1 finite Grobner basis. Note that lex(f) E k[y].1.e. Then l has a We will develop a method for computing Grobner bases over rings in the next section.16. ç R[XI.5..gt} is a Grobner basis in k[y.. .ltx(gt)). We will denote the leading term of a polynomial a E k[y] by lty(a)..1. COROLLARY 4.10 is not necessarily unique.xn ].9.xn ] with respect ta an elimination arder with the x variables larger than y.gt). 0 We further note that the Noetherian property of the ring R..12 that for G a Grübner basis and f E (G). . . where Xi is a power product in the X variables alone and ai E k[y]. and leading coefficient of f in (k[yJ)[XI"" .1. X" . since G is a Grübner basis. the only possible remainder for f is 0 with respect ta G. we see that if r cf 0 then T cannat be minimal by Lemma 4. we see froID the next theorem that we can compute a Grobner basis over R using the theory presented in Chapter 1.5. the leading term.1. and hence of the ring R[XI. . lt x (f). . leading power product. THEOREM 4. and lcx(f) the leading term ideal of l.. That is. for a field k and a single variable y. Since f E (G) we have that r E (G) and sa.. let f E J. Let f E 1 = (g" . Xl. We write gi = aiXi+ lower x terms. Since Gis a Grübner basis in k[y.5). PROOF.9 and its proof) to write (4.2) . If G = {g" .18." ... 0 We note that the remainder r obtained in Theorem 4.. then r = O.1. This result will not be used until Section 4. One inclusion is trivial. and leading coefficient of f with respect ta the elimination arder <. sa ltx(gi) = aiXi.1.).6).Xn ] we can apply the Division Algorithm (see Theorem 1. here the arder is the one on the x variables alone which we will denote by <x). then G is a Grobner basis in (k[yJ) [Xl. We have. IfG is a GTobner basis and f E (G) and f r is minimal. tbe leading term.. lt(f)...1. in the special case where R = k[y]. However. .1. .Xn] be a non-zero ideal. BASIC DEFINITIONS 209 PROOF. and by Ltx(l).4. whose proof we leave ta the exercises (Exercise 4.. however. We need ta show that Ltx(l) = (ltx(g. leading power product. .. lp(f). . in Theorem 4. COROLLARY 4. and le(f) the leading term ideal of l.17. For the reverse inclusion. We will denote by Lt(l) .12. yields the following Corollary. -S+ T.lpx (f) .xn ] (i.1. Finally. the polynomials y(y + l)x and y 2x are both terms.18 is not true as the following example shows. in that ring..)Xi .1. for 1 S j S N and. j S.. Thus jo (4. looking at the first io terms and canceling TjXij = T1Xi1 we get yV. In 21x. Exercises 4.1. Show that 2x 3y ~ _x 3 . lpy(ai. But from Equation (4. yJ with respect to the lex order with x > y (or any other order for that matter). Show that x 3y2+5x 4 +X 3y ~+ _xy3_x2y_2xy2_y3_x2 _2xy_y2 In each case note that the remainder polynomial is minimal with respect to F.210 CHAPTER 4. y2X} is not a Gr6bner basis in klx.".1. .1. and F = {f" h h}.1. 0 The converse of Theorem 4.)Xi" (4. > X for al! other x power products. since the polynomial yx is in l but its leading term (itself) is divisible by neither lt(y(y + l)x) = y 2x nor It(y 2x) = y 2x.) jo oF O. X ip j=l provided that h = L. lpy(ai. Consider the ideal l = (y (y + l)x. Show that x 3y2 ~ x 4 + x 3y + x 2y2 _ xy3 _ xy2. However. Consider deglex with x > y. let J. smee Ip(a. {y(y + l)x. > yVjo lpy(aijo)' and so lpy(h) = yV.) > . and T . E k. h = lOxy . > YV'T2 Ipy(ai.2). > yVNTN Ipy(aiN )XiN" Since we are using an elimination order with y smallest. b. Then T1Xi1 = TjXij for 1 ::.1.yVjTjgij ) = yV'Tj Ipy(ai. the reductions above are just one possibility. c.x. jo.19.) > y"' lpy(ai.4) we see jo ltx(f) = I:ajTjlpy(aij)Xij = I:ajTJltx(gij)' j=l j=l as desired. a. GROBNER BASES OVER RlNGS o. Then {y (y + l)x. There are many ways one can reduce the above polynomials. X i . > "..3) yV'T. EXAMPLE 4.)Xi .2xy2 . y 2x) C. from Equation (4. h = 15y 2 + x.1.3).x 2y .4) ltx(f) = (~= ajyVj aij )T. lpy(ai.y .2xy. ajyVjaij oF O. Rix]' where R = klyJ.1. y].'. we must have Choose io least such that TjoXijo > Tjo+lXijo+l.1. = 6x 2 + Y + 1. X.j where T j is a power product in the x variables alone. appearing in the right side of Equation (4. y2X} is a Gr6bner basis in Rix]' since. . . .x.1. Consider lex with x > y.. and for any power product X E R[Xl. [Hint: Either set is a Gr6bner basis with respect ta deglex with y > x. we have a.10.. ) lp(j.. . 4.8. z] is a Gr6bner basis with respect ta deglex where x > y > z and is not a Gr6bner basis with respect ta lex where x > y > z.4. . + h 2h + h3 h + r where r is minimal with respect to F and lp(f) = max(lp(h . where r is minimal with respect to F and lp(f) = max(lp(h .1) on f = -3xy + 2xy2 2yz + 2z2 + 3x 2y'z' ta write f = hd.) . Prove Corollary 4. y. 3y2 + x') = (2x 2 . In ZlO[X.. then Xf --->+ Xg. ) lp(j. 4. BASIC DEFINITIONS 211 4..{g}. 4. h = 7yz-z-7y-2. z]...3. let h = 3x 2y+3x. then fg --->+ O. a.{r}) op J. and F = {f" 12.xn ] and let G be a Gr6bner basis for 1. let h = x'yz+yz+1. Let 1 " be a non-zero ideal in R[XI' . If f E F. try reducing 6x'y2 .g.g. Cali a generating set F for an ideal J ç R a minimal generating set provided that for ail r E F we have (F . Prove that every Gr6bner basis for l contains a minimal Grôbner basis for J.. 4.xn ] and 0 op h E R[XI.) .4.. Prove that {gl.1. .1.} is a Gr6bner basis if and only if {hg . Show that if G F F F 4.h + h 2 h + r. If f --->+ g.g E R[XI. In Z[x. . We say that G is a minimal Gr6bner basis provided that for ail 9 E G. .7.1. where R is an integral domain.. Also. lp(r )). hgt} is a Gr6bner basis. Let 1 be a non-zero ideal in R[Xl... . . . .9. y. b.. 12 = 7xy' +y. what would be the significance of the equality of ideals in Z[x. Show that the set F is not a Gr6bner basis. [One answer is r = -x + 1.1. 3y2 + 3x 2)? Alternatively think about this example for the idea of uniqueness of reduction.1. Consider deglex with x < y < z. try them both. Use the Division Algorithm (Algorithm 4. h}.1. Ip(h 3 ) lp(h).xn].1.1.1. b. 2Z2 . .X n ].. Let {g" . . 4.17.1) on f = x 3 y 3 + 5x 2y2 + x 2 y + 1 to write f = h.) b. .1. Ip(h 2 ) lp(h).] b. lp(r)).. y]. Show that the subset F = {3x 2 y . Show that the set F is not a Gr6bner basis. a. 9 is minimal with respect to the set G . . y].1. 4. 12 = 5yz'-xy+z. Prove that G n R generates 1 n R. for any finite set of non-zero polynomials F in R[Xl1'" ) x n ]. (There are two obvious ways to begin this exercise. x n ] and let G be a Gr6bner basis for 1.xn ]..2.x 4 ] Show that for any polynomials f. 4. a.'.6. Use the Division Algorithm (Algorithm 4. If you attempted ta mimic the definition of reduced Gr6bner basis of Chapter 1.5.x} ç Z[x.} ç R[XI.1. (2x'. and F = {f" h}. Ip(h 2 ) Ip(f2). 9).. and is the direct analogue of Theorem 3." . Given power products Xl.9) if and only if GnR is a minimal generating set for InR (see Exercise 4. . Let G c:: I be given.5.7. by Theorem 3.12 for R[XI..4. Syz(L) has a finite generating set of homogeneous syzygies.1.. and hence any submodule of it is Noetherian and finitely generated.1.2. Use the ideas in this section and those of Section 3.. . 4. We will then give the analogue of Buchberger's Aigorithm (Algorithm 1.. .212 CHAPTER 4.1. DEFINITION 4.xn]modules. With the notation above.) E Syz(L) c:: (R[XI. . a. We will conclude this section by giving an iterative algorithm for computing the syzygy module needed for this generalized Buchberger Algorithm (see Aigorithm 4.1. for a power product X. where R is an integral domain.. Prove that Gis also a Grübner basis for J with respect to the same order.. xnJ is Noetherian.. .1).J b.14. . is a minimal Grübner basis for I (see Exercise 4. Then.3. lt(h i ) = hi for aIl i) and Xi Ip(h i ) = X for aIl i such that hi # O.2. . Prove Lemma 4. Moreover.. In this section we will give another characterization of Grübner bases (Theorem4. Xn ].5 to obtain a theory of Grübner bases for R[xl.1. . xnJ-submodules of R[XI. ..2.Xs and non-zero elements Cl. Computing Grübner Bases over Rings. now our syzygy modules are submodules of (R[XI.. .1. 4.2..2. LEMMA 4. .. andlc(g) ct InR forallgEG-GnR.. . xnJm. We have the following easy lemma. Let K be the quotient field of R.2.2.2.10).1. Let G be a Grübner basis for a non-zero ideal I of R[XI"" .xnW. xnJ and let 71" denote the quotient map R[XI.12. 4. we cali a syzygy h = (h" .xn ])' homogeneous of degree X provided that each hi is a term (i.3) which is similar to the S-polynomial criterion in Theorem 1.13. write f = fo + ft where 71"(10) = 0 and 7I"(lc(j. Let I be an ideal in R[XI.11. by Theorem 1. Of course. .e. ••. .1. Let J be the ideal of K[XI"" .2).. xnJ . We note. 7I"(G-GnR) is a minimal Grübner basis for7l"(I). . In particular state and prove the analog of Theorem 4.1) for the case of rings (Algorithm 4.4 (R/(I n R)[XI.. then G n R is a minimal generating set for I n R. Show that G is a minimal Grübner basis for I (see Exercise 4. Show that if G is a Grübner basis for I then 71"( G) is a Grübner basis for 7I"(I). f ct In R. that R Noetherian implies R[XI. xnJ generated by I.'" ..1. .Cs in R set L = [ C 1 X 1 csXs ].1.h. (R[XI. xn])' is Noetherian. 4. GROBNER BASES OVER RlNGS 4.3.1.. [Hint: For all f E l.. .)) # O.1. .. . x n ]. 7.1.'" . XI'.2}.'" ...... 2.. . . Let X. + h. we see that for any power product X we may collect together ail the terms in this last SUffi whose power product is X and these must also add to zero. On the other hand. .lt(gt)).. Now we know that h. .s} we cali the saturation of J the set J' consisting of all j E J such that X j divides XJ.+ O...1). and non-zero elements CI. 2. . set X J = lcm(Xj 1 j E J).hs) E Syz(L) we may write h as a SUll of homogeneous syzygies. Then expanding the polynomials hi into their individual terms.2. s}.2. . if X j divides XJ. For any subset J ç {l. Thus it suffices to show that given any syzygy h = (h" .. We say that J is saturated with respect to X" . given power products Xl.2. Let G = {g" . In view of Equation (4. COMPUTING GRbBNER BASES OVER RlNGS 213 PROOF.. . sinee aU the CiXi are terms. .5. c. . + .csX.... This immediately gives the desired representation.2. gt} be a set of non-zero polynomials in A. . gt) if and only if for all (h" . + h. + . Let B be a homogeneous generating set for SYZ(lt(g. 3} we see that XJ = x 2 y and J is saturated since 4 is the only element of {l. + . . we have h..X. X s provided that for aU j E {l.. 3. We first must find a hornogeneous generating set for the syzygy module of the leading terms.. Thns we see how we will go about generalizing Buchberger's Algorithm. The next question we must answer is how do we go about constructing a homogeneons generating set for SYZ(lt(g. Then G is a Grobner basis for the ideal (g" .Cs ER how do we construct a homogeneous generating set for SyZ(C. Or in general. ..g ... ht) E B... . lt(gt)). then j E J. 4} not in J and X 4 = x 4 does not divide X J = x 2 y. and X 4 = x" Then if J = {1.4. .4. s}.. G One can view the expression h. D We can now give the following theorem whose proof parallels exactly the proof of the similar Theorem 3.2. we add the reduction to our set and repeat the procedure. if J = {l.. (Note that X J = XJ'. .g .2} we see that XJ = x 2 y and J is not saturated since 3 ct J.1) " the following and the surrounding discussion.X2 = X 2 . . If o~e of these does not reduce to zero. .).2.. since in that expression the leading terms cancel (this is the basic idea that was used in Section 3. . Clearly {l. For any subset J ç {J.X .c.3).) EXAMPLE 4.. + htgt ----.. = O.X3 = y.3.gt above as a generalized Spolynomial.).X. = xy. THEOREM 4. . we make DEFINITION 4. We then fonn the generalized S-polynomials and reduce each one of thern using the reduction presented in the previous section. 2. 3} is the saturation of {1. As noted above Syz(L) is finitely generated. while X 3 = y divides XJ = x 2 y..5 (Exercise 4..).1.. dj = L 1. . .X . let h = (h" .X . GROBNER BASES OVER RINGS We recall the notation for the standard basis vectors e.. Now. d. .'=1 vJ .0... since c. which is saturated with respect ta X" . .. e. JEJ J jEJ by the definition of bvJ .2. . e2 = (0. J I>jvJ ~J ej jEJ J L bjvJ ~J CjXj = XJ L bjvJcj = 0. . For each set J ç {1.Y. E R and power products Yi.) Then the set of vectors SvJ. Since.. that is..1.0). Set (Note that each of the vectors SvJ is in AS. .2.2.. where IJI denotes the cardinality of J).. (Note that each of the vectors bvJ is in the R-module RIJI. since h is homogeneous of degree Y Then. we have L. Y. It is first of all clear that each of the vectors SvJ is homogeneous of degree XJ.0. We have that YjX j = Y for ail j E J. . .X..6.s} and 1 :s: v :s: VJ.jOdjcj = O.0).. by bjvJ . . = (1. say of degree Y We write h = (d. . d. = (0. . Moreover. .. .) is generated by homogeneous syzygies. .. Thus (dj 1 jE J') is a syzygy of [ Cj 1 j E J' J and so by hypothesis (dj li E J') = L rubuJ" 1/=1 vJ . for J ranging over aU such saturated subsets of {l. Set J = {j 1 dj of O} and let J' be the saturation of J. .. .. .jEJdjYjCjXj = YL. . TvbjvJIJ .. forms a homogeneous generating set for the syzygy module SyZ(C. . for jE J. bvû } be a generating set for the R-module of syzygies SyzR(Cj 1 j E J)..X.... Given the above notation we are now prepared to state THEOREM 4. denote its jth coordinate. csXs ). SYZ(C.. For each such bvJ .. s}.1) for AS. " PROOF. . SvJ is a syzygy of [ C1 X 1 csXs J. c.0. . by Lemma 4.··· . ..).... X" let BJ = {bu.X. c.214 CHAPTER 4... hs) E SyZ(C.0. since h is a syzygy of [ C1 X 1 csXs ] .. Y" .) for d" . .X. for each j E J'. it " suffices to write h as a linear corobination of the svJ's in the case that h is a homogeneous syzygy. y).. as is readily checked. IJ R is a Noetherian ring and linear equations are solvable in R then Algorithm 4. (an easy task) and cornputing syzygies in R (see Definition 4. D We note that we have reduced the problem of computing a generating set for SyZ(CIX . For J = {1. . THEOREM . +6b4 = 0 and one finds a generating set for the solutions of this equation to be {(-3. 85y2. we need to solve 5b2 + 85b3 = 0 . for any set of polynomials {ft. a homogeneous generating set for Syz(lt(ft).-z.4}. 22.2. by Theorem 4. Now ·that we cau compute. -2.. s} which are saturated " with respect to X" . + {2. Finally for J = {l. .. (-40. .2. and (-5.2. 2. 0..::e. -17. and C4 X 4 = 6x 2z.1).. COMPUTING GROBNER BASES OVER RINGS 215 for sorne rv E R.3}. We will assume that the reader can solve the elernentary linear diophantine equations that occur in this example (i.0). the algorithrn for computing Grübner bases for ideals in A = R[XI.2.O) and -5xx~:el + 2 x: :2z e2 Syz(2xyz.3}.. that the reader can "solve linear equations in R = Z"). 3} gives the cliophantine equation 2b . {3}. {3}. namely.O).5). .7. 0.2.0. xz.0.2.4} we need to solve in R = Z the equation 2b . (-5. 1) ..2z. 0).3.. . 0)).1)}. - xx'": e4 . we obtain the syzygies -40"xYy:e.}. Xn]. Now for J = = (-5y.3}.O). = (-3x. Since X J = x 2 yz. as desired. .8. . (0. with XJ = xy2z. J. {2. x.1 produces a Grobner basis Jor the ideal (f"". 0. 4}.2.lt(f.X.0). 1.0) and (-3. {4} do not give rise to any non-zero syzygy. It now follows that VJI = ~)L rvbjvJ' )Yjej jEJ' v=l = L jEJ' djYjej =L j=l S djYjej = h. c. So we obtain that xx~2ze2 + x~~ze3 = (-40y. . -l. O. X. and so are not needed in our generating set. 4}. 6x 2z) = (( -3x. 0). ..x. This may be solved to yield two generators for the solutions.y). J. 5xy2.).. 4. 0.. = 2xyz.2. These will give syzygies that have already been obtained. .xz. One readily checks that the saturated subsets of {1. C2X2 = 5xy2.6.X.l. {1. We consider R = Z and let C. Since R =Z is an integral domain.) to computing the subsets of {l.e.4} are {l}.)).1)} and the syzygy -17~e2 + 'f}re3 = (O. {1.0.· .O. we get the generators (-40.-17. {4}. 3. we can give Algorithm 4. + 5b2 + 85b3 = O. C3X3 = 85y2. 2.4. the corresponding syzygy is -3X. 3. and {l. Then. the singletons {l}. EXAMPLE 4. (-5y. . The set J = {l.2. -l. which gives us {(-17.2.1. (-40y. Now YjXj = y for all jE J implies that X J = XJ' clivides Y. 2.. this method gives an efficient way ta avoid duplication in the computation of syzygies (see Algorithm 4.... . by Theorem 4. . .). By Lemma 4. G' := F 7' G DO for SYZ(lt(g.) E B DO Reduce h. gt Compute B. ..2. Thus.12. with r minimal with respect ta C' IF r 7' 0 THEN G':= G'U{r} ALGORITHM 4. a homogeneous generating set := 0. j. minimal with respect ta C. Grobner Basis Algorithm over Rings.xnl with ji 7' 0 (1 <:: i <:: s) OUTPUT: G={g" . .) INITIALIZATION: G WHILE G' G:=G' Let the elements of G be g" . At each stage of Algorithm 4. ta obtain a new set G'.2 ta compute these examples (see Examples 4. it clearly.2. has given a method for computing the syzygies that arise in Algorithm 4. j.2. We will incorporate this method for computiug syzygies iuto our Grübner basis Algorithm and so we will actually use Algorithm 4.. .lt(gt)) FOR each h = (h" . aGrübnerbasisfor (J" . PROOF. . The idea of Müller is ta use the computations of syzygies already done in arder ta compute the new syzygies.. ... G' + . the algorithm stops..).3. since the ring A is Noetherian.2. ... and 4.. .2.11. after we have giveu a more efficient method for computing the relevant syzygies. .1 we add one or more polynomials ta the generating set for the ideal (!J. In particular.gt}. gives a Grübner basis for the ideal (J" .13) Müller [MoSS]. D We will give examples of computing Grübner bases shortly. If the algorithm stops.l..). 4.g.. .. h.l.216 CHAPTER 4. .} ç A = R[Xl' . lt(r) rf Lt(G) and sa Lt(G) ç Lt(G').j...2. GRbBNER BASES OVER RINGS INPUT: F = {!J. + htgt ---->+ r. Sa we have ta compute a generating set for the syzygy module of this increased set of leading terms..2. j. .2.9.l.2).2.. As the algorithm progresses we add ta a set of polynomials G a polynomial or. . .. be a homogeneous syzygy in Sa of degree Y.) is a homogeneous syzygy in Sa-l.. (Cj 1 j E J. there are two possibilities. .j #cr)R: (Ca)R and so da = L. or ba # 0 ba = 0 and (b.. . If da = 0 it is clear that d is a linear combination of (a" 0). . (5} saturated with respect to X" .. X a with cr E J. . by induction.JJ denote a generating set for the ideal in R.4. We note that a homogeneous generating set of S. ba-lY 2 " and ba E (Cl. for d" . SA assume that da # o.baYa) in Sa. Let d = (d.X . Define for 1 -<: i -<: m..tJ f.Y ..da ER and power products Y l . . Either " a-.Y .. saturated with respect ta Xl)'" . .... and we define the homogeneous syzygy in Sa by Sj.Xa1 sueh that (j E J..J ç R the ideal quotient. 0) to be the homogeneous syzygy in Sa with the coordinates of ai in the first cr . The syzygies PROOF. .. ..L=1 2Recall that for two.. 1 Yu.ideals I.cr}... Let Sa = SyZ(C. NotethatXJ =XJ' dividesY. X s and non-zero elements Cl. . which we assume. consists of a generating set of the ideal {r E R 1 rCl = O} of R (called the anni~ hilatorof Cl and denoted by ann(cl)) viewed as a subset of A. .. . is defined ta be 1: J {r E RlrJ ç I}. We consider all subsets J of {l.O) together with the syzygies for J c:: {l. .J"X-ea' a XJ j#-<7 (The bj's may not be unique.. COMPUTING GROBNER BASES OVER RINGS 217 Note that any syzygy of a set of leading terms is automatically a syzygy of a larger set of leading terms (using zeros for the extra leading terms). ribiJ. Also.Ca-l)R: (Ca)R.bj JE] ~ XJ x.bp. 1 am for 50'-1.O). . . (a=.. has been computed. j # cr) R: (ca) R. We will compute a homoge~ " Ss = SYZ(C1X11 • • • 1 csXs ) by inductively constructing neous generating set of generating sets for the Sa. For each sueh J let b1J . (a=.Jca = 0. .~. Then it is easily checked that ~J Y XJ = d. cr}. ifwe take a homogeneous syzygy (b.. .2. Thenda E (Cj Ij E J.2.Cs E R.. . 0).9. Again consider power products Xl) . Let J' be the set of all j such that dj # 0 and let J be the saturation of J' inside {l.1 coordinates and 0 in the last coordinate. So we proceed as follows.) We also consider a homogeneous generating set al. We nQW can state s~J (al.. . 1: J. . . ..tl = L-. 'caXa ) for 1 -<: cr -<: s. (ai.LT/-LSj. ..Yl . .. THEOREM 4.daY a ).. . J ej + bp. N ow for each b~J there are bj E R such that 1 L JE] j#-o- bjcj + bp.. but any choice will do.··· .. form a homogeneous generating set for the syzygy module Sa.. x) in S3. Sinee the polynomials generated by the algoritbm are scattered throughout the text of the example. we need only consider the saturated subsets J ç {1.3. 3.-z.5). where 1 j. 3}. 0. we note that (5). . We now tum to S3 and note that the saturated subsets of {1.2. 0).3xh = 15x2 + 28y 2. 5xy2. but was included in Example 4. x. We close this section by giving sorne examples of Algorithm 4. 2. Therefore S3 = ((-5y. 2z. D is a linear EXAMPLE 4.218 CHAPTER 4. 0) is not in this list.4 z: (15)" = . 0).9 as Algorithm 4. we note that (2. 85y2. We leave the easy proof that it is correct to the exercises (Exercise 4.(0. For {1. Working with the first one. we have put boxes around them for easier reference. 0.2z. sinee 0 z: (3)z = {O} (we note that. So we obtain S4 = Syz(2xyz.-17.: (85)z = (l)z giving the syzygy (-17.1 To compute new syzygies.1 which makes use of Theorem 4. We continue with R = Z and use lex with x > y in Z[x. (am. The only saturated subset of {1. = Syz(2xyz) = (0). We note that (2)z: (5)z = (2)" giving the syzygy (-5.7.2.: (15)z = (l)z giving the syzygy (5.2.2. 2. is minimal with respect to Gand so we add it to G as 1Jo = 15x 2 + 28y2. Finally for {1. 0.. 2z)). 85)z: (6)" = (l)z and we may use the same syzygy as we did for {1. (-3x. 2} that need be considered is {1. The corresponding S-polynomiaI5j. We start with S. 2.-17.4xy2).11. 0). -z.5xy 2). 2.0) = z(0.: (6)z = (l)z giving the syzygy (-3. We finally tum our attention to S4 and note that the saturated subsets of {1.2.0) + 8( -5y. Thus S2 = (( -5y. 0.-3x) in Syz(3x 2 y. For the second set. 3. we never need ta consider singleton saturated sets). (0. 0). -y) in Syz(3x 2y. We now consider S2 = Syz(2xyz.2.9)..: (85)" = (1)" and we may use the same syzygy as before.2. It is not needed.4}. 3}. we compute (3). EXAMPLE 4.2. y) in S4. 0.3} and {l.4} and {1.2) of [2 5]. = 3x 2y + 7y 1 and 12 = 4xy2 . We compute (3)z: (4)" = (3)z which gives the syzygy (4y. The case CT = 1 in the algorithm does not generate new polynomials. xz. 2. -17. 0. since (-40y. We also compute the syzygies corresponding to {l.5x. 4} containing 4 are {l.0).. . 2. 6x 2z) = (( -5y.3} and {1. 2z. .x)). GROBNER BASES OVER RlNGS 80 is a homogeneous syzygy with a zero in its oth coordinate and combination of (a" 0).-17. 3} containing 3 are {2.. 4yj.10. yi.2.2.1) of [5 85] and so the syzygy (0. y)). 3} by computing 3. -yJo = -28y 3 +35y is mininlal with respect to G and so we add it to G as f4 = -28y 3 + 35y. giving the desired result.x.4}. The corresponding S-polynomial.. Let l = (f" 12). sinee R = Z is an integral domain.4} we compute (2).0.2} itself (actually {2} is saturated and contains 2 but will only yield the 0 syzygy). {1.5.2..2. 15x 2).xz. We note that the syzygy (-40y. We give the improvement of Algoritbm 4. For {1.7 using this method. Now consider the case CT = 2. 2. 3}.3}. 4xy2.2}.1) of [2 6] and so the syzygy (-3x. 3} containing 3 and they are {1. 4} we need to compute (2. The only non-singleton saturated subset containing 2 is {1. We will redo Example 4.2. As before the set {1. fa.. .+ r.1. i'J DO Compute bj E R. 7y... . . 7' 0 (1 ::. Grobner (l)z and we may use the same syzygy (5. J.x). xnl using MOlier's Technique ALGORITHM 4. 4}.4. (J}. For the first we compute (4)z: (-28)z = (l)z which gives the syzygy (0..3..0. 3.Ml for (lc(fi)lj E J.j:f=a bj le(fj) + biJ leUa) = XJ G 0 " XJ L.} ç Rlxl. .. The corresponding Spolynomial is 7Yh + XJ4 = -35xy + 35xy = O. COMPUTING GROBNER BASES OVER RINGS 219 INPUT: F = {f" .2. j E J. . m := s m DO Compute S = {subsets of {l. The saturated subsets of {l. xnl with Ji OUTPUT: Ga Gr6bner basisfor (f" . .2. P 0' J where r is minimal with respect to G IF r7' 0 THEN m:=m+l (J := (J + 1 Basis Algorithm in Rlxl.. . -y) as before.. 3. (J := 1. .j 7' (J)R: (le(fa))R 7' (J FOR i = 1..). We see now that {f" h.. 2. . j such that R.jj::.4} gives no new polynomial.2. 2.2. . .0.educe L jEJ.. ..a P + biJ~(f ) Ja ---.. . bj 1 (J') fi jEJ. 4} containing 4 are {2..9 for the definition of .4} and {l. J4} is a Gr6bner basis for (f" hl· This is a minimal Gr6bner basis (see Exercise 4... s) := F. i = 1. .lp(fa). . Js) INITIALIZATION: G WHILE (J ::. i ::. which contain (J} FOR each J E S DO XJ:= lem(lp(fi)lj E J) Compute generators biJ .. saturated with respect to lp(j. {l. -4yh = 5x ~ O. 4. {l. We first consider 0' = 1. 0). These are {4}.5}.0. Is}. 5} and {l.2. -x. 3} we see that (4.3.2. and {1. lt(h).2} we compute (4): (3) = (4) which gives rise ta the polynomial -3xj.{1. 5} do not give rise ta any new polynomials. The only saturated subset of {1} is {1} itself. This gives rise to the polynomial 5j. 3} we compute (3): (5) = (1) which gives rise to the polynomial 1512 .220 CHAPTER 4. We next compute the saturated subsets of {1.0. = 4xy + x.3}. -x.0. 0.). This polynomial cannot be reduced so we let h = 5x. EXAMPLE 4. ·)z. It(h). 2. For the set {4} we compute (0): (4) = (5) which gives rise to the polynomial 5/4 = 5y ~ O. 2. 2.5. (0. EXAMPLE 4. [ We use the lex term order with x > y.2. h. 2 which contain 2.).2. 0.3. {4. We now compute the saturated subsets of 1. 5} wc compute (4): (15) = (0): (15) = (4) which does not give rise ta a new polynomial. {1.3. For the set {2. 5}.3): (5) = (3): 5 = 1 and this gives rise to the same polynomial as the set {2. 0. 3. For the set {1.4. = 5x.5): (4). Wc let R = . since (4. 4} does not generate a new polynomial. 5} which contain 5. These are {2}. We consider the ideal J in Z20[X. 4y. GROBNER BASES OVER RlNGS minimal Grübner basis). ) for ( . 0). We will now give one more example where the coefficient ring does not have the property that every ideal is principal. (5. {2. We compute the saturated subsets of {1. (0. 3. (-3x. (0): (3) = (0). For the set {5} we compute (0): (15) = (4) which gives rise ta the polynomial 4/5 = O.0. For the set {1.3. (y. Sinee 3 is a unit in Z20. + 4yh = -3x 2 + 4y2 ~ 4y2 + y. (0. 3.0.5): (4) = (1) which gives rise to the polynomial yj. for the set {4. 0).5): (4) = (4. 0. -x)). 1 and [ 12 = 3x 2 + y. We compute (0): (4) = (5) (throughout this example we will use (. Therefore we let 1 ~s )= I~Y) 1 and add it ta G. (0.xh = 15y which cannot be redueed.. 14. .3} we compute (4): (5) = (4) which gives rise ta the polynomial 5j. 0.3. 0. 2.5}.0).2.. 15. and we add it to G.. Note then that {l. (0.4}. These are {5}. {1. 3. {1.2 to compute a Grübner basis for J. 2.3.0. i. 3. 3. 4.13. 0). The set {1.2.0. This polynomial cannot be reduced so we let 14 = 4y2 + y. 3}.{1. and we add it to G. -4y. 3}..5}. This will be used later.12.3 which contain 3. For the set {1. 3}. Note that we have also shawn that Syz(lt(j. These are {3}. 3.2. Finally. 4} which contaln 4.5}. 3.4}. 0. and {1.5} and {1. Thus the algorithm stops and a Grübner basis for J is {j" 12.e.0. 0. 0. 4. 2. 3y.4. {2} does not give rise ta a new polynomial. 2}. For the set {3} we compute (0): (5) = (4) which gives rise ta the polynomial 4h = O. 4. !t(f4). 5} wc compute (4. For the set {l. 4} we compute (4. 0).Xl4 = O.5): (15) = (1) which gives rise ta the polynomial 3yh -xfs = O. It(fs)) = ((5. We next compute the saturated subsets of {1. 2. We again follow Algorithm 4. 0).0. For the set {1.0.4).3. y] (where Z20 is the ring Z/20Z) generated by the polynomials 1 j. A. These are discussed in elementary number theory courses (see Niven. For an alternative method to do these computations. It is well-known (see. + ')'1 A and a = fJ+')'A.2).2}. . . v. 1.A)Yh = (1 .2. {3.5V1Î'1 .Œ.u')' = 0 for the integers {31.cg): (c) for ct. Sinee Z[A] is an integral domain we do not need to eonsider saturated sets which are singletons.. One can do this either by using sorne elementary theory of quadratic fields (see [Ad Go]) to find generators for these ideals.xy. + U1Î'l . According to Theorem 4.. .. one can simply equate the real and imaginary parts and obtain a pair of linear Diophantine Equations. .A are not units and cannot be factored in Z[AI. . . but that 2·3 = (1 + A)(1 . which is in the spirit of this book..2.vfJ .A)xy2 + 3Axy·1 Also. We use the lex ordering with . Indeed it is not hard to show that 2.A)(1 + A ) = 0. Of conrse. 'YI.A . (Cl.fJ. .lz[v'=5]). We follow 19orit We let 1 h = (1 .e E Z[. this method will give a generating set for (Cl. [AdGo]) that Z[AI is not a UFD and so not a PID.2. in order to compute a Grabner basis in such a setting we will need to be able to compute a homogeneous generating set for the syzygies of the leading terms of polynomials.l. 'Y are all integers. . We need to find all '" E z[AI such that 00 = CIal + . the second generator.1 + A.2.A). {3. . vEZ}. where Ul. for example. Many computer algebra systems have the facility to solve such equations.. + Cial for sorne al. We illustrate this more specifically for the case f = 1. Zuckerman. U. and this cannot be reduced any further using fr. We write Cl =U. we need ta solve the pair of equations U. We compute (2): (1 + A ) = (1. Sa in the desired equatioll. Then taking the real and imaginary parts of the equation Cl al .ufJ + 5v')' = 0 and V.(1.3..2.. V. In the computations below we will write down only the non-redundant generators but it is possible to verify that these generators are correct by the above procedure. 'Y. the first generator. of course.A y 1 .4.ca = 0.. . see Exercise 4. +vlA andc=u+vA and "'1 =fJ. y] generated by the two polynomials = (1 + A)x2 . This generating set is then also a generating set for (Cl.3. COMPUTING GROBNER BASES OVER RINGS 221 Z[A] = {u + vAlu.(1. . . 80 in this case we need ta solve CIal = co: for ŒI and Œ. We will follow Algorithm 4.. VI. {311 /1.2. C" cE Z[A] (in this example.. 80 the first saturated set we must consider is {1.Cf) means (Cl.3.c...Cf): (c) as an Z[A]-module. h· fr = 2xy + A y and h x > y. and 1 ..A)xy2 + 3Axy. gives rise to the polynomiaJ (l+A)xfr -2yh We let 1 = 2xy2+A(l+A)xy ~ A(l+A)xy-Ay2 2 f4 = A(1 + A)xy . Since 3·2 . .cg): (c) as a Z-module.. and Montgomery [NZM]). To do this we need to compute ideals of the form (Cl. Let us consider the ideal l in (Z[A])[x.J=5]. . but sorne of the generators might be redundant. gives rise to the polynomial 3xfr . or one may proceed as follows.fJ. ..2 to compute a rabner basis for I. Each of the c's and o's are of the farm u + vyC5 for integers u.. They are {1. 0. For the set {1.R = 3Axy .A ) = (1). O. We let 1Ir = -2Ry2 + 5(1 + A)y. 5} we cau use the same syzygy as we did for {1.15y. 2.4.A)j. They are {1.5} we compute (2. -2Ay2 + 5(1 + A)y. -5 + A). These give rise ta the following polynomials -3yj. 4} we compute (2): (-5+A) = (-1 + A. 4.4} we cau use the same syzygy as we did for {1.. It is easy ta see that for the saturated set {1.1.h .{1. (5 + A)y2 . 4} aud {1. GROBNER BASES OVER RINGS We now consider the saturated subsets of {1. + (-1 + A)/4 (5 . 2.A)y. 0. 0.5}.2h 3A(1 + A)xy . 3} aud {1.A. 2.4. 4} we compute (2. 4. 3} which contain 3. They are {1. y3 + 2Ay2 ·1 We note that 18 cannat be reduced because 3 A rte (2.. 3. 2. 2).3. This gives rise ta the polynomial 3yj. We now consider the saturated subsets of {1.. + 2/4 = = (5 + A)y2 . -5 + . This gives rise ta the polynomial aud we let 16 = -3Rx2y + xy3 + Axy2.15y. 2.4. -2). 4}. These give rise ta the polynomials 3Aj. For {1.3.-5+ A): (5 + A ) = (1).1. The first generator gives rise ta the syzygy (-3y. {1.222 CHAPTER 4. (1 + A)) and the second ta the syzygy ((1. This gives rise ta the polynomial 2yj.5. .1 + A. 3.4}.4.3. 6} and {1. 4} which contain .2. We let 110 = (5 + R)y2 -:. + (1 + A)h (1 . + h + yl4 aud we let 118 = 3Rxy .. 3.. 5}.4}.A)yj.A)y2 J'. O. -6Axy + A ( l . For {1.2. 6} which contain 6. It is easy ta see that for the saturated set {1.. 2. They are {1.4.1 + A): (1 . 2. 3. 3}. 3} we compute (2. 3} we compute (2): (1. 5} which contain 5. 4.A): (-5+ A ) = (1).15y Jo. 6}. 0. 2. 4} we compute (2.3. 3. 4. 2. We now consider the saturated subsets of {1. .. 2. This gives rise ta the polynomial 3xj.3Ay2 J!. 3. 4.1 For the saturated set {1. 6} we compute (2.1 + A): (-5 + A ) = (1). We now consider the saturated subsets of 1.2.1 For the saturated set {1. 4. 5.. 2. 3. 3. .5} aud {1. For the saturated set {1. 4}.Xl5 = 3(5 .A y3 + 2Ay2. 2.3.yfz + Xl4 = (1 . For {1.A)xy + 3Ay2 J!.A) = (1 + A. 2).3.A)xy2 + 3Axy Jo... 6. It is easy to see that for the saturated set {l. 4. 7. 5. 8} which contain 8.u A ) : A(-2) = (-uA): (2) = (3. 4. 4. 3Ayh + (1 2xy 3 + 2Axy2 . We obtain the polynomial AyiJ + xh 15~f. 5. l . They are {l. 7} and {l.1. -5 + A. 3.4. 4. 6. 5. For {l. 3. For {l.1 + A). 5. 2. These generators give rise to the polynomials (1 + A)!5 +3h = 0 and 2!5 + (1. 5. 2. 8}. 2. 4.. 6.7}. -5 + A): (3A) = (2.l5y (5 + A)y2 . These generators gives J. 8} we compute (2. They are {5. 4. 4.l5Ay ~ o. 7} we can use the same syzygy as we did for {l. 8} we compute (2. -5 + A. 5. 4.A y3 -15xy . 6. 8} we compute (2.A. -5 + A).A)y2 . For {l. 5. 7}. For {l. 4. 1 + A): (3A) = (2. 3. 3. 4. {l.2.l5y 0. 3.1 + A. We now consider the saturated subsets of {l. 7.l5xy 2Axy2 .A)h = O. 5(1 + A)xy . 2. We now consider the saturated subsets of {l.'. 6. 3. 7}. 4. 4. COMPUTING GROBNER BASES OVER RlNGS 223 A): (3A) = (2. 3. and so we may use the same syzygy as in the previous case. -5+A. For {5. 4. 2. -2A) :(3A) = (2. 4. J'. -5 + A). 2. -5(1 -Y!!!2jY/7 + A)y3 + 5(5 + 2A)y2 O. 8}. l . 3. 1 + A) and so we can use the same syzygies as in the previous case. 8}. 5+A) :(3A) = (2.1 + A): (3A) rise to the polynomials = (2. 3. 6. 2.7} we compute (5 + A): (-2A) = A ( . We obtain the polynomials 2 Ay3 .. {l.A . 7. 7} which contain 7. -3A): (3A) = (1) . 6. 8} and {l. 7} we compute (2.1A).. (1 + A )xy 3 - + A)!6 y' ~fs (5 + 2A)xy2 (-5 + A)xy2 _ A O. 2. -5+A. 3.l5xy . {l. 7. 6} we compute (2. 5. 5+A. 4.A y3 + 5y2 -(5 + A)y3 + l5y2 0. 5 + A): (-2A) = (1). 5. 3.4Ay2 . y3 ~ For {l. 5.5y2 -5(1 . H A . 5.A . 2. 19 is also a Grübner basis for l Exercise 4.0. 9z).} be a set of non-zero polynomials in A.3..5. 3. 7. such that and the set J has been used.such that for each J E S there is a jv E J. [An answer: {( -5z. (-3z 2. 0. 5x 2z. -7xy.2..2..5. x 2. if we consider the case where R = k is a field then we may improve the algorithm as follows: For each 0.2.) Ip(gd. and so tbis polynomiàl is 8h + 3/4 - 18 . with 1 ~ jv < 0.0).) if and only if for all (h" . Show that in Müller's method of computing syzygies of terms (i. 14 and fs form an ideal which is equal to (1) = 2[A]. 0). . 4. Then in the remainder of the WHILE loop we need only compute the reduetions of SUj" .(3z 2. Let B be a homogeneous generating set of Syz(lt(g.(0.)). 0.0. 6. We note that the leading coefficients of h.)).2.g.2. lu) for 1 ~ v < r. 0. 8·2 + 3(-5 + A ) + (-1)3A = 1. h. We now have that the polynomials h.0).3.. 12. then the set J' may be ignored. we have h.3. 0. . 7xy2Z)... It is easy to see that for the saturated set {1..) E B. + . h.0).2). Compute generators for the following syzygy modules. .xy + A y 3 . be shown that ExercÎses 4. 5.J . 2.) Ip(g..(0. .3.h... Compare this result to the use of crit2 in Example 3.0. _y2. [Hint: See the proof of Theorem 3. 16. (-yz. so we can add to the Grübner basis a polynomial whose leading power product is xy and whose leading coefficient is 1.J 4. Indeed.2.) = max(lp( v..e. in Theorem 4. 9z)}. 13'/4'/5. Show that in Algorithm 4. 0.5. (0.in the main WHILE loop find the minimal number of distinct j" . 4. + h.It(g.5. [An answer: {(0. 4. 7.Ip( v.5Ay2 + SAy = 19.9 and Algorithm 4.2. . + .2. y2.224 CHAPTER 4.je.0. Prove Theorem 4.2. 18 form a Grübner basis for I. 3y.3.2. 3. For R = 2 compute Syz(3x'y. 9yz2. 7xy2z). 4.A )xy3 + 3Axy2 A o.o-} with 0.. + .E J. .x2. Use this method to redo Example 3. Then G is a Grübner basis for the ideal (g" . 4. . if J ç J' ç {1.y. .0)..g. Prove the following: Let G = {g" .J b.g. 8} we can use the same syzygy as we did for {1.(5z. 0). GROBNER BASES OVER RINGS We obtain the polynomial 16 + xls (1 .g. .g.1. (-4yz. 9yz'.. x)... + h...g. 8}. 0. For R = 2'5 compute Syz(2x 2y.). 5x 2 z. where Ip( h.2. 7xy. = v..4.g. . 5.5. a. .2. + v. x)}.g. It can then {h'/5.10). 15. .1 how to solve the ideal membership problem and we will give an example of this. Then by Theorem 4.i)y2.. Recall that a Grübner basis for the ideal generated by fI = 3x'y + 7y and h = 4xy' . Applications of Grobner Bases over Rings.. We are interested in applications similar to the aues in the previous tWQ chapters. xn]~submodules of (R[XI.11. For R = Z[i] (where i 2 = -1) compute Syz(3ix 2y.. fg} forms a Grübner basis as asserted at the end of Example 4. 0. . Let J be a non~zero ideal of A. c. b. 12. Also. xn]. 4. (0. x).x 2 and lex with x < y. fz = 3y . fz = lOx 2y + 2xy and lex with x > y. 4. 4.. 5yz)). g. [An answer: {( -(2 + i)z. Sa let R be a Noetherian ring in which linear equations are solvable (see Definition 4.} be a Grübner basis for J.8.3.14).11. -7xy. .2.5) and let A = R[XI. if 1 E J.2. For the ring R = Z[i] (where i 2 = -1) compute a Grübner basis for the ideal generated by the polY!1omials fI = 3ix 2y + (1 + i)z and fz = (2 + i)x 2 z + y with respect ta deglex with x > y > z.] For the ring R = Z compute a Grübner basis for the ideals generated by the given polynomials with respect to the given term order.. 3iy. . For the ring R = Z15 compute a Grübner basis for the ideal generated by the polynomials fI = 2x2y + 3z and fz = 5x 2Z + y with respect to deglex with x > y > z. .g..5yz2. (2 . APPLICATIONS OF GROBNER BASES OVER illNGS 225 4.10.) = TU"". .2.9. .1. -(2 .1. xn])m (see Exercise 4. explicitly. a..x.1. Generalize the results in this section to the computation of Grübner bases for R[XI. 7xy2z). (4iyz. .6. We will close this section by showing how ta compute a generating set for the syzygy module of an arbitrary set of polynomials.. (0. We go back to Example 4. Thus we can determine whether or not 1 E J (ideal membership problem). Praye that Algorithm 4. x 2. 4. 0.1.. . Wc can then substitute in for the g.'s to obtain 1 as a linear combination of the l/s.). y] with respect to the lex ordering with x > y is {fI.1. . 4. (2 + i)x 2z. provided 1 E J. 14} where fa = 15x2 + 28y2 . + . we can obtain 1 = h. c. fI = 3x2y .}.1.5x in Z[x.2. Moreover as we saw in Algorithm 4. . .3.. We have basically seen in Section 4. l.1.2 terminates and has the desired output.7. . Show that {fz. set F = {f" . ideal quotients and ideals of relations.2.2. 0).12 we know that 1 E J if and only if 1 -S+ O. we can find a matrix T such that (g" .8z 2 and deglex with x > Y > z. and let G = {gI.. fI = 6x 2 + y2.13.2. We will then explore the use of elimination in this context to compute ideal intersection.2.3.g . These applications are very mueh the same as before and do not require any sedous modifications in their statements or proofs.. 0)..g.2.i)yz.). This requires more effort than it did in the case of fields because now one has to take into acconnt the ideals in the ring R. using Algorithm 4. Suppose that J = (fI.3yz + y..2.j. h. . 4.2. fI = 2xy . + h.. We will then show how to compute a complete set of coset representatives modulo an ideal. 0. EXAMPLE 4. h. fz = 5x 2Z .4. In arder ta do this we must assume that given any ideal 'S of R. an element c E C such that a =' c (mod 'S). Such rings include the integers Z. First we verify that f E (ft.lt(g.3)h We nQW cOllsider the problem of determining a complete set of coset representatives for Aj I.[~ 4y [h f4 _4y2 + 5 T . then IJ = {O}). . .g..56y4 + 70y2 -:'0..2. .)} of leading tenns of G. GROBNER BASES OVER RlNGS - and f4 = -28y 3 + 35y. With respect to the set {lt(gl). let J x = {i IIp(gi) divides X}.4. .12 xy2 + 5xy + 15x . We let f = 2x2y2 .56y4 + 14y 2 -:!c!.226 CHAPTER 4. If this is the case we say that R has effective coset representatives. -3x· 3xy ~] [ l· ff 1 2 Thus f = (2y .' -56y4 + 70y2 ~ o. We adapt the method given in Zacharias IZa]. . We consider a Grübner basis G = {gl.y .. the finite rings ZjnZ. In Example 4. . the polynomial rings over fields.' -12 xy2 + 15x .3..' -4xy3 . 12).t} as defined in Definition 4.1)ft + (-x .' f-(-7xh+2y 2h) -3x 2y . However here we also consider J = 0 to be saturated.. where I is a non-zero ideal of A. and of course J x could be equal to 0 for sorne X and that is why we had to add 0 in our list of saturated subsets).} for the ideal I.7y. and ZIA].(7x + y + 3)12 + (2 y2 - 2)h + 2yf<. ZIyC5I.3 we will need CJx for aU power products X.30x 2 . .4xy 3 - - 12xy2 + 5xy + 15x .ft .3x 2y + 5x 2 .56y4 + 70y2 -:!c!. Also.. we can determine a complete set C of coset representatives of RICZS and that we have a procedure ta find. It is clear that J x is saturated for ail power products X (in the proof of Theorem 4. for ail a E R. Thus we have f (-7xh+ 2y 2h)-ft-2h-Yh-3h+2yf4 .56y4 + 14y 2 - 7y 12xy2 + 5xy + 15x . we let I J denote the ideal of R generated by {ic(gi) 1 i E J} (if J = 0.2. for each power product X. We assume that 0 E C J . ..4xy3 -30x 2 .4xy 3 15x + 14y2 . we consider the saturated subsets J ç {1. We have 12xy 2 + 5xy + f ~ -:2:1. Let C J denote a complete set of coset representatives for RjIJ..11 we saw that j~ ]. Then for each saturated subset J ç {1.t}. . 3. we have rI . .. satisfies Ip(r.c = I:iEJ Ci !e(gi) (this can be done effectively by our assumption that linear equations are solvable in R).3. (We note that this part of the proof does not depend on the fact that G is a Gr6bner basis for I. So suppose that Tl. We now have that = . If f = 0 then the result is cleaL Thus we assume the result for ail polynomials whose leading power product is less than Ip(f). as above.r2 E l and. f = ri (mod I) and f = r2 (mod I). . there is a polynomial ri which is totally reduced. .) iEJ P 9t = clp(f) + ri (mod I). c E R may very well be 0)..t}. We will first show the existence of a normal form for f. Then every f E A has a unique normal form.ri " Ip(f) +' L. T2 are totally reduced. The normal form can be computed effectively provided linear equations are solvable in Rand R has effective coset representatives. PROOF.3. We will use induction on Ip(f) to obtain a totally reduced polynomial r with f = r (mod I) satisfying Ip( r) S Ip(f).) S Ip(f{) < Ip(f). Finally. a complete sets of coset representatives CJ for the idealIJ . then c E C Jx' For a given polynomial f E A. Assume that for each saturated subset J c:: {l. in particular. We next show the uniqueness of normal forms.+ (clp(f) + r.2. Then we have ri .r2 ~+ 0.~P(f) gi· iEJ P(gi) Write fI = clp(f) + ff· Then we have that Ip(f{) < Ip(f). Let G be a Grobner basis for the non-zero ideal l of A. where we assume that f oF O.)g. To simplify the notation we set J = J lpU )' Wc may choose c E CJ such that le(j) c (mod I J ). Noting that Ip(r. We call a polynomial r E A totally reduced provided that for every power product X. f = fi . but also on the choices of the sets of coset repre-sentatives CJ for the set of saturated subsets J. since G is a Gr6bner basis for l. if eX is the corresponding term of r (here.) S Ip(j{) and has the pro perty that ff ri (mod I). of course.) Then we may write !e(f) . Iph) < Ip(f) implies that r = clp(f) + ri is totally reduced and 80 r is a normal farm for i.4. by induction. APPLICATIONS OF GROBNER BASES OVER RINGS 227 DEFINITION 4. we have chosen. The proof will be constructive. We consider = fI = f - LC. and the induction is complete. Thus..) The proof will be similar in nature to the reduction algorithm. The complete set of coset representatives is given by the following THEOREM 4. Let r = clp(f) + ri. Ci~( . we call a polynomial r E A a normal form for f provided that f r (mod I) and r is totalZy reduced. (Note that c = 0 if and only if f is reducible. = It should be emphasized that the definition of normal form depends not only on the set of polynomials G. we sec that Ip(r) S Ip(f).3. 12.5AyZ + 8Ay. We choose C Jx = CJy = {O. Let d" d 2 be. = 2 20 .. . First note that if X is any power product not 1.3. We go back to Example 4.. we have 1 ~ 1-(3xyh-3xfI)=3x z +2xy+13y-5 J'-. Recal! that a Grabner basis for the ideal I = (4xy + X. We wish to describe a complete set of caset representatives for 2 zo[x. Theorem 4. and 19 = xy + A y3 .13. We wish to describe a complete set of . Thus we obtain Y 1 x 1 x x 3y y x y x y So.1. dl. h = 5x. h 14. GROBNER BASES OVER RINGS if r.dJ! y2 + 14 + yl5 = x 1 x y y. 15 = (5+A)y2 -15y. Thus if X = lp(r. Recall that a Griibner basis for the ideal I = (2xy + Ay. 1. we have xy ~ xy-yh + fI 3y (mod 1).2. CJx = {O}.2. where fI = 4xy + x. . 4}. EXAMPLE 4.1. (2 + 4x + 2y)(3 + 4x + 4y) = 6 + 14y + 16x 2 + 8y z + 4xy == 6 + 14y + 8y + 8y + 4x == 6 + 4x + lOy == 6 + 4x (mod 1). c f/. x. x 2 -7h = -7y == i:!. Namely. 15 = 15y.3.3. EXAMPLE 4.3.228 CHAPTER 4.5. the coefficients of X in TIl T2.3. . 1. we have that r. for example. (1 + A)x2 .5 ~ 12y-2x-5 2y + 3x + 15 (mod 1). as desired. respectively. Thus. 15. and I Jx = h y = (5). D We note that in the case where R = k is a field. y] with respect to the lex ordering with x > y was computed to be {fI. We now choose a complete set GJx of caset representatives for Z2o/IJx for each power product X. Therefore a complete set of coset representatives for 2 20 [x. We fol!ow Theorem 4. x 2 J'-.3 asserts that a k-basis for k[XI. .12.2 Ay2+(5+5A)y. and C J.dz i' 0 we have that d.r2) we have that c E h x .5. We also have h. and dz represent distinct cosets mod IJx and so their difference cannot represent the 0 cosetj Le. Thus c = 0 and r. 14 = 4y2 + y. Then since Tl and r2 are totally reduced. y]/ Iis the set {a+bx+cy 1 a E 2 20 . b. fg}. . and y2 As we did in Example 2. = r2.XV) in (2[yC5]) [x. h = .2. cE {O.rz is reducible.3 and we use the notation given there. for example if 1 = 3x 2y + 2xy + 13y . where 12 = (1 + A)x2 -xv.4. then I Jx = 2 zo . or y. 3x 2 + y) in 2 zo [x.xn]/I consists of all power products X such that J x = 0. 12 = 3x 2 + y.3. = {O}.2. i' r2. This is precisely the statement of Proposition 2. h.6.l J x' This Îs a contr. y]/ I.rz) and c = lc(rl . 220[X. 4}}. We go back to Example 4.adiction. Clearly for those ideals I Jx equal to 2 zo . 15}.8 we can also construct a multiplication table for = X.yJ/I. d2 E C Jx ' Thus sinee c = d. y] with respect to the lex ordering with x > y was computed to be {J2. 2xy + 12y . let G be a Grobner basis for l with respect to an elimination order on A[Yj.5a2 (mod 5 + A ) .Ym].Ym] with the y variables larger than the x variables. We now turn to applications of the method of elimination (see Sections 2. = {D}.1 the variables YI. Note tirst that for every power product X not equal to l. We follow Theorem 4.v = {D. 4.2.3. The proofs are very similar to the ones in those two sections.4. . We choose C J . and JJ.5+ A) for 1-' > 1. THEOREM 4.. So if D of f E InA then lt(J) E LtA[y](I) = LtA[y](G) and so lt(J) is a linear combination of elements lt(g) such that 9 E G with coefficients in A[Yj.Yj.Ym] = R[xj.6 and a generalization of Exercise 4.Ym] respectively. v.l-'::> l. 1.. hv = (1 + A ) for v > l. .. . or Y~.Ym be new variables.··· . since we are using an elimination order with the y variables larger than the x variables.. and e~ E {D. . 3. 5..·. each term in such a 9 involves only the x variables. . 6. we wish to compute generators (and a Grübner basis) for the ideal J n A.YI!J is n m {a + bx + cy + Ldvx + L 1/=2 J1=2 v e~y~ 1 a.4. 7." = {D.4.10 a. . We wish ta "eliminatel. yI!J. 3. . 2. and 10 E JJ. We will use the notation LtA and LtA[y[ for the leading tenn ideals in A and A[Yj. 5} for v ::> 2 (note that 6 E (1 + A ) and that aj + a2A ~ aj .3 and 3. Then G nAis a Grobner basis for InA.3A and 3.3. 8.4. Let YI. With the notation set above. The next result is the analog of Theorems 2.e. First. APPLICATIONS OF GROBNER BASES OVER RINGS 229 coset representatives for (Z[Al)[x.l..3 and we use the notation given there.1. 9} }. 0 . 1 Yrn. 7. and consider a non-zero ideal J ç A[Yj. as desired. l. Hence for cach 9 E G appcaring in the snm for lt(J).3)..xv. Since lt(J) is a term involving only the x variables.6).v = (5 + A . = h. 2. We now choose a complete set C Jx of coset representatives for Z[Al/JJx' for all power products X. We also have JJ.6.Ym]. lt(g) involves only the x variables and thus.. we have JJx = ZIA]. . We choose CJ.3. . 6.Xn. . 9} for 1-' ::> 2 (as above we have aj + a2A ~ aj . 8. .6. = JJ.c E Z[H]. Le.l.. i.v)' Therefore a complete set of coset representatives for (Z[ADJx. 5.b. 5}. .. 80 most of them will be omitted and left to the exercises. .. Clearly. 9 E G n A.. -2A) = (10. 2. we may assume each summand is a term and involves only the x variables. 3. we choose an elimination order on the power products of A[Yj..3.a2 (mod 1 + A)).Ym] with the y variables larger than the x variables (see Section 2. d v E {D..· . PROOF. for those hx's equal to Z[Al we have CJx = {D}. Thus lt(j) E LtA(GnA). Then we need to show that LtA (G n A) = LtA(I nA). 230 CHAPTER 4. GROBNER BASES OVER RINGS EXAMPLE 4.3.7. We go back to Example 4.2.12. Recall that a Gr6bner basis for the ideal l = (4xy + X, 3x' + y) in Z20[X, y] with respect to the lex ordering with x> y was computed to be {4xy+x, 3x 2+y, 5x, 4y2+ y , 15y}. Using Theorem 4.3.6 and Exercise 4.1.10 a, we see that In Z20[y] = (4y2 + y, 15y), and that InZ20 = {O}. EXAMPLE 4.3.8. We go back to Example 4.2.13. Recall that a Gr6bner basis for the ideal l = (2xy+,/=5y, (1+,/=5)x 2 -XV) in (Z[,/=5]) [x, y] with respect to the lex ordering with :r > y was computed to be {2Xy + Ny, (1 + N)x 2 - xv, (1 - N)xy2 + 3Nxy, + Nxy2, N ( l + N)xy - N -2Ny2 y2, (5 + N)y2 - 15y, -3Nx2y + xy3 N y3 + 5(1 + N)y, 3Nxy - + 2 Ny2}. + 5(1 + N)y) Using Theorem 4.3.6 and Exercise 4.1.10 a, we see that In (Z[N]) [y] = ((5 + N)y2 - 15y, -2Ny2 and In Z[N] = {O}. Our first application will be to compute the intersection of two ideals of A and the ideal quotient of two ideals of A. First, as in the ideal case over fields, Proposition 2.3.5, and as in the module case, Proposition 3.6.8, we have PROPOSITION 4.3.9. Let l = (J" ... ,j,) and J and let w be a new variable. Consider the ideal L = (wj" ... ,wj" (1- W)g" ... ,(1 - w)g,) = (g" ... ,g,) be ideals oj A ç A[w] = R[XI, ... ,Xn, W]. Then In J = L n A. As a consequence of this result we obtain a method for computing generators for the ideal l n J ç A: we first compute a Gr6bner basis G for the ideal L in Proposition 4.3.9 with respect ta an elimination order on the power products in A[w] with w larger than the x variables; a Gr6bner basis for In J is then given by GnA. EXAMPLE 4.3.10. In Z[x, y], we wish to compute (3x - 2, 5y - 3) n (xy - 6). So following Proposition 4.3.9 and Theorem 4.3.6 we consider the polynomials j, = 3xw - 2w, h = 5yw - 3w and 13 = xyw - 6w - xy + 6 and compute (J" h, 13) n Z[x, y]. We will outline the computation. We consider the deglex ordering with x > y on the variables x and y and an elimination order between w and x, y with w larger. We follow Algorithm 4.2.2 (observe that mueh use of Exercise 4.2.3 is made in this computation). The first saturated set to consider is {l, 2} which gives rise to 5yj, - 3xh ~+ O. The only saturated set containing 3 is {l, 2, 3} and this gives rise to 13 - 2yj, + xh ~+ 4wy - 8w - xy + 6 = j4. For the saturated sets containing 4 we need to consider {2,4} which gives rise to -5j4 +4h = 28w+5xy-30 = j5 and {J, 2, 3, 4} which gives rise to -Xj4 +413 = 8wx-24w+x 2y-4xy-6x+24 = j6. For the saturated sets containing 5 we need 4.3. APPLICATIONS OF GROBNER BASES OVER RINGS 231 to consider {1, 5} which gives rise to 3xJs -28h --++ 15x 2y-lOxy-90x+60 = h, {2, 4, 5} which gives rise to yJs - 7J4 --++ 5xy2 - 3xy - 30y + 18 = Is, and {1, 2, 3, 4, 5} which gives the salhe result as the last set (Exercise 4.2.3). For the saturated sets containing 6 we need to consider {1, 5, 6} which gives rise to J6 - 12h + xJs = 6x 2 y - 4xy - 36x + 24 = J9 and {1, 2, 3, 4, 5, 6} which gives the same result as the previous set. For the saturated sets containing 7 we need only consider {1, 2, 3, 4, 5, 6, 7} which gives rise to wh - 5xyh --++ O. The remaining cases (only 4 need to be computed) ail go to zero and we see that {f"h,h,J4,JS,J6,h,Js,fg} is a Grübner basis for (J"h, 13)· Thus (3x - 2, 5y - 3) n (xy - 6) = (5xy2 - 3xy - 30y + 18, 15x 2y - 10xy - 90x + 60, 6x 2 y - 4xy - 36x + 24). We also note that ho = h - 2J9 = 3x2 y - 2xy - 18x + 12 is in (J" h, 13) and we can replace h, J9 by ho so that {h,!2, 13, J5, J6, Js, JlO} is also a Grübner basis for (J" h, 13)· We thus obtain the simpler generating set (3x - 2, 5y - 3) n (xy - 6) = (5xy2 - 3xy - 30y + 18, 3x 2y - 2xy - 18x + 12). The computation of ideal quotients is almost the same as in the previous cases except that one must be careful of possible zero divisors in Rand hence in A. PROPOSITION 4.3.11. Let l = (J" ... , Js) and J = (YI, ... , gt) be ideals oJ A. Then t I: J={fEAIJJç:I}= nI: (Yj)· j=l M oreover, if In (g) = (h,g, ... , h,g) and 9 is not a zero divisor in A, then we have I: (g) = (h" ... ,hl). PROOF. The proof is exactly the same as in the ideal case over a field, see Lemmas 2.3.10 and 2.3.11, until we must verify that In (g) = (h,g, ... , h,g) implies that I: (g) = (h" ... , h,) which requires that 9 must be canceled and this requîres that 9 nat be a zero divisar. D As a consequence of this result we obtain a method for computing generators for the ideal 1: J ç A, provided that none of gl, ... ,9t are zero divisors. EXAMPLE 4.3.12. In Z[x, yi, we wish to compute (3x - 2,5y - 3): (xy - 6). From Proposition 4.3.11 we need ta compute (3x - 2,5y - 3) n (xy - 6) and divide the generators of this ideal by xy - 6. This intersection was computed in Example 4.3.10 and so dividing the polynomials obtained there by xy - 6 we obtain immediately that (3x - 2, 5y - 3): (xy - 6) = (5y - 3, 3x - 2). 232 CHAPTER 4. GROBNER BASES OVER RlNGS In Section 2.4 we used elimination to compute the kernel of an algebra homomorphism. Sometimes Rlxl, ... , x n ] is called an R-algebra, in order to emphasize the special role played by the ring R. In a similar vein, a ring homomorphism between tWQ polynomial rings QVeT R q,: RIYI, . .. , Ym] ---> Rlxl, ... , x n ] is called an R-algebra homomorphism provided that q,(r) = r for aU r E R. It is then dear that q, is uniquely determined by q,: Yi f----> 1;, for h, ... , Jm E RIXI, ... , Xn]. That is, if we let h E RIYI,." , Ym], say h = Lv cvyr 1 • • • y~tn, where Cv E R, v = (VI, ... , V m ) E Nm) and only finitely many cv's are non-zero, then we have v Given this setup we have the analogue of Theorem 2.4.2. THEOREM 4.3.13. With the notation above, let J = (YI - h,··· , Ym - Jm) ç; RIYI, ... ,Ym,XI,··· ,xn ]. Then ker(q,) = JnRIYI,'" ,Ym]. Another way to view Theorem 4.3.13 is that J n RIYI, ... , Ym] gives the ideal of relations among the polynomials h, ... , J,. EXAMPLE 4.3.14. Consider the Z-algebra homomorphism q,: Zlu, v, w] ---> Zlx,y] defined by q,: u f----> 3x-2, q,: v f----> 5y-3 and q,: w f----> xy-6. We wish to find the kernel of q,. Following Theorem 4.3.13 we let h = 3x - 2 - u, h = 5y - 3 - v, and 10 = xY - 6 - w and compute (h, h, 10) n Zlu, v, w]. We consider the term orders deglex with u > v > 'UJ on u, v, w and deglex with x > y on X, y and an elimination orcler between X, y and u, v, w with X, y larger. We follow Algorithm 4.2.2. The first saturated set to consider is {1,2} which gives rise to 5yh - 3xh --->+ O. The only saturated set containing 3 is {l, 2, 3} and this gives rise to -10 + 2yh - xh = vx + 3x - 2uy - 4y + w + 6 = J4. For the saturated sets containing 4 we need to consider {l,4} which gives rise to 3J4 - vh --->+ -6uy - I2y + uv + 3u + 2v + 3w + 24 = J5 and {l, 2, 3, 4} which gives rise to yJ4 - vh --->+ -2y 2u - 4y2 + yu + yw + 8y + vw + 6v = J6' There are three saturated sets containing 5 but we need only consider (Exercise 4.2.3) {2,5} which gives rise to 5J5 + 6uh = -uv - 3u - 2v + 15w + 84 = h and {l, 2, 3, 5} which gives rise to XJ5 + 6uh --->+ O. Ali remaining polynomials arising from saturated sets containing 6 or 7 recluce to zero. Thus we see that ker(q,) = (uv + 3u + 2v -15w - 84). We conclude this section by giving a method for computing a set of generators for the syzygy module, Syz(/J, ... , J,), for a set of non-zero polynomials {h, ... , J,} in A. Wedo this byfirst computing SYZ(YI,'" ,y,) fora Grobner basis {YI, ... ,g,} for the ideal (h, ... , J,). The theorem describing Syz(gj, ... ,y,) 4.3. APPLICATIONS OF GROBNER BASES OVER RINGS 233 is the analogue of Theorem 3.4.1. In our case, we will have to replace Theorem 1. 7.4 with Theorem 4.2.3. Let {g" ... , g,} be a Grübner basis in A. We let lt(gi) = CiXi for Ci ER and power products Xi. Let B = {h" ... , hg} be a homogeneous generating set of the syzygy module SYZ(lt(g,), ... ,lt(g,)) = SyZ(CIX ... , c,X,). Assume that " for dji E R and power for 1 S j S C we have that hj = (djlY,I, ... , Cj'Y,,) products Yji, where for each j we assume that hj is homogeneous of degree Zj; i.e. for all i,j such tbat dji of 0, we have XiY,i = Zj. (we assume that Y,i = 0 if dji = O.) Then by Theorem 4.2.3, for each j, 1 S j S C the generalized S-polynomial L:~l djiY,igi has the representation , , Ldji0i 9i = i=l Lajvgl/l 1/=1 where 1 <v<t - - , max lp(ajv) lp(gv) = lp('" dji}jigi) L-t i=l < max Y,i lp(gi). 1 <i<t -- We now deline for 1 S j S C, Sj = hj - (ajl, ... , aj') E At We note that Sj E SYZ(g" ... ,g,). With the notation above, the collection {Sj Il S j S C} is a generating set for SYZ(g" ... , g,). THEOREM 4.3.15. PROOF. Suppose to the contrary that there exists (UI, ... ,u,) such that (UI, ... ,u,) E SYZ(g" ... ,g,) - (Sj Il S j SC). Then we can choose such a (UI, ... , u,) with X = max'<:i",,(lp(u;) lp(gi)) least. Let S = {i E {l, ... ,t} IIp(ui)lp(gi) = X}. Now for each i E {l, ... , t} we deline , u; as follows: {UiifirjeS Ui = Ui - lt( Ui) if i E S. Of course we have that for all i, lp( u;) lp(gi) < x. Now, for i E S, let lt( Ui) = C;X;. So for all i E S we have XiX; = X. Since (UI, ... ,u,) E SYZ(g" ... ,g,), we see that L C~CiX:Xi iES = 0, and 80 L <X;ei E SyZ(C1X ll ··. ,CtXt). iES 234 CHAPTER 4. GROBNER BASES OVER RINGS Thus, we may write , Lc~X:ei iES = LVjhj) j=l for sorne A. We note that, we may assume that either Vj = 0 or that Vj = bj ~ , bj E R. We can see this sinee, for each i E S, we have <XI = L~=l VjdjiYji Vj E from which we obtain, aiter multiplying through by Xi, C;X = ~~~, vjdjiZj ; and thus the power products involved are independent of i. Then we have (Ul, ... ,Ut) = L<X:ei+(u~"",U~) iES L bj : j=l J , , hj + (u;, ... , u;) ,u;) + Lbj Z(aj" ... , aj'). j=l X J Lbj ZBj j=l + (u;, ... ' X J We define , (w" ... ,w,) = (u;, ... ,u;) + Lbj :(aj" ... ,aj'). j=l J We note that (w" ... ,w,) is in SYZ(9', ... ,9,) - (Sj 1 1 :s; j (u" ... ,u,) and each Sj are in SYZ(9',··· ,9,) and (u" ... ,u,) j :s; i). We will obtain the desired contradiction by proving that :s; i), sinee rt (s,Il :s; For ead v E {l, ... , t} we have lp(u~ , j=l X J + Lbj ZŒjv)Xv l::;J~l :s; max(lp(u~), But, by definition of u~, we have Ip(u~)Xv max (ZX Ip(ajv)))Xv . j < X. Also, Therefore lp(w v ) lp(gv) < X for each v E {l, ... , t} violating the condition that X = max'<i<,(lp(Ui) Ip(9i)) is least. 0 We now turn our attention to computing Syz(f" ... ,J,), for a collection {J" ... , j,} of non-zero polynomials in A that do not necessarily form a Gr6bner basis. We first compute a Gr6bner basis {9',· .. ,9,} for (f" ... , j,). Set F = [ J, j, and G = [ 9' 9, We can compute a t x s matrix S and an s x t matrix T with entries in A such that F = GSand G = FT. (Recall 1 1. 4.3. APPLICATIONS OF GROBNER BASES OVER RINGS 235 that S is obtained using Algorithm 4.1.1 and T is obtained by keeping track of the reductions in Algorithm 4.2.1 or Algorithm 4.2.2.) Now using Theorem 4.3.15, we can compute a generating set {SI, ... ,se} for Syz(G). Then exactly as we did in Theorem 3.4.3 we have (Exercise 4.3.12) THEOREM 4.3.16. With the notation above we have Syz(h,· .. ,l,) = (Ts I where Tl) ... , ... ,Tsf, Tl,··. T S. ,T s ) ç; A', ,r 8 are the columns of 18 ~ EXAMPLE 4.3.17. We go back ta Example 4.2.12. R.ecall that a Grübner basis for the ideal l = (h,h) in Z20[X, y], where h = 4xy + x and 12 = 3x 2 + y, with respect ta the lex ordering with x > y, was computed to be h, 12, 13 = 5x, 14 = 4y2 + y, is = 15y. We wish to compute a generating set for Syz(h, h h!4, Is) and Syz(h, 12). As in Theorem 4.3.15, we first compute a homogeneous generating set for Syz(lt(h), lt(h), lt(h), It(4), ItUs)) = Syz( 4xy, 3x 2, 5x, 4y2, 15y). In Example 4.2.12 we found that the syzygy module is generated by (5,0,0,0,0), (-3x,4y,0,0,0), (0,0,4,0,0), (5,0,-4y,0,0), (0, 15,-x,0,0), (0,0,0,5,0), (y,O, 0, -x, 0), (0,0,0,0,4), (0,0, 3y, 0, -x). These syzygies give rise to the following polynomials 5h -3xh +4yh 413 5h -4yh 5x=h -3x 2 + 4y2 ~ 4y2 + Y = 14 ° 5x=h 15y = Is 5y = 31s 1512 -xh 514 yh -X14 41s 3yh - xls Therefore ° ° O. ((5,0, -1, 0, 0), (-3x,4y + 1,0, -l, 0), (0, 0, 4, 0, 0), (5, 0, -4y -1, 0, 0), (0,15, -x, 0, -1), (0,0,0,5, -3), (y, 0, 0, -x, 0), (0,0,0,0,4), (0,0, 3y, 0, -x)). To compute Syz(h, 12), we first compute the matrices Sand T such that [h 12 13 14 Is] = [h 12] T, and [h h]=[h 12 13 h is]S 15. (15y + 5x2. 0. TS = [~ ~]. 0) T(O. 3y. 4y + 1.3.0) (0. . l+x.0) (0. for example.x) (0..0) (0.0. 0.0) (15y+5x 2. 0.5x).12.x 3 y . .-1) T(O. We note that {1 + x.x 2 +5.IOx 3 + 4x 2y2 4x 2y . 0. 0. 0) T(5.2. 4. 0.H).6x Eland write f as a linear combination of fI and h . 0. Therefore Syz(fI .3.0. give a method for computing generators for ideal quotients in R[x" . where R is a commutative ring and 1 is an ideal in R[Xl. x 2 + 5} is a Grabner basis and we can compute a generating set Syz(l+x.0) T(O.1. (1x. In this exercise we give another method for computing the ideals in Exam~ pIe 4. 5.-2)} which yields (2): (1+ H) = (-2. that is. -3) T(y. -4xy . -4y .12) = ((y + 3x 2 .0) (0. Exercises 4.x n ].2.-x. -1.0. 4. 5x)). Generalize Exercise 4. (2): (1 + H).1. 4.. .13.4xy2 + Uxy . 0. 0).xnJ/J.3. 0. 2. 0. Show that f = 12x 3 y 2 . As in Example 4. Verify the statements made so far and then go on to use this method to compute ail of the ideal quotients in Example 4.3. Wejirst observe that Z[HJ '" Z[xJ/(x 2 + 5) under the map H c--> x+ (x 2 + 5). yJ generated by fI = 4xy + x and fz = 3x 2 + y.2. 0. -x.4xy3 . Thus to find.1..3. consider the ideal J ç.x).xnlj I.-3. 0) T( -3x. -4xy .(0. 0.2. Z20[X. GROBNER BASES OVER RlNGS It is easy to see that We have T(5.0) (0.1 to the case where Z[xJ/(x 2 + 5) is replaced by R[Xl1'" .13. -1. -x) Finally we have J2 - (0. 4) T(O. 0. . 0.1). we need to find the syzygies of the matrix [ 1 + x 2 x 2 + 5 J and read off the coefficients of 1 + x. 0) T(O.0) (y + 3x 2. 2.236 CHAPTER 4. x 2 +5) = {( -2. : u ---.xnl. we must consider rings of fractions of the ring A. Theorem 3.4. w] ---. a.. x n ] with J9 E P. Algebraically.6. xn]m. Trager and G. Prove Theorem 4.6. Prove Proposition 4. A PRlMALITY TEST 237 4. 4.2.4. [Hint: Look at Exercise 4.13. 3x .2 and q. xn]-submodules of R[x1. In Example 4.7.5y. A Primality Test. Compute generators for ker( q. Tasks (i) and (ii) at the beginning of Section 3. prime ideals in k[Xl1 . .6. 4.5. 4. Geometrically..3.7.3. it is a long. xy . yI! J.3.3. since that is an that is needed in .3.13.16). we will assume when discussing rings of fractions that R is an integral domain. In Z[x. Zacharias [GTZ].3. Consider the Z-algebra homomorphism q. Use this to construct the syzygy module for the original polynomials (of course.] 4..: u c--> 3x . Recall that an ideal P is called prime if and only if given any polynomials J.4. 10y2 .5.2. y] defined by q. 4. Although this is not strictly necessary.3. Compute generators for the syzygy module of the Grübner bases constructed in Exercise 4. q. y] defined by q. J9) for the Grübner basis in Example 4. .7.3.2. by unique factorization.13 (although this problem is quite doable. messy computation).3. Before we give the primality test. 4.3. 4. b. Z[x.] 4. 4.3.3.. . d.3 and 3.: v ---.12. .8. 9 E R[x1. Find the totally reduced form of 5x3 y 3.9.3. v] ---. Generalize the following to the case of R[x1' . 4. Compute SYZ(h.14 and 4. Gianni. v..: Z[u. Compute generators for ker( q. c. 4.J5.15.3..).3. The test for primality that we present in this section is taken from the paper of P. B.2. Consider the Z-algebra homomorphism q. Prime ideals are basic building blocks like prime integers are building blocks in Z. The following exercise depends on Exercises 4.. 4.11 compute a complete set of coset representatives for Z[x. x n ]/ Pis an integral domain.48xy + 27x.6. 3x 3 y 3 and 4x3 y 3.1.9.8 as energy permits. .8.3.).3. y] compute (3x-2.4.3. In this section we will give an algorithm that will determine whether a given ideal l in A = R[Xl 1 • • • l Xn ] is prime or not. [Answer: (15xy-9x-lOy+ 6. Prove Proposition 4.: w c--> x .14.11. 17.11.. this latter problem is trivial).7 and 4.3. . xy-6): ·(5y-3).: Z[u.6 (of course use the ideas in Theorems 4. correspond ta irreducible varieties (varieties which are not the union of two proper subvarieties) (see [CLOS]). In Z[x. e.: v ---.6.6. 2y .16.2.6.2.11. any radical ideal is the intersection of prime ideals. Theorems 3. we have J E P or 9 E P.96y + 54).5 and q.y] compute (3x-2. Z[x. where k is a field.. Repeat this exercise for Exercises 4. This is easily seen to be equivalent to the statement that the ring R[x1. Prove Theorem 4.xy-6)n(5y-3). As much of Section 3.10.15 and 4. ..5xy2 .2. . Task (iii) at the beginning of Section 3. 238 CHAPTER 4. f E (J.gM(X1"" .4. Then w"g"f E JA[w] ç (J.xn )= I. . {/v f where S-l J = 1 f E J and 9 ES}. let + (1 m w" g")f E (J. if S = {g E A 1 9 '" A}.wg-l) and so f = w" g" f as desired. It is easy to see that S-l A is a subring of K. 1 E S. 9 E A. Our main concern is the saturation of a non-zero ideal l of A with respect to S. wg . PROOF. S = {g E A 1 9 ri' P}. 9 E S then f 9 E S. then S-lA is called the local ring at P and is denoted by A p .w) f=f(X1"" . (X1. Then J Ag nA = (J.1) nA.4. Conversely. since for ail f E A. Consider the ideal (J. then S-lA = K. where P is a prime ideal in A. 9 '" 0 } of A. Let R be an integral domain. Given a multiplicative set Sc A we define the subring S-l A of K by S-l A = {~ E Kif E A and 9 ES} . Let S c A be a multiplicative set. Then there is a non-negative integer v such that g"f E J.1. in fact. since every element of S-l J is the product of an element of J and an element of S-l A.xmw) E A = R[X1. In the three examples above. and if f.xn ]. Then. We show that for two multiplicative sets S we can easily compute generators for the saturation of an ideal l with respect to S.xn ) -1)ho(X1. Examples of multiplicative sets are S = {g E A 1 9 '" O}. wg .xn )hM (h=1 +(Wg(X1"" .1) nA. and S = {g" 1 vEN} for a fixed 9 E A. 0 ri' S. whlch we call the ring of fractions of A with respect to S.xn. Then A is also an integral domain. It is often also written as J(S-l A). we have f = and 1 E S. the ideal in S-l A generated by the set J.1) and S = {r E Rlr '" O} (Proposition 4.1) of A[w]. We have the field of fractions K = {~ 1 f. wg .1. whlch is defined to be S-l J n A.4). if S = {g E A 1 9 ri' P}.'" . by assumption. We consider certain subrings of K.4. 9 '" O. that is. and finally if S = {g" 1 VEN} then S-l A = 1 f E A and VEN} is denoted by Ag. We note that A is a subring of S-l A. GROBNER BASES OVER ruNGS order to obtain the primality test (even in the case where R is not an integral domain).'" .1). } (Proposition 4. wg . namely for the cases where S = {g"lv = 0. Let f E J Ag n A..'" . It is readily seen that S-l J is an ideal in S-l A and is. Let w be {f a new variable. .. PROPOSITION 4. Let 9 E A. X n ] = 8.{a}. We will give an example which illustrates Proposition 4.X n ]) (this is an easy exercise).1 R)..gt of the desired type showing that G = {g" . Thus. let 8 be any multiplicative subset of R.12. We compute a Griibner basis.4.6). when combined with Proposition 4.. + .. LEMMA 4. . A PRlMALITY TEST 239 where g~ Eland h~ E A[w] for all IL..1 R) [Xl.1 (R[X1. ...g..4. which shows that f E lAg. allows us to compute generators for the saturation of an ideal 1 with respect to 8 = R .X n ..6.. Then we see that J is a (infinite) Gr6bner basis for 1.1 A. = {:" 1 r E R..1.e. We note that (8. Now let 8 = R..4.)gt. w]. 'X n1 W with w bigger than the x variables.xn ) we rnay substitute w = }.11.2. let f E 8.. Then G is a Griibner basis for the ideal 8. .. Then s f = h ..t(lp(hi) Ip(gi))' Thus f = (~h. t).xn ] with respect to 8 = {g" Iv E N} we eonsider an elimination order on the variables Xl. This follows easily from the third eharacterization of Griibner basis in Theorem 4.} is a Gr6bner basis for 8. .4. and let k denote the quotient field of R (i. 0 Reeall that for s E R we denote by R. we need two preliminary results. Then there is a s E 8 c R such that s f E 1. Since Ris an integral domain.1 in Example 4.gt}.l. + htg. We observe that the proof of Theorem 4. In order to compute generators for the saturation of l with respect to S. .l.4. 0 We note that in the following lemma we do not need that R be an integral domain. we have Ip(sf) = Ip(J) and Ip(~hi) = Ip(h i ) (1:. Namely. But J is an ideal and so generates itself. Let 8 c R be a multiplicative set and let 1 A be a non-zero ideal of A.{a}.)g. for the ideal (1. We now give the result which. such that Ip(sf) = max''. Of course. PROPOSITION 4. . G.1) in A[w] = R[X1"" .12 did not require that the set G be a finite set. 0 Proposition 4. that is.. J generates 1. . Thus we have a representation of f in terms of 91.. In this situation we see that Griibner bases are well-behaved. since J 1.2. . wg .4. k = 8. i:.'" . . Let R be an integral domain.i. c: PROOF.15. Suppose that G = {g" . Sinee the variable w does not appear in f = f(X1. g .3. we have that Lt(J) = Lt(l).1 1. Let J J=l.. For an integral domain R. . .3. VEN}. in order to compute generators for the saturation of an ideal l ç A = R[X1. c: 1 be ideals in A c: and assume that Lt(I) c: Lt(J). Then PROOF.4. Let G be a Griibner basis for 1 with respect ta some term order.l.. Then G nAis a Griibner basis for this saturation (Theorem 4.. . and hence by Corollary 4.11 in 8.4.1 should be eompared with Theorem 2. + . + (~h. J = 1.13. xn ] n Rs [x" .. E Lt. . We see that It(s~ f) = 2)t(s~hi) It(gi) E Lt(I)k[x]. Let IcA = R[x].2... Thus... We note that it suffiees ta show that (4.).. reealling that Lt(I) = Lt( G).Xn]. ..xn ].g.xn]n R[x" . ~S!/i i=l ... lndeed.Xn] iEV n R[x" .[x" .4. ..4..2) holds.} be a Grobner basis for l with respect ta sorne term ordering.. The desired result (4.g.4. xn].. that Ik[x...xn ]. We will show that bath sides are equal ta the ideal. GROBNER BASES OVER RlNGS PROPOSITION 4..3. .. . .. We need ta show that It(f) is in Lt s (IR. ... ...(IRs[x" ..xn ]. . . PROOF. . We let f E Ik[x" . + ..xn]nR[x]. if Equation (4..xn ] nR.[x" .... . . Henee It(f) = ai L -.. .[x" .1) Ik[x" .. where hi E k[x" .xn]nR[x" .. [X" .Xn].. ..[x" .4. . whieh we denote by Lt..xn ]. and in Rs [x" . We first note that Rs ç k and sa we have immediatelythat Lt(I)R. . .x n ]. It remains ta prove the Claim. (lp(gi) [1 ::. . by Theorem 4.[x" .xn]nR... Assuming the Claim. we prove (4.xn ]).xn ]...xn ].. applied ta the ring Rs[x" .. .. .. sueh that Ip(f) = Xi Ip(gi) for ail i such that ai # O...4.xn ]. ...... + h. .xn ] ç Lt(I)k[x" . .. . . . .. CLA]M: Lt(I)k[x" . p.[x" . we have from Lemma 4. as desired. and sa from the Claim It(s~ f) E Lt(I)Rs[x" .Xn]).. .. ... .2) as follows. .xn]nR[x" .. Then (4. ..xn]nR[x" .4. . .xn ] be a non-zero ideal and let G = {g" .. We write It(gi) = CiXi for Ci E R and a power product Xi' .4.xn]nR[x" .12... .4. ....xn ]).Xn] =IR. .. We will need leading term ideals in R[x" . .xn ] n R[x" .. G is a Gr6bner basis for Ik[xl. for ai E R.2) Lts(Ik[x" .. 1 xnJ... . t) in R[x" . .4.1) follows by interseeting this last equation with R[x].Xn]) ÇLts(IR.xn ] = IR. i ::. and power produets Xi.g. . Since. Set s = lc(g.xn ].xn ]. non-negative integers Vi.[x" .. we may write It(s~f) =" ~Xilt(gi). f = h. Let R be an integral domain with kits quotient field.. ..xn ] there is an non-negative integer l' sueh that s~ f E R[x" .240 CHAPTER 4.Xn]).. whieh we denote by Lt s . .. i=l since It(gi) E Lts(IR.[x" .. we may write..xn ] and Ip(f) = max'99(lp(hi )lp(gi))' Let V = {i [ Ip(f) = Ip(hi)lp(gi)}' Then It(f) = I:iEV It(h.1.) It(gi)' Since f E R.) lc(g2)" ·lc(g.+-Xilt(9i) sv.xn ] = Lt(I)Rs[x" ... by Proposition 4.. . xn ] as desired.4. This follows from Proposition 4.. . We consider Example 4. 0 EXAMPLE 4.. 4y3 .. begin with the polynomials fI. . 420w 1) n Z[x. since s = aiCi for sorne ai E R.xn ].6.J3. Then. we compute IQ[x.11 and compute the saturation of I = (3x 2 y + 7y. 4.xn ]. . of course..4.xn ] be an ideal. provided we use the lex term ordering with w > j. by the definition of Xi = ~(C. x n ] n R[x" .h = 4xy2_5x.. we need to compute IZ 420 [X. h.35y.J4. y]. h. Finally we are ready to discuss the issue of primality in rings. Let R be an integral domain in which linear equations are solvable and let k be its quotient field.xn ] ç (Xi 1 1 ::.4xy2 . i ::. t) ç Lt(I)Rs[x" .. 980wy+x 2y.5x) with respect to Q[x. J4 together with J5 = 420w ... we are adopting the notation of this section and 1. where c E R and X is a power product. and hence {f" 12. .420 does not ruean the integers mod 420).. Then. from Proposition 4. > y. 980wx+x 3 . A PRlMALITY TEST 8. h...xy2.1..xn ] n R[x" . i ::. . i ::. Then we can compute. .5y.2.15. t).4 and Proposition 4.28x} is a Gr6bner basis for IQ[x. . 4y3 -5y.Xn ] as a linear combination ofthe Xi with coefficients in k[x" .Xi) E Lt(I)Rs[x" .4.4. That is.xn ] n R[x" . t) (Exercise 4.Xn] n R[x" . 3x3 +28xy2_28x} is a Gr6bner basis for (fI. PROOF.xn ] n R[x" . 241 Then. We.2.. . .xn ] and so we see that we have that (Xi Il::.4. ho 420w -1). and J4 = 28y3 . ç (Xi Il::. We need two lemmas from commutative algebra which give criteria for when an ideal in R[Xl. ..525wx . Then in order for c to be non-zero we must have an Xi dividing X and then we see that Lt(I)k[x" . J5. .2. to be fI = 3x 2y+7y.5.4..1).4 can be replaced by s = lcm(lc(gi)11 ::. y]nZ[x.. 784wy2 + x 2. following Proposition 4. we may start with (J = 5 in Algorithm 4.J4 is a Gr6bner basis.4. y]. 12. i ::.J4.'" .6. y] n Z[x.4.. h = 15x 2 +28y2. ... It remains to show that Lt(I)k[x" .4.. y] n Z[x.3. y]. 3x 3 + 28xy2 . sinee fI.4 and Exercise 4. with respect to lex with x> y. .1 and observe that. 0 We note that if R is a UFD. generators for the ideal Ik[x" . which we do using Theorem 4. Let eX be a term in J. of course. .. y].4.. We are no longer assuming that R is an integral domain. We will not go through the computations. t). 525wy_ y3. Let IcA = R[x" . . since lcm(3.xnl is prime. we need to compute (J" 12.4.1. Since Lt(I) = Lt(G) we can express each element J of Lt(I)k[x" . but will note that {f"h. . y]. then the element s in Proposition 4. In that example we computed a Gr6bner basis for I.28) = 420 (here. . COROLLARY 4..xn ].h.1. . this follows immediately from Lemmas 4. .~! aiXi = 0.Xn] and assume that fg E Ik[x!... . .xn ] and 1 = Ik[x!. ... PROOF. is onto and ker( q.Xn] n R[x!. .. .. Let f. We now assume that the ideal Ik[x!. By the hypothesis.. . e E R be sueh that df.: i=l .4. .8. It is then easy to see that q. and l' = l'k'[x] nR'[x].. Let 1 ç R[x!.4. f E Ik[x!. .xn] and assume that fg E 1.Xn]. .. .Xn ] = 1 and so wc see that 1 is a prime ideal. D l '" 1 '" COROLLARY 4.. . . For f = I::~! aiXi E R[x!. . then l' R'[x] is a prime ideal of R'[x].. as desired. It is now immediate that 1 is a prime ideal if and only if the image of 1 in (R/ 1 n R) [X!..land 80 ~ E I.xn ] be an ideal such that 1 n R = {O}... .Xn].. . The reverse inclusion is trivial.. .. That the map is well-defined follows since if I:.Xn].... Moreover.xn ] n R[x!. .4.. . . Then fg E Ik[x!.xn ] or g E !k[x!. ..xn ]). eg E R[x!. PROOF. .xn]->R[x!. .. then ai E 1 n R for all i and so clearly I:.7..8.~! aiXi E 1.... . .Xn] is prime. Then 1 is prime if and only if (i) 1 n R is a prime ideal in Rand (ii) if we let R' = R/lnR. Noting that l' D n R' = {O}. ..xn ]. . .. we write 7 = I:i~! aiXi. r of O. . Then fg = ~ sueh that h Eland r E R. D LEM MA 4.Xn] is prime in k[x!. Then 1 is prime if and only if Ik[x!. . ....9. .4.xn ] is prime. . Let f. Then we have r(df)(eg) = deh Eland so r(df) E 1 or eg E l. . Then 1 is a prime ideal if and only if the image of 1 in (R/I n R)[x!.xn ] or g E Ik[XI ..xn]/I .Xn] be an ideal..xn ] is prime. . .xn ] is prime in k[x!. . Finally.... its image in (R/I n R)[x!..xn ]. . in this case 1 n R is prime... . InR = {O}.Xn] nR[x!. . with ai E R and power products Xi. . k' be the quotient field of R'. Let 1 ç R[x] be an ideal.7 and 4. r fi. i=l LaiXi>--+ LaiXi+ I . Let d. ...) is the image of 1 in (R/lnR)[x!. it is trivially seen that if 1 is prime then so is 1 n R. and 80 Ik[XI . We assume that f E Ik[x!.... . . .. .xn ].: by (R/lnR)[x!. q. .xn ] with h Eland r E R.242 CHAPTER 4. GROBNER BASES OVER RINGS LEMMA 4. ...g E R[x!. .. . g E k[x!. . q. Let R be an integral domain with quotient field k and let 1 ç R[x!.xn ]..xn ].Xn].. . For r E R we let r = r+lnR be the image of r in R/lnR.. Then f E Ik[X!... We define a ring homomorphism PROOF. .. since 1 is prime in R[x!. .Xn].. .xn] and 1 = Ik[x!. . . We then have immediately that f E Ik[x!. . and l' be the image of 1 in R'[x]... . .. let ~ E Ik[x!. . . We first assume that 1 is a prime ideal in R[x!. Moreover... . .xn ] n R[x!.Xn ].. .4...xn ].. ..xn ] and so by hypothesis. Then r( ~) Eland so rEl or ~ El (reeall that ~ E R[x!. X n].10. and 3 k ' = Q(z).z4 We have n = 3. h h f<. . . By Corollary 4. and the ideal l'.Z2 As we saw in Exercise 2. and J. we first need to determine whether J n R is prime.4. we can determine a single generator f of l'k'[x].1. this is the ideal of relations among the tillee polynomials t3. its quotient field k' . . using the Euclidean Algorithm (Algorithm 1. In this case we will consider the integral domain R' = RjJ n R. z]j J is isomorphic to a subring of the integral domain Q[t]). We first compute a Grübner basis G for J using the lex ordering with x > y > z and get G = {fI. z].4. R' = R3j J 3 = <Ql[z]. .yz and h = x 2 y .t4. h = (y5 . We use "brute force" to see that f is irreducible over Q(z): It cannot have a root in Q(z).4.4. the image of Jin R' [x].. Corollary 4. y.4. Then J'k'[y] = (y5_ Z4)Q(z)[y] and so f = y5_ z 4 Here f is viewed as a polynomial in y with coefficients in Q(z).5.1. we then need to determine whether l'k'[x] is a prime ideal for one variable x.5 we can determine whether or not l' = l'k'[x] n R[x].4. = J. where f4 = xy3 .t5 and hence must be a prime ideal (sinee then Q[x. Let J c Q[x. we must assume that we can determine when an ideal in R is prime.9. We first note that the case i = 4 in the algorithm is trivial. given a prime ideal P of R[XI" . Since x is a single variable. fs}. Now.. we can determine whether J n R is prime. EXAMPLE 4. Again by Corollary 4.1.3.z4). In this example we will show that J is prime using Algorithm 4.Xn-I])[Xn].'" . Moreover. we can determine whether l'k'[x] is a prime ideal or not.. The idea is summarized in the next paragraph..4. z] = (y5 .. For a single variable x and an ideal J ç R[x] we want to decide if J is a prime ideal. We may compute a generating set for J n R from Theorem 4. by Corollary 4. Thus we see that in order to have an algorithm determining the primality of ideals in R[Xl1'" .xn ] is prime.5.3.9. We use the notation established in that algorithm. .4.4.. Finally.9 gives us an inductive technique of determining whether a given ideal in R[Xl) .2) on the known generators of l'. sinee there cannot be a rational function in <Ql(z) whose fifth power is the fourth power of the prime z E <Ql[z].Z4).xn]jPare irreducible or not. the quotient field of Q[z]. we must assume that we can do similar things in ~. J4 = h = (0). because of 3Here Q(z) denotes the field of rational functions in z. . that is. A PRIMALITY TEST 243 Sinee R[XI' . k'[x] is a PID and. Thus. The next case to consider is i = 3. whether polynomials in one variable with coefficients in the quotient field of R[XI.. h = X3 . y.y2. assuming that we can determine whether f is irreducible or not in k'[x].4. Specifically.6 and then. We have h = J n <Ql[y. we must assume that we can determine. J' = J 2R' [y] = (yS _Z4)<Ql[y.Z3 and fs = y5 . The algorithm for determining the primality of an ideal is given as Algorithm 4.Xn] = (R[XI. We now give an example of Algorithm 4. For example Z and Q have these properties. z] be the ideal generated by fI = xz .xn ]. assuming we can determine whether ideals in R are prime. 244 CHAPTER 4. GROBNER BASES OVER ruNGS INPUT: J, an ideal of R[Xl"" ,xn ] OUTPUT: TRUE if J is a prime ideal, FALSE otherwise Set Ri := R[Xi'" . Compute Ji := J ,xn ] for i = l, ... ,n, and = l, ... , n Rn+1 := R n FI.; for i +1 IF J n +1 is not a prime ideal of R THEN result:=FALSE ELSE result:=TRUE i:= n+ 1 WHILE i > 1 AND result:=TRUE DO ~ ~ R' .·-R·/J· J' := Ji-1R'[Xi-l] k' := quotient field of R' Compute the polynomial IF f such that J'k'[Xi_l] = (I) f is not zero or irreducible over k' THEN result:=FALSE ELSE Compute J'k'[Xi_l] n R'[Xi-l] IF J'k'[Xi_l] n R'[Xi_l] oF J' THEN result:=FALSE ELSE i:= i - 1 RETURN result ALGORITHM 4.4.1. Primality Test in R[Xl, ... ,xn ] unique factorization in Q[z]. Moreover, it cannot be the product of a cubic and a quadratic in Q(z) [y]. One way to see this is as follows. If f = y5 _z4 were the product of a cubic and a quadratic in Q(z)[y], then we would have a system of 5 polynomial equations in 5 unknowns which would have to have a solution in Q(z). 4.4. A PRlMALITY TEST 245 If one computes a Grübner basis for tbese polynomials one finds that there would . have to be a rational function e(z) E lQI(z) satisfying eS = Z8 (use lex with e > z being the smallest variables where e is the constant term of the quadratic) and again this is impossible. It is then trivial to check that (y5_ Z4)IQI(Z) [y] nlQl[y, z] = (y5 _ z4)IQI[y, z] (if( y 5 - Z4) ~~~o C>v(z)yv E (y5 - z4)IQI[y, z] with C>v(z) E lQI(z), then a simple induction shows that C>v(z) E lQI[z]). Thus we have completed the WHILE loop for i = 3 with "result=TRUE". We now consider the case i = 2. In this case J, = J. Wehave R' = lQI[y,z]f(y5Z4), J' = J((IQI[y, zl/ (y5 - z4) )Ix]) and k' = quotient field of R'. For polynomials f E lQI[x,y,z] we denote by 7 the polynomial in R'[x] = (lQI[y,z]f(y5 - z4))[x] obtained by reducing the coefficients of the powers of x (these being polynomials in lQI[y, z]) modulo (y5 - Z4); i.e. we have a homomorphism lQI[x, y, z] -. e---> (lQI[y, z]f (y5 - z4)) [x] f f. We also denote the image of an ideal K C lQI[x, y, z] in R'[x] by K. We first must find a generator for J'k'[x], where J' is the ideal generated by 7,,72,73,74 E R'[x]. It is, in fact, easy to see that 72,73 ,74 are multiples of 7, in k'[x]. For example, ignoring the "bar" notation for the moment, viewing the following equation as being in k'[x], and noting that z, z2, and z3 are non-zero in R', we see that x 3 - yz = (zx - y2)( ~x2 + z'o-y 2 x + z'o-y4); that is, 72 is a multiple of 7,. Thus f = 11 and it is irreducible over k' , sinee it is of degree 1 in x. It remains to show that J'k'[x] n R'[x] = J'. We will apply Proposition 4.4.4 and Proposition 4.4.1. So we first need to compute a Grübner basis for the ideal J' = (f" 72' 73 , 74) C R'[x]. We will use Algorithm 4.2.2. We make the following general observation which is easily proved: if K is an ideal of lQI[x, y, z] containing y5 - Z4 and g E lQI[x, y, z], then K: (g) = K: (g). We consider the saturated subset {1,2} for which we need to compute (z): \il = (z) and the corresponding syzygy is O. Now the saturated subsets of {l, 2, 3} containing 3 are {l, 3} and {l, 2, 3}. For {l, 3} we compute (z): (y) = (z, y5 - Z4): (y) = (z, y5 - Z4): (y) = (2, y4) (note that this latter is just a computation in lQI[y, z] and so can be done using Lemma 2.3.11). The two syzygies are (yx, -z) and (Z3 x , -y4). The first gives the Spolynomial yx7, - Z73 = _y3 x z3 x (x 2, -z) which gives the S-polynomial x 27, - Z72 = -ïlx 2 + yz2 l.!.., 7, - y473 = (z4 and the saturated subsets of {l, 2, 3, 4} containing 4 can be handled in the same way. After this computation, we see that a Grübner basis for the ideal J' is just the original generating set {f" 72' 73 , 74}' Then, as in Proposition 4.4.4, we set s = lea,) le(2) le(3) le(74 ) = ZÏyy3 = y4z, and we need to show that J' R;[x]nR'[x] = J'. We do this using Proposition y5)x 2 - + Z3 L., O. The second gives the S-polynomial y2 z 3 x + y4z2 1>.., O. The saturated set {1,2,3} 246 CHAPTER 4. GROBNER BASES OVER RlNGS 4.4.l. Thus we need ta compute (J', wzy4 - 1) n R'[x] = (J" h, h, J., y5 - z4, wzy4 - 1) n Q[x, y, z]. This latter computation can be done in Q[x, y, z, w]. Using lex with w > x > y > z, we compute that a Griibner basis for the ideal (J" 12, 13,14, y5 - Z4, wzy4 - 1) consists of the 6 polynomials listed above together with wy3z3_X2) wyz4-x, and wzS_y. Thus (J"h,h,14,yS-z4, wzy4-1)nQ[x,y,z] = (J"h,h,f4,yS-z4) and sa (J', wzy4-1)nR'[x] = (J"h,h,J., ys-z4, wzy4-1)nQ[x,y,z] (J" 12, 13,14) = J'and the algorithm terminates with "result=TRUE." Exercises 4.4.l. Prove that if R is a UFD then in Proposition 4.4.4 we can let s = !cm(lc(g,), !c(g2),'" , !c(g,)). 4.4.2. Compute the saturation of the following ideals in Z[x, y] with respect ta Q[x, y] using Proposition 4.4.4. a. (6x 2 + y2, lOx 2y + 2xy) . . b. (3x 2 y - 3yz + y, 5x 2z - 8z 2 ). 4.4.3. Using Algorithm 4.4.1 and lex with z > y > x, show that the ideal (xzy3, yz - X2) ç Q[x, y, z] is not prime. 4.4.4. Using Algorithm 4.4.1, show that the ideal (y4 - Z3, y2 - XZ, xy2 _ Z2, x 2 _ z) ç Q[x, y, z] is prime. 4.5. Grobner Bases over Principal Ideal Domains. In this section we specialize the results of the previous sections ta the case where the coefficient ring Ris a Principal Ideal Domain (PID). Recall that an integral domain is a PID if every ideal of R is principal, that is, if every ideal of Rean be generated by a single element. We note that sueh rings are aIso Unique Factorization Domains (UFD) (see [Go, He, Hun]). Wc will make extensive use of this fact in this section and in Section 4.6. Examples of such rings include Z, Z[\I'2], Z[i], where i 2 = -1, and k[y], where k is a field and y is a single variable. Of course the theory that we have developed sa far in this chapter applies to these rings. But, because of the special properties of PID's, we will show that we may construct Griibner bases using S-polynomials as we did in the case of fields (Algorithm 4.5.1). We will then define strong Griibner bases which are similar ta Gr6bner bases when the ring R is a field, and we will show how ta compute them in Theorem 4.5.9. We will also describe the structure of strong Griibner bases in the case of a polynomial ring in one variable over a PID. This will give us a lot of information about the given ideal as we will see in Section 4.6. We will also specialize the ring R to k[y], where k is a field and y is a single variable, and we will describe the relationship among the different notions of Gr6bner bases in this case. Recal! that in order to compute a Gr6bner bases for (I!, ... , l,) in A = R[xj, ... , x n ], we need to compute a homogeneous generating set for the syzygy 4.5. GROBNER BASES OVER PRlNCIPAL IDEAL DOMAINS 247 . module, Syz(lt(j,), ... ,lt(j,)) (Theorem 4.2.3). When R is a field, we saw that a generating set B for Syz(lt(j,), ... ,lt(j,)) exists such that every element of B has exactly two non-zero coordinates (these syzygies correspond ta S-polynomials; see Proposition 3.2.3). DEFINITION 4.5.1. A generating set B oJ Syz( (lt (j,) , ... ,lt(j,)) is caUcd an S-basis if every element of B is homogeneous and has exactly two non-zero coordinates. In general) when the coefficient ring is not a field, there is no S-basis for Syz((lt(j,), ... ,lt(j,)) (see Example 4.5.4). However, when R is a PID such a generating set exists as the next proposition shows. We assume that s > 1 in this entire section) because the case s = 1 is trivial, since any single polynomial is automatically a Grübner basis (note that this is not the case if R has zero divisors; e.g. {2x + 1} is not a Grübner basis in Z6[X]). We first prove the following identity4. LEMMA 4.5.2. Let R be a PID and let a, a" ... (a" , ... ,a')R: (a)R = L.: ((ai)R: (a)R). i=l ,a, be in R - {O}. Then PROOF. Since R is a PID, we have (see Proposition 1.3.8 for the case where R = k[x]) (a" ... ,a,)R=(gcd(a" ... ,a'))R and (see Lemma 2.3.7 for the case of polynomial rings) (ai)R n (a)R = (lcm(ai, a))R for ail i = l, ... ,R. Moreover, it is easy ta show that in any UFD we have lcm(gcd(a" ... ,a,), a) = gcd(lcm(a,, a), ... ,lcm(a" a)). Thus (a" ... ,a')Rn(a)R = (gcd(a" ... ,a'))Rn(a)R , = (lcm(a" a), ... ,lcm(a" a)) R = L.: (lcm(ai, a)) R = L.: (a,) Rn (a)R. i=l i=l = (lcm(gcd(a" ... ,a,), a))R = (gcd(lcm(a,, a), ... ,lcm(a" a)))R , The result now follows easily as in Lemma 2.3.11 and Proposition 4.3.11. 0 PROPOSITION 4.5.3. Let R be a PID and let J" ... ,J, be non-zero polynomiais in R[x" ... ,xn ], with s > 1. Then Syz((lt(f,), ... ,lt(f,)) has an S-basis. 4For ideals h, ... ,le in a ring R we define the ideal L:=1 Ii to be the ideal in R generated by the ideals Il, ... ,Il- That is, 2::=1 Ii = = (Il, ... , le). It where i8 easy to see that every element f E :2::=1 Ii can be written as f /1 + ... + li li E Ii for each i = 1, ... ,e. 248 CHAPTER 4. GROBNER BASES OVER RlNGS PROOF. For i = 1, ... ,s let lt(Ji) = CiXi, where Ci. E Rand lp(Ji) = Xi. For any subset J of {l, ... ,s} we deline, as before, XJ = lcm(Xj 1 j E J). We use the notation and technique presented in Theorem 4.2.9 to construct the desired generating set. Let Su = SYZ(CIXl, ... ,cuXa ) for 1 :": (J :": s. We will show by induction on (J that every Sa has an S-basis. Our induction starts at (J = 2. Then it is easy to see that S2 is generated by the syzygy ( -'XX , - -'- XX ), CIl C22 where c = lem(cl,c2)' and X = lcm(Xl ,X2 ). Now let (J > 2 and assume by induction that we have computed an S-basis 80--1 for Bd - l . \Ve now construct an S-basis Bu for Sa. Recall from Theorem 4.2.9 that Bu consists of two groups of elements. First, to each element a of Ba-l corresponds the element (a,O) in Ba. Clearly (a,O) has only two non-zero coordinates by the choice of Ba-l. The other elements in Ba_l are obtained from the ideal (Cj 1 jE J,j op (J)R: (Ca)R, where J is a saturated subset of {l, ... ,(J} containing (J. From Lemma 4.5.2 we have (Cj 1 jE J,j op (J)R: (Cu)R = L(Cj)R: (Ca)R. JEJ ji-u Nowfor jE J, j op CT, let dj be a generator of the ideal (Cj)R: (Ca)R. Therefore Moreover, as in Theorem 4.2.9, associated ta each dj we have the element S"J J = dje u X ju e· _ d X Ja e Cj Xj J J Xa a, since djc a E (Cj). Clearly the element sp has only two non-zero coordinates. By Theorem 4.2.9 the vectors SjJ, where jE J,j op (J, and J ranges over all saturated subsets of {l, ... ,(J} which contain (J, together with the vectors (a, 0), where a ranges over aIl elements of a generating set for 8 u- l , forrus a generating set for Sa. Therefore Sa has an S-basis. 0 We now give an example which shows that S-bases do not exist for rings which are fiot PID's. EXAMPLE 4.5.4. Consider the ring R = Z[zl. Ris not a PID, sinee, for exampIe, (2, z) is not principal. We note that Ris a UFD. Now consider the following polynomials in (Z[z]) [x, yi fI = 2xy2 + y, 12 = zx2y + x, 13 = (2 + z)xy + 1. We use the lex order with x > y. Then we have lt(JIl = 2xy2, lt(j,) = zx 2y, lt(h) = (2 + z)xy. The homogeneous elements of Syz(lt(fI), lt(J2) , lt(h)) with exactly two non-zero coordinates are multiples of SI = (-zx, 2y, 0), S2 = (-(2 + z), 0, 2y), or S3 = 4.5. GROBNER BASES OVER PRINCIPAL IDEAL DOMAINS 249 (0, -(2 + z), ZX). However, the vector (-x, -y, xy) is in Syz(lt(ft),lt(h), It(J3)) , and sa, if SI, S2, S3 generated Syz(lt(ft), lt(h), lt(h)), we would have (-x, -y, xy) = h l (-zx, 2y, 0) + h 2( -(2 + z), 0, 2y) + h3(0, -(2 + z), zx), for sorne hl, h 2, h3 E (Z[z]) [x, y]. Then xy = 2yh 2 + ZXh3' Note that h 2 = a2X and h3 = a3Y, for sorne a2, a3 E Z[x, y, z]. But then 1 = 2a2 + za3, and this is impossible. So Syz(lt(ft), lt(h), It(J3)) does not have an S-basis. In fact, it can be shown that if R is a UFD and if Syz(lt(ft), ... ,lt(J,)) has an S-basis for every ideall = (ft, ... ,/,) in R[Xl,'" ,Xn], then R is a PID (see Exercise 4.5.14). However, there are rings which are not UFD's for which Syz(lt(h), ... ,lt(J,)) has an S-basis for every ideal l = (fI,'" ,/s) in R[Xl"" ,Xn] (see Exercise 4.5.15). We note that the syzygies obtained in Proposition 4.5.3 are the analog of the syzygies used to deline the S-polynomials in the case where the coefficients are in a field. Indeed, iflt(Ji) = ",Xi,lt(Ii) = CjXj , where Ci,Cj ER, then the syzygy with exactly two non-zero coordinates corresponding ta these two polynomials is where c = lcm(ci,Cj), and X and!j as (4.5.1) = Icm(Xi,Xj ). We define the S-polynomial of fi cX cX S(j" Ii) = -;;; X/' - Cj X j k Proposition 4.5.3 can then be used to modify Algorithm 4.2.1 to obtain an algorithm for computing Grübner bases in R[Xl"" ,xn ], where R is a PID. This algorithm is presented as Algorithm 4.5.1. We note that Algorithm 4.5.1 is similar to the algorithm given in the case of R = k, a field (see Algorithm 1.7.1). We now give an example of how Algorithm 4.5.1 is applied. EXAMPLE 4.5.5. We go back to Example 4.2.11 and use Algorithm 4.5.1 to recompute a Grübner basis with respect ta the lex term ordering with x > y for 1 = (fI, 12), where ft = 3x2y + 7y, 12 = 4xy2 - 5x and R = Z. We initialize G = {JI, h} and 9 = ft,h . Since 9 '" ,we choose {JI, h} E g, so that now 9 = 0. We compute C = lcm(3, 4) = 12, X = lcm(x2y, xy2) = x 2y2, so that the corresponding S-polynomial is S(Jl, 12) = 4yft - 3xh = 15x 2 + 28y 2 This polynomial cannot be reduced and we add h = 15x2 + 28y 2 to the basis, so that now G = {Jl,hh} and 9 = {{ft,h , h h . Since 9 '" 0 we choose {JI, h}, so that 9 = {{h h}}. In this case we have c = 15 and X = x2y, sa that the corresponding S-polynomial is S(Jl, h) = 5ft - yh = -28y3 + 35y. This polynomial cannot be reduced and we add !4 = - 28y 3 + 35y to the basis, so that G = {ft,hh!4} and 9 = {{hh}, ft,!4 , h!4 , h!4}}' Since } ç R[Xl. In the case when the coefficient ring is a PID. {ho 14}}. Mimer gave an analogue of crit2 whicheliminates many S-polynomial computations. Grobner Bas.= F. 9 := {{f.. We note that this computation required more steps than the computation of this Grobner basis did in Example 4.. In this case. and X = X2 y 3. sa that 9 = {{h 14}. Therefore we get (as we did in Example 4.g} E g.1t(g) = CgXg g:= 9 . lS mmunal wlth respect ta G IFhiOTHEN 9 := 9 U {{ u. h ho 14} is a Grobner basis for J. It is easy ta see that the other two elements of 9 also give rise ta polynomials which reduce ta zero and sa do not contribute new polynomials ta the basis. Let It(f) = cf X f .}. where h eX G Cg X g . {h 14}.. h} G:=GU{h} 1 for all u E G} ALGORITHM 4.11 nsing Algorithm 4. a Grobner basis for (fI. ~ s) OUTPUT: G = {g" .1..13 because of the problem of dividing by elements of the coefficient Ting R. g} WHILEgi0DO .2. 80 that 9 = {{f" 14}.5. sa that the corresponding S-polynomial is Since 9 i 0..Xn] with li i 0 (1 E G} ~. In arder ta improve the efficiency of Algoritbm 4.2. . The interested reader should consult [Mo88].1.s Algorithm over a PJD 9 i 0 we choose {h h}. we choose {f" 14}.5. GROBNER BASES OVER RINGS INPUT: F = {f" . C = 84.11) that {fI.g.{{j.'" .250 CHAPTER 4.1.) Iii 9 Choose any {f. sa that the corresponding S-polynomial is S(f" 14) = 28 y2fI + 3x2/4 = 105x 2y + 196y 3 A 196 y 3 - 245y~ O.. {ho h}}· In this case we have C = 60 and X = x2y2.g ---->+ h.g}} Compute e = !cm(cf' Cg) and X = lcm(Xf' X g ) cX --1 cf Xf .2.xn] the way we did in Definition 4.l.. .1.2.. Recal! that we defined Grobner bases in R[Xl"" .··· INITIALIZATION: G . there is a stronger version . Let lt(fi) = CiXi. every non-zero ideal in R[Xl1 . z in J.6).. there exisls an i E {l... x n ].s} 1 Xi divides lp(f)}. .))..5.xn ]. f. DEFINITION 4. fs) = (lt(ft). We also have lt(f) E (lt(/j) 1 j E J) and so le(f) = LjE] djej for sorne dj E R. .5.. .g. ç R[XI.9. 80 let {ft. Then the set {!J 1 J is a saturated subset of {l.} is a Grobner basis for J.. Jf G = {gl. from which we conclude that C] divides le(f).. . Let R be a PJD. D . . Then we say that Gis a strong Grobner basis for J = (gl. Let R = k[y. . PROOF. .5. This is because there are an infinite number of non-associate irreducible polynomials in the two variables y.5. . .. then each of these irreducible polynomials would have leading term (which is the irreducible polyuomial itself) divisible by the leading term of an element of G. .8.} then it is a Grobner basis.16).. . . .s}} is a strong Grobner basis for 1. GROBNER BASES OVER PRINCIPAL IDEAL DOMAINS 251 of Gr6bner bases which is similar ta the one we gave when the coefficient ring was a field.. .) if for each f E l. let C] = gcd(cj 1 j E J) and write c] = LjE] ajcj (any such representation will do)... Also. . Assume that {ft. .} be a Grobner basis for an ideal l in R[XI.g. . . This would force an infinite number of elements in G. LEMMA 4.5. However strong Grobner bases exist onIy in the case when R is a PID (Exercise 4. Therefore lt(!J) divides lt(f). . z]. . let X] = lem(Xj 1 jE J). s}. Let 0 # f E J. . If a strong Grobner basis G existed. Let G = {gI. which violates the definition of a strong Grobner basis. Consider the ideal J = (x.. . f.} be a set of non-zero polynomials in R[XI. y. . and J be an ideal of R[XI. It is clear that J is saturated and that X] divides lp(f).. . lt(f. Note that this definition does not in itself require that R be a PID. The following result is immeruate and its proof is left to the reader (Exercise 4...xn ]. .. x n ] is a strong Grobner basis. Consider the polynomial THEOREM 4..xnl has a strong Grobner basis. .. z) in R[x]. where Ci E Rand lp(/i) = Xi' For each saturated subset J of {l.6. Then lt(f) E Lt(ft..4. .. In particular.. We now show how ta construct a strong Gr6bner basis frOID a given Grobner basis.7. where k is a field. .... .5. We say that G is a minimal strong Grobner basis if no lt(gi) divides lt(gj) for i # j..··. Let J = {i E {l.5. .. .g... This ideal does not have a (finite) strong Grobner basis. EXAMPLE 4. t} such that lt(gi) divides lt(f). 15x 2 y . 4. One can show that for a set G of non-zero polynomials in R[XI. = xy3.3. = xy12 .28xy3 + 35xy . Thus J = 7h.xn ].)) (see Definition 4. = gcd(3. . h. we could also define the concept of strong reduction. Together with the concept of strong Grübner bases. .28y3 + 35y. and J. Then we have + J -5. -28) = 4.4}. J3 = {4}. a strong Gr6bner basis for l is {4xy2 . J6 = {1.9 we first compute all the saturated subsets of {l. h. In this setting we have three concepts of Grübner bases: Grübner bases in k[y.1).28y3 + 70y2 28y3 + 70y2 .. = {2}.I)J4 and since h. = h. We note that the first reduction above could not have been done with a single polynomial from fJ. give rise to the original polynomials h.yfJ = x 2y2 .28xy 3 -7y. 12 = 4xy2 . J4' We now consider the case where R = k[y]. where fJ = 3x2y + 7y. = gcd(4. and h. FoIJowing the proof of Theorem 4. Consider the polynomial J = 7x 2y2 .252 CHAPTER 4. The sets J" J 2. h = 15x2 + 28y2.5x. J 7 = {l..5. J4 = {l. J5 = {2. ha = J4.1)f<.5.5. Grübner bases in (k[Y])[XI. We saw that a Grübner basis for the ideal l = (3x2y + 7y.5.28y3 . and strong Grübner bases in (k[y]) [XI.15) = 1.'" . and X J . .2.10. we have that every element in (G) strongly reduces to zero if and only if G is a strong Grübner basis (Exercise 4..56y4 .35x 2 28xy3 + 35xy .'" . 15) = 3.9).3}.'" . 15x2 + 28y2. 3x 2y + 7y.4. EXAMPLE 4. = x 2y2 Therefore h.15. To conclude this example we give an exarnple of determining ideal membership using strong reduction. 12.x 2y2 .3}. h.28y3 21y2. and therefore.y2 fJ = yh. .5x) ç Z[x.. J4}.h + 35xy 4 - 56y4 . h. GROBNER BASES OVER RlNGS We leave it to the exercises to show that every ideal has a minimal strong Grübner basis (Exercise 4.). 4xy2 . Theorem 4.. . and X J . = gcd(3.1. Associated with J4 we have cJ. Therefore h. = x2y. = x12 .il ~ -15x 2y . and J4 = -28y3 + 35y.9 gives us a method for computing strong Grübner bases.· We see we do not need h. We go back to Example 4. . and X J . 2. Therefore.lt(f.28 y 3 + 35y ----> + 70y2 + 35y + 35y -:y.yfJ we have J = (-7y . .xn ] (as defined in Section 1. 12.XI. so h.} is performed if lt(f) E (lt(f. Reduction of J modulo a set F = {fJ. J. = fJ. Associated with J 5 we have cJ. We give an illustration in the next example.5)fJ + (7x .xnl (as defined in Section 4.1).5x 2 _ 7y2}. = x12 .5.4}. = 12. = y12· Associated with the set J 6 we have cJ.2.5fJ . with k a field and y a single variable.6). 4}: J. 3.7y2 Associated with the set J 7 we have Ch = gcd(3. Strong reduction would require instead that an lt(fi) divides lt(f).5.Xn] (as . . J 2 = {3}. yi with respect to the lex term ordering with x > y is {fJ.5x 2 .7y)12 + (2y .-28) = 1 and X h = x 2 y 3.56y _56y4 .7y12 + (2y .1)..5..5x. O. .} is a Gr6bner basis with respect to an elimination order with the X variables larger than the y variable in k[y. GROBNER BASES OVER PRINCIPAL IDEAL DOMAINS 253 defined in Definition 4.. x" . . . The next theorem strengthens this result considerably.Xn]. ..gt} is astrong Gr6bner basis in (k[Y])[XI"" . .XI.)"" . . and we will use It x1 1px'lcx for corresponding abjects in (k[Y])[XI'''' .x n ] and (ii) for every J ç {l.. which we may assume are terms with respect ta the variables Xl.Xn ]..'" . but. G is a Gr6bner basis in R[x!. by definition of J. . We use It. X" . We need a notation to distinguish between leading terms in k[y. Then J is saturated. Let c = gcd(lcx(gj) 1 j E J) and write c = 2: jE J dj lcx(gj).xn ]. let G be a strong Gr6bner basis in Rix!.Xn with coefficients in k[y].'" 1 X n ]. Let J be a saturated subset of {l. satumted with respect to {IPx(g. ..xn ] is a strong Grobner basis if and only if (i) G is a Grobner basis in Rlxl. . .XI.xn ]. Let f E J... for sorne .X n ]... g. x n ] and (k[Y])[XI"" . .. we have Itx(f) E (Itx(g. 1 xnl. X" . Let R = k[y]. .1.g. for f E k[y. Let J = (g" .xn ].. .xn] then it is a Gr6bner basis in (k[Y])[XI'''' . . we get ltx(gi) divides \tx(f) as desired..5..t} IIPx(gj) divides IPx(f)}..5. Sinee G is a Gr6bner basis in R[xl..9). Note that because of the chosen order. . We will first need a lemma.. . Then.18 that if {g". then it is a Gr6bner basis in (k[y])[x". if f E k[y. there exists i E J such that lcx(gi) divides lcx(gj) for ail jE J. we saw in Theorem 4. . JEJ for sorne aj E R[Xl1 . Clearly the ring k[y. . . by Lemma 4. Moreover..t}.'" ...xn ].5.8.11..8 that if {g" . . We have seen in Lemma4. . 80.. ..4.. We first prove that (i) and (ii) imply that Gis a strong Gr6bner basis in R[xl. ..".} ç R[XI. Xl. for example. then lex(f) is a polynomial in y. ." . in the latter ring.5.. Let J = {j E {l. Ip.5.. .. Then lcx(f) = L: lcx(aj) lex(gj). . . x n ] (this proof is very similar to the proof of Theorem 4.xn ].t}. PROOF. .IPx(gt)}.6).x n ].)"" . Finally for a E k[y] we use Ity(a) for the leading term of a.5. polynomials are viewed as polynomials in the variables x" . Then G = {g" ..gt). le for their usual meanings in k[y. JE] Now choosing i E J as in (ii) we get lcx(gi) divides lcx(f) and sa.Itx(gt)).xn ]. Moreover we have It x (f) = L: aj Itx (gj). In both results we assume we have an elimination order with the x variables larger than y. Conversely..xn ] we have It(f) = Ity(lcx(f)) IPx(f)· LEMMA 4...... .Xn with Ipx (f) = Ipx (aj) Ipx (gj) for ail j E J such that aj # O.xn ] is the SaIDe as the ring (k[Y])[XI"" . We can write e = 2. . Thus by Lemma 4. Let us first assume that G = {gl. Since G is a Grübner basis in k[y. lcx(g. .)·divides lcx(gj) for all j E J.Xn] with respect to an elimination order with the X variables larger than y. It(f) = lty(e)X.: jE s bj lcx(gj).xn ].. 0 We now go back to the case of an arbitrary PID Rand conclude this section by giving a eharacterization of strong Grübner bases in R[x]. which immediately gives the desired result.Lbj~( .. Therefore G is a Grübner basis in k[y. let X = lcm(lpx(gj) 1 j E J). Therefore i E J since IPx(gi) divides X. Conversely. AIso.} is a Grobner basis in k[y..)gj.. PROOF. . Now consider the polynomial f =.254 CHAPTER 4. Let f E J. saturated with respect to {IPx(g.. Moreover.g.xn ].. .. This will be used in Section 4.. there exists gi E G such that It(gi) divides It(f) = Ity(e)X. By Theorem 4.13.X n]. . . From Lazard [Laz85] (see also Szekeres [Sz]).gi} is a minimal strong Grobner basis for an ideal J of R[x] and . Therefore. let us assume that G is a strong Grübner basis in (k[y]) [Xl.5. . But then i E J since IPx(gi) divides IPx(f) = X.11 it suffices to show that given a subset J of {l.. Ity(lcx(gi)) divides Ity(lcx(f)) = Ity(e)... X" .18 we see that Gis a Grübner basis in (k[Y])[Xl"" . So let J be such a saturated set. . we have THEOREM 4. .12. Xl.t}.).) divides e and we are done.'" .xnl with respect.xl.5..IPx(gt)}. since we have an elimination order with the X variables larger than the y variable. Xl..xn ]. Then there exists gi E G such that Itx(gi) = lex(g.5. Since G is a strong Grübner basis there is a gi E G such that Itx(gi) divides Itx(f). .1.gi)' Note that Itx(f) = eX.'" .xn ]. deg(lcx(gi)) S deg(e) and so the fact that c divides lcx(gi) implies that e is a non-zero constant in k times lcx(gi) . . . But then It(gi) = Ity (lc x (gi)) IPx(gi) divides It(f) = Ity(lcx(f)) IPx(f). Let e = ged(lcx(gj) 1 j E J) and X = lcm(lpx(gj) 1 j E J). 0 THEOREM 4. Let R = k[y].g. .} is a Grübner basis in k[y. Then G = {g" ......'" . where X is a single variable. there exists i E J such that lcx(g.ta an eliminaiion order with the x variables larger than y if and only if G is a strong Grobner basis in (k[y]) [Xl. Assume that G = {g" . GROBNER BASES OVER RlNGS dj E R = k[y].)gj JES X Px gJ E (gl. . for sorne bj E k[y]. .. Moreover. Let R be a PJD and let X be a single variable.. Now consider the polynomial X f= Ldj~( . jEJ Px gJ Note that Itx(f) = eX and f E J.) IPx(gi) divides Itx(f) = lcx(f) IPx(f)· So IPx(gi) divides IPx(f) and lcx(gi) divides lcx(f). AIso.6 to give the primary decomposition of ideals in R[x]. 1 PROOF. Thus we may assume that gCd(Y1"" . for i = 2.+1 divides c.L Ci+l C· Let h = -'-Yi+1 .9i)' and hence CLAIM 4. ci+d. Thus Vj :'0 Vi+1. .... 1 :'0 i < t. such that Vi = Vi+l..+1) and write c = biCi + bi + 1 Ci+1. .t}. Let c = gcd(Ci. . . there exists aj E {l.. Therefore C.. at hi9 at h t-19 ht9..Yi). such that lt(Yj) = CjX Vj divides lt(h). there exists a j E {l. and so. we have that j = i + 1. .. Ci+l divides Ci. Note that lt(h) = ex v"" Since G is a strong Gr6bner basis..1. . l t . Therefore hE (gl)'" . for i = 1. '" . and since h ~+ 0. Let i E {l. . and lt(Yi) divides lt(Yi+1)' This contradiets our assumption that G is minimal. we have that i = j. .. and so Ci+l = C = gcd(Ci. Sinee Gis a strong Gr6bner basis.Yt} is a strong Gr6bner basis if and only if { ~ ...-v'Yi. we have that {YI" ' . for sorne bi1 bi + 1 E R.. and ej divides c. .~ } is a strong Gr6bner basis.5. where Ci E R (1:'0 i:'O t). t. Now consider the polynomial h = bixv. for sorne bi. Ci+l) and write c = biG + bi + 1Ci+l. We further assume that we have ordered the Yi 's in such a way that VI :'0 V2 :'0 .lp(Yt) are larger than lp(h). .+.5.Yt).-v'Yi + bi+1Yi+1 E J. .. . let e = ged(ci. CLAIM L VI < V2 < ' .'" . Then Vj :'0 Vi' But then lt(Yj) divides lt(Yi). Note that h E J and that lp(h) < CHI e· lp(Yi+1)' Sinee lp(Yi+1).t} sueh that lt(Yj) = CjX Vj divides lt(h) = ex"'.. . 80 is ~9i+l' Ci+l YI E R.. CLAIM 2. the only polynomials that cau be used to recluce h to zero are 9b . . GROBNER BASES OVER PRlNCIPAL IDEAL DOMAINS 255 let lp(gi) = xv'. PROOF. < Vt. . byour assumption that G is minimal.t -1}. Henee c = Ci. Then YI Y2 Yi Yt-1 Yt a2a3'" a3'" at9 ath29 ai+l . Assume to the contrary that we have an i. for i = 1. From Exercise 4.xVH.11. PROOF. PROOF. :'0 Vt· Let Y = gCd(Y1.C. and so lt(YJ) divides lt(Yi+1)' Since G is minimal... As in Claim 1.t . 19i. . bi + 1 E R... . CLAIM 3.Yt) = L Let lt(gi) = Ci XV . and so j :'0 i + 1. Consider the polynomial h = biYi + bi+1Yi+1' Then h E J and lt( h) = ex v.4.. -'-Yi+1 E (YI.'" . Ci+l The result follows after dividing by aH1 . . Therefore YI = c(y.and hi = ---=-.t-2 Ct-l ct = at t-l = at_lat t-2 h h ct 9i Y2 YI CLAIM a3'" at_lath2 a2a3' .. at· 6. .Yi) by Claim 3. a.Yi· Since ~9i+1 E (91.. Then we have CHI Ci y. We use induction on i.5. ..12). GROBNER BASES OVER RINGS PROOF. any set of polynomials YI.256 CHAPTER 4. . The case i = 1 is clear. PROOF. Then. Then 91 = C(91)P(91).'s have no factor in cornmon. . Now set ai+l = . .) E R. The case i = 1 is clear. ..y.9i' Since ~9H1 E (YI. We show by induction on i that P(91) divides Yi. . . C· y.t and ct = 1. . CLAIM 5. Now. where p(91) E R[x] and the greatest cornmon divisor of the coefficients of P(Y1) is 1. by Claim 2. By Claim 3 we have ~Yi+1 E (91. l Ci divides 9i. hi+l E (hil aihi-l. in R[x] which satisfies the conditions in Theorem 4.Y2. Ci+l But gcd (~.... we see Ci+1 that Ci divides ~gi+l' Therefore Ci+l divides 9i+l.9i)' that is.'" .. sinee 91 = CI. since P(Y1) divides every Yi...' . That is. P(91) must be 1 because of the assumption that the y. .··· .P(Y1)) Ci+l = l.. 0 We note that the converse is also true. we have that Ci divides Y1.Yi) by Claim 3. The Theorern is now completely proved... Y'-l cth. since any cornmon factor of ~ and p(91) would have to be in Rand would have to be a factor of the coefficients of powers of x appearing in P(Y1)' Therefore P(Y1) divides Yi+1' Now.. Ci divides Yi for ail i = l. .5. C2 divides 92. = h.'" . . 92. sinee Ci divides gi Ci+1 for each i.13 is a strong Grübner basis and is minimal if no ai is a unit (Exercise 4. .a3'" aih21 a2'" ai}. and sinee Ct divides Ci for each i. we see that Ci+1 P(Y1) divides Ci+l ~YH1. Now assume that P(Y1) divides YI. ct-l h t-l = C'_l h ct-t-l Y'-2 ct-2 h t-2 = C'-l h cC'_2 t-. Assume that Cl divides 91. Therefore Ct = 1 because of the assumption that the Yi '8 are relatively prime. Let C(91) E R be the greatest cornmon divisor of the coefficients of the powers of x that appear in 91. we see that Ct divides 9i for each i. PROOF.. < V" and for i = 2. Let G = {g" .. PROOF.1) = a2a39 70x 2 + 70x . It can be easily verified nsing Lemma 4.5.).5. We consider the ring R = lQI[y] and the ideal l = ((x + y)(y2 + 1). .12) that {J" h h} is a strong Grübner basis.5.13. 91 92 a2a3'" at9 a3'" ath29 9i ai+l . 5 .5.ht . h} is a strong Grübner basis. . . x]..70x -210 (x 3 + 6x 2 + 20x + 15) . x] and let IPx(gi) = We assume that we have ordered the 9i 's in Buch a way that Vt· Let 9 = ged(g" .5. 0 EXAMPLE 4.. .5. VI :S V2 :::.4.t. y] witb respect to the lexicographie arder with x > y (Theorem 4. athi9 9t-l 9t a. 14(x .16.1 9 h'9.5. We also note that the fi's have the form required in Theorem 4.5. We find + 91 92 93 = = = y4 + 2y 3 + 2y2 + 2y + 1 = (y + 1)2 (y2 + 1) = a2 a3 xy2+x+y3+y=(y2+1)(x+y)=a3h2 x 2 .1l that {J" 12.. We assume that we have the lex term ordering with y < x. lt(hi ) = xv.1) = h39..630 = 9 . x be variables and set A = k[y. Consider the three polynomials in Z[x] fI 12 h = = = = 630x .gt} be a minimal Grobner basis for an ideal lof k[y.15..5.13 (Exercise 4. 14(x .X + Y + 1 = h3· EXAMPLE 4... GROBNER BASES OVER PRINCIPAL IDEAL DOMAINS 257 COROLLARY 4.140 = 5(x + 2) . Let k be a field and let y.13. 14(x . We compute a strong Grübner basis for l by computing a Grübner basis for l viewed in lQI[x..5. x + 2) and so it also follows from the COnverse of Theorem 4. VI < < . since h3 = x 3 + 6x 2 + 20x + 15 = (x + l)(x + 2)(x + 3) + 9(x + 1) E (9. We have from Theorem 4.12.. that G is a strong Grübner basis in (k[y])[x] and so the result follows immediately from Theorem 4..5. x 2 .12).14.1) = a3h29 14x4 + 70x 3 + 196x 2 .. Then XVi. 1/2 where ai E k[y].X Y + 1) in R[x].. .g. . ::::. . (see Equation (4.5x 2y 3 .. . 4. where R is a PID. -30x 3 y2 + 6x 3 .2. as lisuaI.. when we iterate the preceding. Prove that every non-zero ideal of Rix" . g.'" .5. x n ] where R is a PID and 9 = gCd(gl.5. Xn]. . GROBNER BASES OVER RlNGS Exercises 4.7 (for Exercise 4. 4. Prove that if {g" . has a minimal strong Gr6bner basis.lt(~)) = 1. Let J. g. h = 3y .] Prove the converse of Theorem 4. . G is a strong Gr6bner basis for J. h = lOx 2y + 2xy and lex with x > y. . 9 E Rix" . g.9.5.x.2.. Ir = 6x 2 + y2.5.3yz + y.). We deline strong reduction in Rix" .258 CHAPTER 4. 4. i ::.6. g.. s we have lt(fi) divides lt(f) and 9 = J . xn].. x n ] as follows.5 part cl. then {gl.} is a strong Gr6bner basis if and only if {~. with J.) IHint: Follow the proof of Lemma 3.5.. J.g E Rix" .g).8z 2 and deglex with x > y > z..5.} of non-zero polynomials in Rix" . b. .. prove that if the set of !J delined there is a strong Gr6bner basis for J then {Ir. ....8.1 to compute a Gr6bner basis for the ideals generated by the given polynomials with respect to the given term order. .5. Compute Ji strong Gr6bner basis for the ideal in Zlillx.5. For ail J E J we have J -->+" O. Prove Lemma 4. h = 5x 2z .5.5.)) k We write f ~+. x n ] we write J --"--" 9 provided that for sorne i. . Show that in Exercise 4. 4.9. (This is the analog of crit1.4. with respect to a set F = {J" . Prove that {J.l~tU. 4.5.10. a:} . J.6x 2 y2 + 16x2y + 5xy3 + xy2 + 2xy .. . 4. and let d = gcd(f. b. Show that for the strong Gr6bner basis constructed in Exercise 4.1 ::. Ir = 3x 2y . .5.5.8..} is a Gr6bner basis for J..11.1.} is a set of non-zero polynomials in Rix" .. y.5y 5 + y" strongly reduces to zero.. 4. IHint: Look at the polynomials Ir = 2x + 1 and h = 3y + x in Zlx. x n ] where we set J = (G)..3. For the ring R = Z use Theorem 4..xn]. is a strong Gr6bner basis.s g.. a. .5. . for R a PID.] For the ring R = Z use Algorithm 4.5. ...1 the following statement: e 4.3.5. Show that the following are equivalent for a set G = {g" . That is.5.g} is a Gr6bner basis if and only if gcd(it(~).1. y].9 to compute a strong Gr6bner basis for the ideals generated by the given polynomials with respect to the given term order in the exerCÎses in Exercise 4. .7. .5. a.g oF 0.5. 4.5..5. Ir = 2xy ..} of non-zero polynomials in Rix" . z] in Exercise 4.1)) does not imply that G is a strong Gr6bner basis. For J.5. .x 2 and lex with x < y...5. c. 4. Assume the facts that if every prime ideal in a UFD is principal then it is a PID.2. Recall that an ideal P in a commutative ring A is prime if f g E P implies that either f E P or g E P.5.7. .3 (note that more than one dj may be needed).. x n ].12. . Let A be a Noetherian ring and let S be any non-empty set of ideals of A..4 that . We have . Verify that the Grobner bases for (_x 2 . In our setting.P is a multiplicative set (see Section 4.5.14.] 4. and that every non-principal prime ideal in a UFD cantains an infinite number of non-associate irreducibles.} ç: R[x" . then R is a PID.5.xn ] we have that Syz(lt(!J). prime ideal .5 to "decompose" ideals in R[x]. b).lt(f.)) has an S-basis. . The decomposition we have in mind is one similar ta the decomposition of natural numbers into products of powers of prime numbers. then there exist a. e. the analog of a product is an ideal intersection..6... .] 4. . where R is a PID and x is a single variable.4). We will need the following elementary fact about Noetherian rings.. Prove that if R is a UFD and iffor every non-zero ideal l ç: R[x" . or equivalently. Prove the converse of Theorem 4. Then S contains a maximal element. 4.14. -5y + 5xy3xy2 + 3x 2 y2) ç: Q[x. i. . then R is a PID. . and the analog of a prime integer is a prime ideal.lt(f.. 4. then for every {J" . 4.5. In this section we follow Lazard [Laz85] and use the results of Section 4.5.] 4.Ji= p lep n P. f.Ji for an ideal l in A.13. ...5.5. bER such that gcd(a. l has a strong Grobner basis.. there is an ideal lES such that there is no ideal J E S such that l ç J.6. a + b). [Hint: Prove the identity in Lemma 4.5.6..13..5. .)) has an S-basis.xy + x 2 y2 + xy3.6. .4. [Hint: If R is not a PID. x n ] we have that Syz(lt(fIl. [Hint: Use the idea in Example 4. l. Prove that if R is a UFD and iffor every {!J. Let l be an ideal in a Noetherian ring A.2. b.l.13 as stated immediately aIter the proof of Theorem 4.Ji = LEMMA {f E Air El for sorne vEN }. -1) E Syz(a.2 and follow the proof of Proposition 4. y] with respect ta the lex ordering with x > y has the form stated in Corollary 4. Note that (l. if the set S = A .. . 4. Primary Decomposition in R[x] for R a PID.16. fs} ç: R[x" . Show that if R is a Dedekind Domain. PRIMARY DECOMPOSITION IN Rix] FOR R A PID 259 4.5. . b) = 1 and 1 rt (a. LEMMA We first consider the decomposition of the radical Recall from Definition 2.5.15. if E Vi for sorne v E N. if Q were the intersection of powers of prime ideals. DEFINITION 4. we see that powers of maximal ideals are primary. ConsidertheidealQ = (4.X2) inZlx]. EXAMPLE 4. Anyprimeideal which contains Q must contain both 2 and x.Vi. Now consider the collection S of all ideals of A which contain VI and have empty intersection with S. n ' prime ideal and we 0 btain a contradiction. . one might think that any ideal in Rix] ean be decomposed as the intersection of powers of prime ideals. We now consider the reverse inclusion. since VI E S.3. By Lemma 4.13.1. which is a contradiction. which is a contradiction. VI E N. We now prove that M is a prime ideal. Therefore. and a. and hence must be equal to M = (2. D P In view of the above lemma. Note that n [CP prime ideal M E I. But Mn S = 0. there exists an ideal MES which is maximal in S. Now we have a prime ideal M which contains VI and henee I. Clearly S is non-empty. Therefore there exist v. x). In particular. Let gh E M and assume that neither 9 nor h is in M. VI ç M and M n S = 0. for otherwise.1. The inclusion Vi ç nP ICP follows from the fact that if a prime P prime ideal ideal P contains I. a' E A Bueh that ag +m But then = rand a'h+m' = r'. using Lemma 4.260 CHAPTER 4. This is not the case as the following example shows. Consider the set S = {Iv 1 v = 0. EM EM -'. Sn Vi = 0. M) n S of 0 and (h.. m. But M3 ç: Q ç: M 2 80 Q cannot be a power of M.6. The correct analog to powers of primes is the following. However 1 powers of prime ideals need not be primary (see IAtMD]) although. sa f E P ç M byassumption. it also cantains Vi. .6. and this implies that f E VI. then (JV)M = for sorne i" E N.4. An ideal Q of A is called primary if Q then either f is in Q or some power of 9 is in Q. x) "" Z2).6. and 80 M is prime. of A and if fg E Q It is easy ta see that prime ideals are primary. By maximality of M we have P r r (g.6. GROBNER BASES OVER RINGS PROOF. }. it would be a power of (2. x). M) n S of 0. ICP ES r+ v ' = (ag +m)(a'h + m') = (aa') (gh) + (ag + m)m' + (a'h)m E Mn S.. Sa let us assume that there is an element f in P . sinee M is a maximal ideal of Zlx] (sinee Zlx]/(2. m' E M. 2. and the obvious statement that we can remove superfluous Qi. Otherwise. where at. we see that any ideal that has a primary decomposition olso has an irredundant primary decomposition. If Q is a primary ideal in a Noetherian ring A. ° ° LEMMA 4.6. where ht.6. and let f cf' Q. We cali n~~l Qi a primary decomposition of 1. ag. Given this. or 3 (notethat rf andrg are the totally reduced remainders of f and 9 as defined in Definition 4. PRIMARY DECOMPOSITION IN Rlxl FOR R A pm 261 EXAMPLE 4. Then (fg)" = rg" E Q and so r E Q or (g")!' = 9"!' E Q. We cau now define what we ruean by decomposition. Also. . First we prove that every ideal in A is a finite intersection of irreducible ideals. and rt = atx + bt . An ideal 1 is irreducible if 1 = Ir n 12 implies that 1 = Ir or 1 = I. The second statement follows from the fact that any prime ideal containing Q also contains yfQ. .8. PROOF. 1 :S i :S we have n#i Qj 't Qi. M oreover yfQ is the smallest prime ideal containing Q.3 is primary. X2). the ideal Qi is said to be the primary component of 1 which belongs to Pi. 4 divides atbg + agbf and bjbg. in Example 4.3.6.5. hg E Q.9. Since fg E Q. rg = agx + bg. and then we show that every irreducible ideol is primary. in addition. and 80 since af and bg are non-zero. In this latter case. and we would be done. Therefore f E yfQ or 9 E yfQ and yfQ is a prime ideol. let fg E Q = (4. the Pi are ail distinct and for aU i. where Qi is Pi-primary for each i. Let 1 = n~~l Qi. DEFINITION 4. we have THEOREM 4. It is easy to prove that if Q" . we caU the primary decomposition irredundant.. Moreover. It is easy to see that we can write f = ht+rf.2 we gave a primary decomposition of VI.5.2). so again we have g2 E Q. since f cf' Q..6. In general.. Let fg E yfQ. The key to the proof is the concept of irreducible ideals. The ideol Q given in Example 4.Qr are ail P-prmary then n~~l Qi is also P-primary. we have that either af or bj is not equol to O. bg = 0.7. we say that Q is P-primary. we have af = bg = 2 giving g2 E Q. bf # and so bj = bg = 2.6.6. So we may assume that bg # O. and Pi is said to be a prime component of 1.6. and thus was its own primary decomposition. IfQ is primary and yfQ = P. We note that in Lemma 4. then g2 E Q. The proof of the theorem is done in two steps. To see this. Every ideal in a Noetherian ring A has a primary decomposition. the ideal Q was primary. r.6. Note that if bg = 0. 1.6.4. then yfQ is a prime ideal.6. PROOF.bt . If bf = then 4 divides afb g. and 9 = hg + r g . 0 DEFINITION 4. If. {O} be an ideal in R[x] and let {g" . sinee M is in S. and hence g' E 1. Let 1 # R[x]. By Lernma 4. M is not irreducible. CLAIM. For the reverse inclusion.fJ E 1. we assume that linear equations are solvable in R..g. Therefore r E 1: (g'+') = 1: (g'). We will be using Theorem 4. Consider the ascending chain 1: (g) ç 1: (g2) ç . Now let 1 be an irreducible ideal. GROBNER BASES OVER RINGS Let S be the collection of ail ideals in A which axe not the intersection of a finite number of irreducible ideals.5. By the maximality of M. Z and (ZlpZ) [y] for a prime integer p. Clearly 1 ç (1 + (l)) n (1 + (f)).. . and M 2 are not in S. The main purpose of this section is ta use Theorem 4. There has been mueh work done on this problem. Now by the assumption that 1 is irreducible. Thus M ç: M" M 2 .5. Namely. R.s E A.6. El El 5For certain PID's. .) = l.13 and the notation set there.7 and Corollary 4.5. But then M is also a finite intersection of irreducible ideals. we have that M. Therefore there exist ideals M. n M 2 . But then '" + g'r E 1.. r. Sa let f 9 E 1 and let f rte 1.1 there exists a maximal element M in S. see [GTZ. Then rg'+' = '-v------' -g'" + gfJ + _ sfg E 1. where "'. this definition is more restrictive than the one usually found in the literature.gt} be a minimal strong Grübner basis for 1. Since A is Noetherian we have that for sorne e. and the Claim is proved. either 1 = 1 + (f) or 1 = 1 + (l). Therefore they are both a finite intersection of irreducible ideals..13 ta show how ta do this explicitly in the ring R[x]. ç 1: (g') ç . that is. For simplicity we will assume that 9 = gcd(g" . Examples of such rings include Q[y]. 1: (g') = 1: (g'+'). PROOF. It follows from Theorem 2. ' . In this case we say that the ideal l is zero-dimensionaI 5 . and that given any prime element uER we can factor in (RI (u) )[x]. We will show that 1 is primary. and therefore S = 0.2. yi.. and M 2 such that M # M" Mol M 2 and M = M. Since f rte 1. where R is a PID in which certain computability assumptîons must be made.14 that this definition coincides with the one given before for the special case of lQi[x. . Now. let '" + rg' = fJ + sf E (I + (g')) n (1 + (f)). Therefore 1 is primary. This is a contradiction. it gives no indication how ta go about computing the primary decomposition of a given ideal in same specifie ring.262 CHAPTER 4. EHV]. Assume ta the contrary that S is not empty. we have 1 = 1 + (g'). 0 We note that the preceding Theorem is purely existential. (I + (g')) n (1 + (f)) = 1... that we can factor in R. hi)' So. .. We also have RlxJl(u.: P if and only if there exists i ::0: 2 such that (ai.i we have 9j E (ai.. hi E P.4. If (ai.. The image hi of hi is in P. hi) for j = i + 1. P be ideals in Rix] with 1 zero-dimensional and P prime. for i = 2. . PIDMARY DECOMPOSITION IN Rix] FOR R A pm 263 That is. since 9j is a multiple of hj we have 9j E (ai. P by the choice of i. First note that for all j = 1. hi)' Thus.... at a3' . Note that the ideal (u) is now a maximal ideal of R.: P then P = (u. Let v E Rix] be a pre-image of il. that is.: P. and sinee u E P. let i ::0: 2 and assume that ai... Since Rix] is a PID. . Therefore (u..: P.. Since il E P..a2' .. . THEOREM 4. athi gt~l atht-l 9t hl. Since hj E (h j . at '1:. and so il is an irreducible factor of hi. P # {O} and so P is a maximal ideal of Rix]. we have 91 92 a2a3 . . with the notation above. and hence R = R/ (u) is a field and Rix] is a PID. Now gi = ai+l .. .. hi) c. (i) 1 c. v). where u is an irreducible factor of ai and v is an irredueible factor of hi modulo u.' t. hi) c. Therefore (ai. .. PROOF. we have 1 c.. We first show how to compute ail of the prille ideals containing 1. hi) c. since (ai. an irreducible factor of ai. . Let l. . To prove (i). We now see that 1 c.: P.: P.. ...6. t we have gj E (ail hi). . athi E P. then P is maximal. P = (il).l . is in P. Let (ai.: P.: (ai. Then.: P.t}. '. aj_ 1h j . (iii) If 1 c. we see that v E P. ath2 9i aHl . so that there exists i ::0: 2 such that ai E P. but ai+l . . .. . hi) c. Let P be the image of P in Rix].. hi)' Now we show that for j = i+ 1. We now prove (ii). let 1 c.a2'" aj_l) c. and 80 hi E P. t.: (ai. v) c..21 •. Then.: P.t we have J (hj_l' aj_ I hj _ 2 . Since ai E P.10.. since hi E P.: P. say u. .v) '" Rlx]/P = Rlx]/(il). For the converse. and we choose i largest with that property. Then 91 = a2a3'" at E P. it is an easy induction on j ta show that for j = i + 1. (ii) Let i E {2. aj-l).6. hi) c. we see that P is generated by an irreducible polynomial il E Rix]. 15 where R = Q[y] and J = ((X+y)(y2 + 1).v) is a field and so (u. Note that the method presented above gives. We give an example to show how this method is used. To do this we find all maximal ideals which contain (ai. X + y) and those which contain (a3' h 3) = (y2 + 1. we have that R[x]/(u.X + Y + 1).6. . Therefore there exists j E {l... . j=l IlMJçnMjçM.t (i) compute the irreducible factors of ai. . j=l .10 and Lemma 4.5. compute the irreducible factors of hi modulo u...e} sueh that Mj M=Mj. Qj be a primary decomposition of J. there exisls j E {l.. 50 we find those maximal ideals which contain (02. h 2 ) ((y + 1)2. to compute the primary decomposition of the zero-dimensional ideal J ç R[x]..5. .13.x 2 -x+y+1). we have . Theorem 4.10 gives an explicit method for finding al! such maximal ideals: given {gl. 0 ç M. . Therefore P = (u. g. 92 and 93 is l.6. hi)' for eaeh i = 2. (iii) each u.t. . namely M = (u. in fact. for each i = 1.} a strong Grobner basis for J as in Theorem 4. (ii) for each u computed in (i). so Therefore. We computed a strong Grobner basis for J to be gl g2 g3 = = y4 + 2y3 + 2y2 + 2y + 1 = (y + 1)2(y2 + 1) xy2+x+y3+y=(y2+1)(x+y)=a3h2 x2-x+y+l=h3' = 02a3 Note that the greatest common divisor of gl. .2). .264 CHAPTER 4. a way to compute the primary decomposition of . v computed in (i) and (ii) respectively gives rise to a maximal ideal containing J. Let J = n j=l . = Mj is maximal. e} such that M = MJ' PROOF.v) is a maximal ideal..6. v). x 2 . Statement (iii) is now immediate.12. v). EXAMPLE 4. where Qj is primary and VQ.'" . Then for any maximal ideal M Buch that J ç M. We have Since l ç M.. GROBNER BASES OVER RlNGS Since R[xl/P is a field.fI for a zero-dimensional ideal J (combining Theorem 4.. We go back to Example 4.1l. 0 LEMMA 4. . we first have to determine al! the maximal ideals containing J.6. But Mj is maximal.6. We find al! maximal ideals containing J. Let J be an ideal in a Noetherian ring A Buch that every prime ideal containing J iB maximal. Since we nQW can compute aU the maximal ideals containing I. We let v = x . nM2 nM3 = (y+ 1. .x-1) n (y2 + l. We first give a criterion to determine whether a given ideal Q is M-primary. Moreover. FurtheT assume that faT each m E M there exists l/ E N such that m" E Q. Then there exist h E A and m E M such that hg + m = l.i)(x + i . we find the irreducible factors of the image of x 2 . We now have the primary decomposition of .x +i + 1 = (x . We do this in two steps.1.x-1).fi:J. Now we find the irreducible factors of x 2 . It is easy to see that this last polynomial can be factored as x2 .fi:J = M. x 2 - + Y + 1): M 2 = (y2 . Also. We first prove that .1).x + i + 1. We find pre-images of these factors in R[x] and we have x2 - X + Y + 1 == + l.fi: .fi:J as weIl.. where i = Fr. X Therefore we have two maximal ideals in R[x] containing (y2 + l. Let A be a Noetherian ring.x + y) is M . Maximal ideals which contain (y2 + l.y). Let fg E Q and f ct Q. where R= R/(u) = Q[y]/(y2 + 1) = Q[i]. x + Y - 1). Let M be a maximal ideal of A and let Q ç: M be an ideal of A.1) (mod y2 + 1). Therefore the only maximal ideal which contains ((y + 1)2 . x (x . Let l/ E N such that m" E Q.fi = M. we have .x + y + 1 in the ring R[x]. We then give a criterion to determine which of the M-primary ideals belong to the primary decomposition of J.6. We show that 9 E M. Then Q is M -primary. PRlMARY DECOMPOSITION lN Rix] FOR R A PID 265 Maximal idealB which contain ((y + 1)2.6.x + y + 1 in (Q[i]) [x] is the polynomial x 2 . PROOF. The image of x 2 . Since M is prime and Q ç: M.x + y + 1 modulo u.. and M3 = (y2 + l. Suppose to the contrary that 9 ct M.x+y-1).fi:J ç: M. since M is maximal. if m E M then m E . and thus M ç: . X + y): The only irreducible factor of (y + 1)2 is u = Y + 1. we need to find the primary ideals which correspond to each maximal ideal in order to compute the primary decomposition of J. or equivalently. Then 1 = 1" = (hg + m)" =h'g+m".x-y) n (y2 +1.4.1 (mod y + 1) is irreducible. x 2 .X + Y + 1): Clearly u = y2 + 1 is irreducible in Q[y].y)(x + y . x + y == x . =(y+1. LEMMA 4. sinee a power of m is in Q.13. and each ofthe factors in the right-hand side polynomial is irreducible in (lQI[i])[x]. It remains to show that Q is primary. 15. Indeed. Let m be the largest integer such that u m divides g. . Moreover g. E Qi.6.. W E R[x] be such that g.Cl. = h. E J: (j) and s ~ M j . and W '= w (mod u).10. '= VW (mod u m ). . since Mi and Mj are distinct maximal ideals. . E Q. . of g" sinee g. . there exists Vi E N such that sr. Let jE {1. Let f E Qj. But since 9 ~ VQ. v) be a maximal ideal which eontains (ai. 0 We now return ta the case where A = R[x] where R is a PID. For the reverse inclusion let f E Qj. 1). Let V. Qi be a primary decomposition of l such that Qi is Mi·primary with Mi maximal. Then f s E S II Qi ç: nQi = J.. sueh that the image {jj of w is not divisible by il. = a2··· a. for sorne w E R[x].. Therefore Qj ç: Q j. (sa m . THEOREM 4. Now. Cl. and fg E J ç: Qj. Then for j = 1. .. Then u is an irredueible factor of ai and so u divides g. Since Mi = VlJi. Let 1= n i=l . Let n be the largest integer such that il n divides ?f.. Therefore i=l i=l . PROOF. Therefore il divides the image ?f.. Then there exists 9 E A such that 9 ~ Mj = VQ. as obtained in Theorem 4. a contradiction. (note that n . GROBNER BASES OVER RlNGS for sorne h' E A..266 CHAPTER 4. we can write g. Sa let M = (u. hi) and u divides ai. We use the notation above. Therefore J ç: Q. We first note that l ç: Q. Since Qj is primary.Mj. hi). E (ai. 1).14. We denote {! E A 1 J: (j) çt M j } by Qj. i '" j. . we must have f E Q j.6. .6..g'-l.. V) is M-primary and is the M-primary component of J. We deline s= II sr. . 0 + m" f E Q. E II Qi· i=l i-:pj i=l i-:pj . Then f = h' gf gEM. We will use the above ta give a method for computing the primary decornposition of J. we know that the image il of v in the ring (RI (u) )[x] is an irredueible factor of the image hi of hi. = VW + hum. for sorne h E R[x].0. .. Then we have g. E Q.0. For each i E {l. . Then Q = (U m . either f E Qj or a power of 9 is in Qj. sa that f E Qj. and sa g. Therefore LEMMA 4. e Qj={!EAII: (j)çtMj }. Note that s ~ Mj by construction.. Let A be a Noetherian ring and let l be an ideal of A.g2. '= vnw (mod u). since u m divides g. PROOF. V '= vn (rnod u). there exists Si E Mi . To see this we note that W ct M._.u m) -g. u g. THEOREM 4.D The last thing that remains to be done. First note that ~ is relatively prime to m u u because of the choice of m. um = E J.16 (HENSEL'S LEMMA). and so u m E Q'.6.!!2. a contradiction.6. Then fg E J ç Q. We now show that Q is the M-primary component of J using Lemma 4. Also. PRIMARY DECOMPOSITION IN Rix] FOR R A PID 267 We now show that Q is M-primary using Lemma 4. for sorne h' E R[x]. hi) ç M). W = hu + h' v for sorne h. hl') E R[x] be two polynomials such that their images in (R/(u))[xl are relatively prime and such that Then for any m there exist polynomials lm) and Mm) in R[x] such that f '" g(m)h(m) (mod u m ). Clearly u m E Q. for otherwise. fj. But this means that divides and this is a m contradiction. h(m) '" h(!) (mod u). .6. VWm u = (g. To conclude. Therefore v n = V . Let u be an irreducible element of the PJD R and let f E R[x]. . and 92.1. Now let j be such that 2 <: j <: t . in order to compute the primary decomposition of J. Then u 1 ct M.h. Now we can write V = v n + h' u. for sorne vEN. . ThenW ~ ct M and VW = g.h. Jf linear equations are solvable in R then these polynomials are computable. and 19j E J.h' u..14.. sinee U E M. ... Since Q is M-primary.g. m m . The interested reader should consult [Cohl. but we will not provide a proof of it. h' E R[x]. and hence v nm = (V . g. = g. Therefore V E Q'. u for sorne h.6. . then 9 E . Let g('). v w. We will state the result and show how it is used in examples. and Q'ÇQ.M. . g(m) '" g(1) (mod u). and u are in Rand that R is a PID).6. Therefore . V. These exarnples will also illustrate how the Hensellifting technique works. We deline Q' = {J E R[x] 1 J: (1) rt M}. M. E R[x] imply m g: g. either f E Q or gV E Q. so that gj E Q'. um we would have 1 E M (recall that g. . We now prove the reverse inclusion. This is always possible and it follows from a result known as Herisel's Lemrna... . But if gV E Q. Clearly Q ç M.jQ = M.h. It remains to show that V E Q'.4..9t-l are in M (reeaU that we noted in the proof of Theorem 4. Let f E Q' and let 9 E J: (1) . is to compute the polynomials V and W required in the last theorem. for otherwise. We wil~ now illustrate the technique presented in this section in three exampIes.10 that g2. .g.13.h' ulm E Q. sinee u m . Therefore f E Q. E (ai. and so w '" W '" h' v (mod u). it is sufficient to show that sorne power of u and v are in Q. u u E J. We have 93 = x .17. x + y .1) + (y + 1) + 1)2) xix . The largest integer m such that u m divides 91 = (y + 1)2(y2 + 1) is m = 2.y). x 2 .6. One obvious solution to this equation is h = 1 VW (mod (y and h' = -1. Q2. We use Theorem 4. By Theorem 4. x .y .268 CHAPTER 4.(y + 1) = x ..6. Q2.l)(y + 1) h' (mod (y + 1)2).l)x 2 2 (mod y + 1). (<QI[y])[x] has the following strong Grübner basis 91 92 = 93 = y4 + 2y3 + 2y2 + 2y + 1 = (y + 1)2(y2 + 1) = a2a3 xy2+x+y3+ y =(y2+1)(x+y)=a3 h2 x2 . Therefore y + 1 "" x(y + l)h + (x .X + Y + 1) C. We now find the primary components QI.1) + x(y + l)h + (x . Primary component which corresponds ta Ml: In this case we have u = y + 1 and v = x -1.6.X + Y + 1 "" x .1) h' (mod y + 1). We go back to Example 4.1) + x(y + l)h + (x . Now we need to factor 93 modulo u. M 2 = (y2 + 1. and that the maximal ideals which contain lare Ml = (y + 1. We also have w = x. and Q3 corresponding to Ml.6. y] such that V = (x . + (y + 1) h'.1)(y + 1) h' +(y + 1)2hh' (mod (y + 1)2) xix .l)(y + 1) h' (mod (y + 1)2).X "" (x . and W = x . M3 = (y2 + 1. and W "" x (mod y + 1). Now we need to find V and W such that VW "" x 2 - X +Y+1 (mod (y + 1)2).1). Therefore the largest n such that v n divides 93 modulo u is n = 1. V"" X - 1 (mod y + 1). or equivalently 1"" xh + (x . x .1).1) + (y + 1) = x + y. To do this we find polynomials h and h' in <QI[x.X +Y + 1 = h3.1) + (y + l)h and W = x and which satisfy the congruence 93 x 2 -x+y+l xix . GROBNER BASES OVER RINGS EXAMPLE 4. and Q3.1. We found that J = ((x + y)(y2 + 1).15 the primary component of J which corresponds to the maximal ideal Ml is .12. Therefore we have V = (x . M2' and M3 which belong to J.15 and the notation set there to find QI. l. Maximal ideals which contain (y4. Let J be the ideal ((x + y)2(y . The only irreducible factor of x + 2y3 + y2 + Y modulo y is v = x.x . n Q2 n Q3 l. We find those maximal ideals which contain (a2. = = Note that the greatest common divisor of 91. x 2 + X + y). x 2 + X + y): It is easy to see that the only maximal ideal that contains (y . (y2 + l)(x + y).y) n (y2 + The next example illustrates how the Hensel lifting technique is applied repeatedly.18.1). x + y . Since m = 1 we may let V = v and W = w. h 3 ) = (y .4. Again. Therefore the largest n such that v n divides 93 modulo u is n = 1. In this example we again consider the ring R = Q[y].1) = M3' Therefore we have J ((x + y)(y2 + 1). = (y. Maximal ideals which contain (y . y] with respect to the lex term ordering with x > y. We have 93 =x 2 -x+y+1 = (x-y)(x+y-1) (mody2+1). Therefore the primary ideal corresponding to M 2 and which belongs to J is Primary component which corresponds ta M 3 : This computation is the same as the previous one and we get tbat the primary ideal corresponding to M3 and which belongs to J is Q3 = (y2 + l. and 93 is 1. x 2 + X + y) is .1. As in Example 4. Therefore the maximal ideal which contains (y4. x + 2y3 + y2 + y) and those which contain (a3. We also have w = x + y . to compute the strong Gr6bner basis for J we compute the Gr6bner basis for J viewed as an ideal in Q[x. We get 91 92 93 yS _y4 = y4(y_1) = a2 a3 xy-x+2y4_y3_y=(y-l)(x+2y3+y2+y)=a3h2 x 2 +x+y=h 3. x 2 + X + y) of R[x].12 we first compute the maximal ideals which contain J. n (y2 + l.1. X + 2y3 + y2 + y) is M.92.6.6. PRlMARY DECOMPOSITION IN Rix] FOR R A pm 269 Primary component which corresponds ta M 2 : In this case we have u = y2 + 1 and v = x-y.l. Now we need to factor 93 modulo u. EXAMPLE 4. x 2 ((y + 1)2. The largest integer m such that u m divides 91 = (y + 1)2(y2 + 1) is m = 1. h 2) = (y4.x + y) X + Y + 1) = Q. x).6.x+y-1).x + 2y3 + y2 + y): The only irreducible factor of a2 = y4 is u = y. (3) sueh that V(3) = V(2) W(3) = X + Y + y 2h(3). We also have w = x + 1. the primary deeomposition of v7 is v7 = Ml nM2 = (y.1 and v = x 2 + X + 1.y + 1.(2) in Ql[x. To lift modulo y3. An obvious solution to this congruence is h(2) = 1 and h.(2) = -1. Therefore we have y '" (x + l)yh(2) + xy h.6. We start with'lifting modulo y2 To do this we must find polynomials h(2) and h.(2) (mod y2).(2). The largest integer m sueh that u m divides gl = y4(y -1) is 17' = 4. Therefore the largest n sueh that v n divides g3 modulo u is n = 1. or equivalently 1 (x + l)h(2) + x h. Now we need to factor g3 modulo u. Thus.(2) (mod y). W(i) such that at each stage we have V(i)W(i) '" x 2 + X +Y (mod yi). W(i) '" x +1 (mod y).1) "" Ql. for i = 2.x(x+l) (mody). We have g3 =x2+x+y".x 2 +x + 1). We now find the primary eomponents that belong to l and whieh correspond to Ml and M 2 • We use Theorem 4.x) n (y -1.1. we find h(3) and h. = W(2) + y 2h. GROBNER BASES QVER RINGS That is. sinee x 2 + x + 1 is irreducible in the ring R[x]. and whieh satisfy the following congruence g3 X2+X+Y V(2)W(2) (mod y2) x2+ X + (x + l)yh(2) + xyh. Thus we need to find V and W sueh that To do this we use the Hensellifting technique three times to find V(i).(3) = X _ Y + 1 + y 2h. x 2 + X + y) = M 2 is itself maximal.(2) (mod y2). 80 the original ideal (y . we have u = y . V(i) '" x (mod y).4. y] such that V(2) = x + yh(2) and W(2) = x + 1 + yh. where R = R/(u) = Ql[yl/(y . This is done inductively by constructing V(i+l) and W(i+1) in terms of V(i) and W(i) respectively.(3) .3.270 CHAPTER 4. Therefore we have V(2) = X + Y and W(2) + y 2h(3) = x .x2+x".15 and the notation set there to find QI and Q2' Primary component which corresponds ta Ml: In this case we have u = y and v = x. PRlMARY DECOMPOSITION IN Rix] FOR R A PID 271 and which satisfy the congruence 93 2 x +x+y V(3)W(3) (mod y3) x2 +X +Y - y2 + (x + y)y2 h.4.1. _ = W(4) = W(3) + y 3h.g2.(4) +(x .(y_1)(x+2y3+y2+y).y2 _ Y + 1)y3h(4) (mod y4).1 and v = x 2 + X + 1.(4) and which satisfy the congruence 93 x 2 +x +y _ 2 y 3 + (x +y2 +y)y3h. We find h(4) and h.(3) + (x . We thus have that Q2 = (y .Y = V(4) = X + 2 y 3 + y2 + Y and W = W(4) = X _ + 1. Canceling y2 we have (x + y)h. Therefore we have 2 y 3 _ y2 .y2 . We canclude this section with an example in Z[x]. A solution to this congruence is h(4) V = 2 and h.V) = (y4.x + 2y3 + y2 + y). = x + y2 + Y and = x .x+2 y 3 +y2 +y) n (y -1. Primary component which corresponds to M 2 : In this case we have u = y . Canceling y3 we have (x + y2 + y)h'(4) + (x _ y2 _ Y + 1)h(4) '" 2 (mod y). Now that we have V and W we can find the primary ideal QI corresponding to MI and which belongs ta l.1) is m = 1.x+2y3+y2+y) (y<.Y + 1.y + 1)h(3) = (mod y). Therefore the primary decomposition of 1 is 1= QI n M 2 = (y4.(3) W(3) = -1. QI = (U m .(4) = X 2 x +x+y V(4)W(4) (mod y4) y2 _ Y + 1 + y 3h. Therefore we have A solution to this congruence is h(3) V(3) 1 and h. x 2 + X + 1) = M 2 .x 2 +x + 1). . Finally we lift modulo y4.(4) such that V W = V(4) = V(3) + y 3h(4) = X + y2 + Y + y 3h(4).(4) = -2.(3) + (x '" 1 y + 1)y2h(3) (mod y3).6. The largest integer m such that u m divides 91 = y4(y . We first compute the maximal ideals wbich contain (a2. Therefore the largest n such that v n divides 93 modulo u is n = 1.272 CHAPTER 4. Also. x + 1). The largest integer m sueh that u m divides 91 = 45 is m = 2.93).19. The polynomials 91 92 93 45 = 9 . We will compute the primary decomposition of l = (91.: In this case we have u = 3 and v = x + 2. The primary decomposition of v7 is + 2) n (5.6. x + 2): The only irreducible factor of 9 is u = 3. Primary component which corresponds ta M. Now we need to find V and W such that VW == x 3 + 6x 2 + 20x + 15 (mod 9). x 3 + 6x 2 + 20x + 15): The only irreducible factor of 5 is u = 5. We have 93 = x 3 + 6x 2 + 20x + 15 == x 3 + 2x == x(x2 + 2) == x(x + 1)(x + 2) (mod 3). = (3.x) and M3 = (5. x 3 + 6x 2 + 20x + 15). x) n (5. M 2 = (5. We find polynomials h and h' in Z[x] such that V=(x+2)+3hand W=x(x+l)+3h'. n M 2 n M3 = (3. h 2) and (a3. Therefore there is only one maximal ideal which contains (9. x + 2). Also. (mod 3). h 3) respectively. 5 = a2a3 5x + 10 = 5(x + 2) = a3h2 x 3 + 6x 2 + 20x + 15 = h3' form a minimal strong Grabner basis in Z[x]. and W == x(x + 1) As before we use the Hensellifting technique. v7 = M. Maximal ideals which contain (5. and it is M. x 3 + 6x 2 + 20x + 15 == x 3 + x 2 == x 2 (x + 1) (mod 5). Maximal ideals which contain (9.x + 2). x Now we find the prirnary component for each of these maximal ideals. namely x and x + 1.x + 1). We also have w = x(x + 1). . Thus there are two maximal ideals which contain (5. V == x + 2 (mod 3). Now we need to factor 93 modulo u. GROBNER BASES OVER RlNGS EXAMPLE 4. v = x + 2 is irreducible modulo 3.92. Therefore there are two irreducible factors modulo 5. 1) = a2a3 (y . Therefore the largest n such that v n divides 93 modulo u is n = 2. and whlch belongs to J. Since m = 1 we may let V = v and W = w. The largest integer m such that u m divides 9' = 45 is m = 1. x + 1). Therefore we have the following primary decomposition of J J = Q. as above 93"" x 2 (x + 1) (mod 5). _ _ VW (mod 9) Therefore we have 3x 2 + 6"" 3x(x + l)h + 3(x + 2) h' (mod 9). . Also. We now can find the primary ideal Q.1)(x2 + y) = a3 h 2 xS + 2x 6 + 3x 4 + x 2 - Y + 1 = h3.6. and so the primary ideal Q3 corresponding to Ma and whlch belongs to J is Q3 = (U m . or equivalently.v) = (5.:1: + 1) = (5. As above the largest integer m such that u m divides 9. x 2 + 2 "" x(x + l)h + (x + 2) h' (mod 3). x 2) n (5.5(x + 2). Now we factor 93 modulo u. Consider the following three polynomials in QI[x. A solution to this congruence is h = 0 and h' = x + 1. yi 9' 92 93 (y2 . and so the largest n such that v n divides 93 modulo u is n = 1. We have 93 = x 3 + 6x 2 + 20x + 15 "" x 3 + x 2 "" x (x 2 + 1) (mod 5). We also have w = x 2 • As above we can choose V and W equal to v and w respectively. The primary ideal Q2 corresponding to M 2 and which belongs to J is Primary component which corresponds to M 3 : In this case we have u = 5 and v = x + 1.1.4. corresponding to M.Y + 1)2(y . = 45 is m = 1. We also have w = x + 1. PRlMARY DECOMPOSITION IN Rix] FOR R A PID 273 and which satisfy the following congruence 93 = x3 x3 + 6x 2 + 20x + 15 + 3x 2 + 2x + 3x(x + l)h + 3(x + 2) h' (mod 9).6. Primary component which corresponds to M 2 : In this case we have u = 5 and v = x.x + 1). x + 2) n (5. Exercises 4. n Q2 n Q3 = (9. Thus we have v = x + 2 and W = x(x + 1) + 3(x + 1) = x2 + 4x + 3.92. 93) ç Z[x]..93} is a strong Grübner basis for J and find the primary decomposition of J.2.92.274 CHAPTER 4.92..92. Verify that {9" 92. GROBNER BASES OVER RlNGS Let J = (9.. . Verify that {9. 4.6. Consider the following three polynomials in Z[x] 9' 92 93 4·7 = 7(x 2 a2a3 x 4 + 1) = a3h2 + 4x + 3 = h3' Let J = (9. 93} is a strong Grübner basis for J and find the primary decomposition of J.93) ç (<Q[x])[y]. 80 errars may occur and the user must keep this in mind. When starting a computation with CoCoA the user specifies the characteristic of the field. MAPLE. CoCoA performs arithmetic modulo a large prime number.. There are other packages which are entirely devoted to computing in polynomial rings and which have an extensive list of commands ta perform sorne of the computations presented in this book. Computations and Algorithms Computations. CoCoA was developed at the University of Genova. . Sorne of the computations could still be done using MACAULAY. we mention CoCoA and MACAULAY. and Lorenzo Robbiano.. Xn].. Italy. but with care (see Exercises 1. the user can choose among predefined term orders or define a custom ordering. Ideal quotients. . any practical implementation of these algorithms requires a lot of material not included in this book and many hours of work.xn])m. The algebraic procedures Înclude the computation of the intersection of ideals and modules. . elimination Ideals and modules. Ta obtain a copy 275 . by Antonio Capani. These systems allow the user to program and the algorithms presented in this book are.X n]. . ideals of k[Xl.9 and 1. and. However. There are many Computer Algebra Systems which have a Grübner basis package. normal forms. In particular. in principle... Most of our examples are non-homogeneous. and MATHEMATICA. . MACAULAY computes only with homogeneous polynomials and foelises on computations applied ta algebraic geometry. syzygy modules. .Appendix A. CoCoA provides commands for the algebraic manipulation of polynomials in k[Xl. MAPLE and MATHEMATICA do not allow computations in modules and have only a limited choice of orders. where k is a field. for example AXIOM.4.18). Many of these computations could not have been done with the other systems listed above. be implemented using sorne of these systems. and submodules of the free modules (k[Xl. free resolutions.6. None of the above systems have an implementation for the computation of Grobner bases over rings which are nat fields.. Gr6bner bases for ideals and modules. of course. Most of the computations in the examples and exercises in this book (except in Chapter 4) were performed using CoCoA. in principle. Of course the algorithms Qver rings could also. Alessandro Giovini. Even when computing ovef Q. Moreover. programmable. Gianfranco Niesi. in many instances. and R-takes on its new value 5. No other instruction is executed until the entire WHILE or FOR loop is completed.276 APPENDIX A. Note that the truth of condition depends on the CUITent value of the variables in the algorithm. Aigorithms.3.2. We also use two loop structures: WHILE condition DO action and FOR each item in a set S DO action In the WHILE loop. The algorithms in this book are presented in pseudo-code. 4}. 2. . then action 1 i8 performed. and the format is as follows. and if the instructions G:=GU{C} R:= C+ 1 are executed. and the FOR loop must be executed in that order. 3}. action is performed once for each item in the set S. Sometimes we omit the ELSE statement which always means that action 2 is simply "do nothlng. We use the following conditional structure: IF condition THEN action 1 ELSE action 2 This means that if condition i8 true. and then the operation 4 + 1 = 5 is performed. an order on the set S is prescribed. In the FOR loop. COMPUTATIONS AND ALGORlTHMS of CoCoA send a message to the developers at cocoa@dima. action is repeated as long as condition holds and is not performed if condition does not hold. When we need to assign an expression to a variable we use the instruction variable . We then always specify how we initialize the variables involved in the algorithm (thls follows the word INITIALIZATION)." The indentation always indicates what action 1 and action 2 are and when the conditional structure terminates.it.2. and G takes on its new value {l.3}U{4} = {l.4} is performed. 2. Note that. The indentation always indicates what the action is and when the loops terminate. if the CUITent value of the variable C is 4 and the CUITent value of the variable G is {l.unige.= expression For example. and if condition is false then action 2 is performed. We always start our algoritluns by specifying the input (whlch follows the word INPUT) and the output (whlch follows the word OUTPUT). 3. then the operation {l. The one we will use most in this book is the set '[n of power products in the variables Xl. We consider a non-empty set T which we assume has a total order ":S. We assume that we are given a set {Pt 1 t E T} of statements to be proved. That is. X n on which we will put various orders (see Section 1. Well-ordering and Induction In this book we use a form of "proof by induction" which may be unfamiliar ta the reader. t ::. we assume we have a relation ":S" on T satisfying the following properties: • the relation :S is reflexive: for an t E T.. }. in N this element is 0). ail of them will be well-orderings. the one familiar ta most people. t E T. the set of natural numbers N = {D. Since S # 0 we may choose sES least.Appendix B. either s <: tort <: s. there is an sES such that for all tES we have s <: t. . Then S contains a smallest element. 1. if s <: t and t <: u then s S u. • We argue by contradiction. Il is the purpose of this appendix ta briefly describe this process. it has a smallest element which we will denote by 1 (of course. Another way ta say this is that if sorne of the Pt 's are false then we may choose a smallest sET for which P s is false.4). t E T. • the relation <: is total: for all s. We say that a total arder <: is a well-ordering provided that we have the additional property • Let S ç T such that S # 0. of course. tj • the relation <: is transitive: for all s. .2. u E T. The idea then is ta find t < s for which Pt is also false and sa arrive at a contradiction and conclude that aU Pt '8 are true... we can let S be the set of all t E T such that Pt is false. We go about this in One of two ways. That is." on it. • The second way to proceed is in the more traditional "induction argu277 . t. • the relation <: is antisymmetric: for ail s. There is. We will use two examples of well-ordered sets. If one of the statements is false. if s <: t and t <: s then s = t. Let T be a well-ordered set. .. Since T is a subset of itself. WELL-ORDERING AND INDUCTION menf' style (commonly referred to as strong induction). . We conclude from this that ail of the PtS are true. Moreover the proof that these two forms of induction are equiva!ent proceeds exactly as it does in the case T = N. The validity of the first method of reasoning is obvious. is true.278 APPENDIX B. for ail s < t implies the truth of Pt. (ii) For al! t E T the truth of P. We assume that (i) P. Springer Verlag. MA. [BaSt90] D.1991. in EUROCAL'83. Maple V Library Reference Manual. Introduction to Commutative Algebra. Goldstein. Stilhnan. Cohen. Berlin and New York. A note on the complexity of constructing Grobner bases.. Bose ed. ed. Ein Algorithmus zum Auffinden der Basiselemente des Restklassenringes nach einem nulldimensionalen Polynomideal. Bayer and M. Prentice Hall. NJ. Berlin and New York. Springer Verlag. Englewood Cliffs. van Hulzen ed. 137~145.). Austria. Weispfenning. Becker and V. Innsbruck. Char. [BaSt88] D. G. Atiyah and LG. 1982. 6 (1988). 72. Stillman. [Bu79] B. 135-147. An International Symposium on Symbolic and Algebraic Manipulation (E. Buchberger. Springer Verlag.J.).D. Bayer. [CoTr] P. Berlin and New York. [BeWe] T. University of Innsbruck. Introduction to Number Theory. A Course in Computational Algebraic Number Theory. vol. B.. Harvard University. 1993. [Ba] D. Thesis. K.A.W. Traversa. Buchberger.A.W.A.F. Sei. 1993. Gonnet. Grobner bases: An algorithmic method in polynomial ideal theory. Springer Verlag. 1985. vol. 184~232. Sei. Bucbberger. European Computer Algebra Conference (J.). [Bu85] B. Conti and C. [Bu83] B. [AtMD] M. A criterion for detecting unnecessary reauctions in the construction of Grobner bases. Inst. Lecture Notes in Comput. 162. Springer Verlag.K.D. Adams and L. 1976. Thesis. Ph. Cambridge. Addison-Wesley. Geddes. Reidel. [Cohl H. 1969. and S. Macaulay: A system for computation in algebraic geometry and commutative algebra. The Division Algorithm and the Hilbert Scheme. M. MacDonald. On the complexity of camputing syzygies. Berlin and New York.M. Leang. Monogan. 3~21. Bayer and M. Watt. Comp. Ng. MA. 1979. J. Ph. Symb. Dordrecht.References [AdGo] W. 1983. Reading. [Bu65] B. Bucliberger. in Multidimensional Systems Theory (N. 1965. [CGGLMW] B. Lecture Notes in Comput. Berlin and New York. 279 . Grobner Bases: A Camputational Approach ta Commutative Algebra. in EUROSAM'79. User's manual. Buchberger algarithm and integer programming. AXIOM: The Scientific Computation System. Algebraic Algorithms. T. [GMNRT] A. [GiNi] A. and Error-Correcting Codes AAECC'9 (H. Faugère. New York. A superexponential lower bound for Grobner bases and Church-Rosser commutative Thue systems. D. Computational Algebraic Geometry and Commutative Algebra (D. 1991. 1993. T. ACM Press. and T. Ideals. [CLOS] D. in Proc. Direct methods for primary decomposition. [Huy] D. . 68 (1986). Gianni. Rensselaer Polytechnic Institute. G. and D. Sutor. Berlin and New York. Grobner bases and primary decomposition of polynomial ideals. Watt ed. and W. Robbiano. 1964. 1 (1985). NJ. 261-270. Aigebra. Niesi. London. McGraw-Hill. Trager. 1992. 196-206. [Hun] T. 1973. Eisenbud. New York. 1990. University of Waterloo. Springer Verlag. Lazard. Mora.). Ann. Berlin and New York. Herstein. [JeSu] R. 79 (1964). [Cz] S. 1988. and G. Ph. Mora.280 REFERENCES in Applied Algebra. Grobner basis methods for solving algebraic equations. NY. Math. Lang. Zacharias. J. [He] LN. MA. and Aigorithms: An Introduction to Computational Aigebraic Geometry and Commutative Aigebra. 329-344.N Rao eds. and T. Efficient computation of zero-dimensional Grobner bases by change of ordering. Vasconcelos. L. Cambridge University Press. Springer Verlag. Comp. Lecture Notes in Comput. On the complexity of computing Grobner bases for zerodimensional polynomials ideaZs. Ideal bases and primary decomposition: Case of two variables. Canada. vol.S. [GTZ] P. [Lak] YN. Compl.). Topics in Aigebra. Giovini and G. Symb. [FGLM] J. Symb. [EHV] D. Englewood Cliffs. Comp. The Theory of Linear Economie ModeZs. Addison-Wesley. Reading. 110 (1992).F.M. Thesis. "One sugar cube. Prentice Hall. Thesis. Berlin and New York. P. Little. Eisenbud and L. 16 (1993). Inf. 1991.W. 130139. J. Gianni. 1990. Contr. Invent. Math. 207-235.R.D. Mattson. Systems of algebraic equations (algorithms and complexity). [Laz91] D. CoCoA user manual. Resolution of singularities of an algebraic variety over a field of characleristic zero. Huynh.R. Berlin and New York. a First Course. Jenks and R. please" or selection strategies in the Buchberger algorithm. J. 149--167. Mora. O'Shea.C. Gale. 6 (1988). [Lan] S. Abstract Aigebra. Lazard. Robbiano. J. Springer verlag. 1960. [Hi] H. Giovini. Sci. C. Goldstein. and C. Cox. Niesi. Huneke. 1992. Lazard. Hironaka. Symb. [Go] L. Aigebra. [Laz85] D. 1970. J. Hungerford. 1974.. 109--326. Troy.T.).D. Ph. B. Czapor. 539. Springer Verlag. New York. eds. Lakshman. [Ga] D. Blaisdell. International Symposium on Symbolic and Algebraic Computation ISSAC'91 (S.D. Varieties. 49--54. Traverso. Bachelor's Thesis. Comp. New corutructive methods in classical ideal theory. On the construction of Griibner bases using syzygies. Proc. 1991. [Za] G. Mishra. 197 (1974). [RDSw] L. J. 267-273. Constructions in algebra. MA. 531-555. Mru::aulay. Berlin and New York. 5th ed. [NZM] 1. 379-386.). Burns and A. Niven. Moller. 59 (1952). [Wo] S. J. 6 (1988). vol. J. Berlin and New York.W. Adv. Diplomarbeit. Szekeres. Lecture Notes in Math. Seidenberg.M. . [Se] A. in Proc. Mon. Algebra 100 (1986). Shannon and M. [Sz] G. MA. Montgomery. 1986.. Trans. Moller and T. Theory of Linear and Integer Programming. Using Griibner bases to determirœ algebra membership. Comp.R Meyer. Soc. Sehrijver. Birkhiiuser Verlag.M. Symb. Wiley.S. Computing syzygies a la Gauss Jordan. A canonical basis for the ideals of a polynomial domain. [MoRo] T. [Mo88] H. Sweedler. London Math. Simis eds. [Schri] A. J. and H. Zacharias. 1980. 6 (1988). Am.. 1990. New York. Progress in Mathematics 94. Die Bereehnung von Syzygien mit dem verallgemeinerlen Weierstrasschen Divisioruatz. 183-208. Some properlies of enumeration in the theory of modular systems. [Sebre] F. Mayr and A. 61-87. Symb. The Griibner fan of an ideal. 1978. The complexity of the ward problems for commutative semigroups and polynomial ideals. 1993. Math. Soc. Reading. MIT. [ShSw] D. Chichester. eds.REFERENCES 281 [Mac] F. split surjective algebra homomorphisms determine bira~ tional equivalence. 26 (1927). Moller. 272-313. [MaMe] E. Springer Verlag. Cambridge. Algorithmic Aigebra. 1991. 335-345. Robbiano and M.). HB. Comp. 345-359. Basel. Mora and C. Addison-Wesley. Wiley.L. in Effective Methods in Algebraic Geometry (T. Robbiano. Mathematica: A System for Doing Mathematics by Computer. [MoMo] H. Subalgebra bases. Math 46 (1982). An Introduction ta the Theory of Numbers.M. [Mo90] H. Traverso. Mora and L. Zuckerman. Wolfram. Symb. Mora. [Mi] B.305329. Hamburg. Math. Generalized Griibner bases in commutative polynomial rings.O. Springer Verlag. Sehreyer. Amer. 1430. 138-178. Commutative Algebra Salvador (W. 6 (1988). Sweedler. 1988. NY. . . J..X n a field k affine space kn ring of polynomials in the variables Xl.·· . xnl by the ideal l coset of J modulo l J+I V(8) variety in k n defined by the set of polynomials 8 variety in k n defined by the polynomials fI.. x~n J reduces to h J reduces to h modulo G in one step modulo G 283 .. ..xnl/I quotient ring of k[x" ..X n with k[x" .. J. . J. .(L) lt(f) leading term of J leading power product of J lp(f) leading coefficient of J le(f) li Z Zn xa J~h G J --4+ h X~l . xnl defined by V ç k n ideal generated by the polynomials in the set 8 (8) greatest common divisor gcd least common multiple lem dimension of the k· vector space L dim..) J == g (mod 1) J congruent ta g modulo l k[x" . . .J.··· . ... . xnl coefficients in the field k ideal (submodule) generated by fI...) I(V) ideal in k[x" ... V(f" ... (f" .. .. . . .List of Symbols natural numbers ring of integers ring of integers modulo n field of rational numbers IQI field of real numbers IR field of complex numbers C 'lrn set of power products in the variables Xl. J.. . .Xn) k(a) k(a" . . } 1 .. .. 1 [ J./. 1 SYZ(f" .... .] kiF] k(Xl. /. J.g) Na(f) VK(S) k -Ji InJ I:J im( <1» ker( <1» k[fI. . .X 9 leading term ideal (submodule) defined by S S-polynomial of / and 9 normal form of / with respect to G variety in Kn defined by S ç k[Xl' . .. . g"... . g.Qn set of column vectors with entries in the ring A quotient module of M by N isomorphic ta standard basis for the free module Am 1 X s matrix whose entries are polynomials h.an ) Am M!N el.N) (U) S-lA Ap Ag / reduces to h and h = / . g2.. . ..) [ fI Syz(F) Im(J) FœG [FIG] ts ann(M) F0G Hom(M. k-algebra generated by the polynomials in the set F rational function field field extension of k obtained by adjoining a field extension of k obtained by adjoining QI..284 LIST OF SYMBOLS /~h Lt(S) S(f.. . .fs m x s matrix whose columns are vectors fI' .fs syzygy module of the matrix [ fI /. .... x n ] algebraic closure of the field k radical of the ideal l intersection of land J ideal quotient of l by J image of the map <1> kernel of the map <1> k-algebra generated by the polynomials fI....··· lem /.. syzygy module of the matrix F leading monomial of the vector J direct sum of the matrices F and G concatenation of the matrices F and G transpose of the matrix S annihilator of the module M tensor product of the matrices F and G set of all A-module homomorphisms between M and N module generated by the columns of the matrix U localization of the ring A at the multiplicative set S localization of the ring A at the prime ideal P localization of the ring A at the set {l. 207 Euclidean. 130 division module case. 37 congruence. 5 Conti. 232 membership problem. 168. 114 standard in Am. 43 . 48 CoCoA. ix. 28 one variable case. 53 effective coset representatives. 250 ring case. 113 commutative ring. 152. 224 deglex. module case. 2 algebra. 113. 12 ring case. 216 ring case using Mûller's technique. 12 module case. 39 algebraic dosme. vii. 5 field case multivariable case. 114 Bayer. 5. 23 direct sum of matrices. 129 PID case. 62 element. 275 Computer Algebra System. 40. 128. 43 Buchberger's. 58 in Am. 14 Grohner basis Buchberger's. 149 improved Buchberger's. . 27 one variable case. 92. 62 algorithm.. 129 Theorem. 217 ascending chain condition. 79. 6 basis forA Tn jM. 155 ring case. 80 homomorphism. 11 division algorithm field case multivariable case. 275 confluence relation. 145 rnultivariable case. 226 285 affine. 39. 53 Dickson's Lemma. 124. 25. 177. 1 computation. 97 geometry. 180 space. 232 algorithm for modules. 258 crit2. 149 improved algorithm. 20 determine. 244 annihilator. 26.. 84. 180 First Isomorphism Theorem. 84. 131 algorithm. 219 primality test. . 141 multivariable case. 227 cost function. 28 one variable case. 207 effective.155 for k[Xl. 15 module case. field case. 102 Buchberger. 275 crit2. 128.xnl/I. 105 critl. 105 coset. 20 degrevlex. 90 algebraically closed field. 174 division for vectors.275 column vector. 145 ring case.Index affine algebra. 57 one variable case. viii.. 13. 267 Hilbert Basis Theorem. 208 minimal. 80 module case.48 universal. 71 Grobner basis algorithm Buchberger's. module case. 32. ring case. 177. field case. 6 Gianni. 62 Hom. 237. 3. 232 image of. 5. 6 Nullstellensatz. 80. 121. 199 Faugère. 102 graph of a map. 15. 62 First Isomorphism Theorem algebra ease. 70. 229 module. 80 kernel of. 195 Gauss-Jordan elimination. 62 algebraically closed. 70 ring case. 80 quotient. 92. 5. 156 order) 69. 34 homogeneous. 3 field case elimination. 254. 147 reduced. 206 monomial. 259 elimination Gauss-Jordan. 91 greatest cornmon diviso!. 34 minimal. 261 primary. 32. 251 for module. 246 radical of. 262 strong. 116 ideal. 79. 219 for ideal. 117 free module. 196 graph. 260 prime. 68 field algebraic dosme. 185 homomorpmsm algebra. 148 improved Buchberger's. 13. 23 of relations.47 reduced. 182 irreducible. -2 ideal field case. 150 Hensel's Lemma.49 component. 72 quotient for modules. 22 polynomial. 2 generating set. 259 principal. 53 monomial. 185 short. 32 membership problem. 201. 232 module. 172. 68. 23 P-primary. 183 homogeneous ideal. 14 exact sequence. 251. 175 leading term. 38 syzygy. 32. 237 global dimension. field case. 121 .286 INDEX for ideal. 49 intersection. 250 ring case. 184 explicitly given module. 157. 211 minimal strong. 62. 229 Euclidean Algorithm. 129 PID case. 216 ring case using M611er's technique. 70 Grobner basis for. 43 Buchberger's. 61. 22. 117 Ext functor) 194. 50 using syzygies. 114 free re~olution. 38 homological algebra. 261 leading term. 212 homogenization. 114 finite. 91 induction. field case. 202 one variable case. 206 module. 238 Noetherian 287 . 147 homomorphism. 182 intersection. ring case.114 Grobner basis for. 143 leading power product. 153 Noetherian. 23.39 ideal over field. 206 membership problem. 94 isomorphism. 143 Mora. 232 quotient. 116 normal forro. 10 vector case. 116 leading term module. 230 leading term. 202 leading term ideal. 225 module. 59. 89 long division. 80 implicitization. 175 :ring case. 148 syzygy. 10 vector case. 261 isomorphic varieties. module. 250 monic polynomial. 97 module. 157. 155 quotient.. 153 minimal Grobner basis. 53 ideal over ring. 21. 211 strong Grobner basis. 201. 47 Grobner basis. 143 leading monomial.178 inverse in k[Xl. 251 with respect to F. 231 saturation of.37 monomial ideal. 254. 204. 225 of relations. 116 kernel of a homomorphism. 232 Lakshman. 178 isomorphism. 116 reduction in. 77 Lazard. 259 leading coefficient multi-variable case. 238 localization. 80. 170. U8 Moller.xnl/I. . 10 membership problem algebra. 141 leading. 230 of modules. 147 membership problem. 217. 19 linear equations are solvable. 202 one variable case. 23 in modules. 68.175. 262 local ring. 277 integer progranuning.INDEX ring case elimination. 32. 21. 238 zero-dimensional. 68 multiplicative set. 175. 117 free. 201. 201. 176. 147 lex. 177. 156 explicitly given. 182 irreducible ideal.176. 21. 143 least cornmon multiple of polynomials. 229 Grobner basis for. 105 inter-reduced. 172. . U3 elimination. 143 S-polynomial. 71 of vectors. 116 ideal quotient. 117 First Isomorphism Theorem. 147 multivariable case. 157. 216. 131 intersection of ideals field case. 157. 64. 205 minimal polynomial. 13 monoid. 208 intersection. 262 image of a homomorphism. 70.. 5. 88 position over terrn (POT). 62. 13. 261 ideal. 140 term over position (TOP). 155 normal selection strategy. 150 polynomial over a field. 18. 238 quotient. 195 primality test. 25. 89 Noetherian. 203. 22. 227 vector case. 144 muItivariable case over a field. 182 quotient module. 27 totally. 259 of fractions. 5 radical of an ideal. 1 local. 20 elirnination. 48 module case. 5 Robbiano. 18 degree lexicographie (deglex). 130 Nullstellensatz. 19 position over term (POT). 1 elementary symmetric. 260 prime component. 62 INDEX ideal. 18. 6. 25. order. 119. 142 power product. 277 P-primary ideal. 57. 2 S-basis. 23. 202 presentation. 143 multivariable case over a field. 97 monie. 75 symmetric. 27 one variable case over a field. 38 minimal. 39 row echelon form. 227 reduced. 116 ring. 18 leading. 261 decomposition. 261 parametrized variety. 261 124 . 40. 19 degree reverse lexicographie (degrevlex). 57 over a ring. 259 normal form polynomial case over a field.246 projection map. 166 lexicographie (lex).13 normal form. 18. 91 polynomial. 142 total. 27 square free. 21. 238 localization. 258 remainder module case. 148 multivariable case over a field. Il ring commutative. homogeneous. 244 primary component. 61. 205 one variable case. 11 strong. 8 module case. 116 ring. 277 weJl-ordering. 6. 259 principal ideal. 117. 252. 144 reduction linear case. 25 27 multivariable case over a ring. 26. 246 Principal Ideal Domain (PID). 121. 247 S-polynomial module case. 1. 13. 69 induced by a matrix. 258 reduced Grobner basis ideal case over fields. 227 vector.288 module. 237. 142 term. 90 pullback. 39 submodule. 262 289 . 1 well-ordering. 238 Schreyer. 102 total order. 171 homogeneous. 2. 277 Zacharias. l.237 transpose matrix. 176 Traverso. 165 Seidenberg. 251. 88 syzygy and Grobner bases.INDEX PID case. 213 SAGBI basis. 118 module of. 25. 201. 246 Szekeres. 262 reduction. 134 of vectors over a ring. 251 minimal Gr6bner basis. 90 isomorphic. 226. 64. 212 of a matrix. 18 polynomial case. 140 order module case. 78 Shannon. 254. 32 strong Grobner basis. 105 universal Gr6bner basis. 91 vector space. 142 three color problem. 90 zero divisor. 88 square free polynomial. 61. 121 applications of. 140 polynomial case. 61 zero-dimensional ideal. 118. 75 standard basis. 39 saturated set. 213 saturation of an ideal. 82. 18 term over position (TOP). 254 tensor product of matrices. 39. 277 totally reduced. 232. 227 Trager. 189 term module case. 88 symmetric polynomial. 213. 121. 212 module of. 237 Zariski closure. 114 Sweedler. 252. 82. 161 of vectors over a field. 18. 18. 94 parametrized. 226 saturation of a set. 258 subalgebra. 50 variety. 249 ring case. 161 module of. 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