Dummit Solutions

Figure 0.1 ABSTRACT ALGEBRA DAVID S. DUMMIT AND RICHARD M. FOOTE Solutions provided by Scott Larson. Contents 0. Preliminaries 3 0.1. Basics 3 0.2. Properties of the Integers 4 0.3. Z/nZ: The Integers Modulo n 6 Part I – Group Theory 8 1. Introduction to Groups 8 1.1. Basic Axioms and Examples 8 1.2. Dihedral Groups 10 1.3. Symmetric Groups 11 1.4. Matrix Groups 11 1.5. Quaternion Groups 13 1.6. Homomorphisms and Isomorphisms 14 1.7. Group Actions 14 2. Subgroups 15 2.1. Definition and Examples 15 2.2. Centralizers and Normalizers, Stabilizers and Kernels 15 2.3. Cyclic Groups and Cyclic Subgroups 16 2.4. Subgroups Generated by Subsets of a Group 17 2.5. Definitions and Examples 19 2.6. The Lattice of Subgroups of a Group 20 3. Quotient Groups and Homomorphisms 20 3.1. Definitions and Examples 20 3.2. More on Cosets and Lagrange’s Theorem 21 3.3. The Isomorphism Theorems 23 3.4. Composition Series and the Holder Program 24 3.5. Transpositions and the Alternating Group 24 4. Group Actions 25 4.1. Group Actions and Permutation Representations 25 4.2. Groups Acting on Themselves by Left Multiplication – Cayley’s Theorem 25 4.3. Groups Acting on Themselves by Conjugation – The Class Equation 25 1 2 DAVID S. DUMMIT AND RICHARD M. FOOTE 4.4. Automorphisms 26 4.5. The Sylow Theorems 26 4.6. The Simplicity of A n 27 5. Direct and Semidirect Products and Abelian Groups 28 5.1. Direct Products 28 5.2. The Fundamental Theorem of Finitely Generated Abelian Groups 29 5.3. Table of Groups of Small Order 30 5.4. Recognizing Direct Products 30 5.5. Semidirect Products 30 6. Further Topics in Group Theory 31 6.1. p-groups, Nilpotent Groups, and Solvable Groups 31 Part II – Ring Theory 33 7. Introduction to Rings 33 7.1. Basic Definitions and Examples 33 7.2. Examples: Polynomials Rings, Matrix Rings, and Group Rings 34 7.3. Ring Homomorphisms and Quotient Rings 36 7.4. Properties of Ideals 39 7.5. Rings of Fractions 42 8. Euclidean, Principal Ideal, and Unique Factorization Domains 45 8.1. Euclidean Domains 45 9. Polynomial Rings 46 9.1. Definitions and Basic Properties 46 9.2. Polynomial Rings Over Fields I 47 9.3. Polynomial Rings That are Unique Factorization Domains 48 9.4. Irreducibility Criteria 48 9.5. Polynomial Rings Over Fields II 50 9.6. Polynomials in Several Variables Over a Field and Gr¨obner Bases 50 Part III – Modules and Vector Spaces 53 10. Introduction to Module Theory 53 10.1. Basic Definitions and Examples 53 13. Field Theory 54 13.1. Basic Theory of Field Extensions 54 13.2. Basic Theory of Field Extensions 54 13.3. Classical Straightedge and Compass Constructions 56 13.4. Splitting Fields and Algebraic Closures 56 13.5. Separable and Inseparable Extensions 56 13.6. Cyclotomic Polynomials and Extensions 57 14. Galois Theory 57 14.1. Basic Definitions 57 14.2. The Fundamental Theorem of Galois Theory 57 15. Commutative Rings and Algebraic Geometry 58 15.1. Noetherian Rings and Affine Algebraic Sets 58 15.2. Radicals and Affine Varieties 58 15.3. Integral Extensions and Hilbert’s Nullstellensatz 58 15.4. Localization 58 3 0. Preliminaries 0.1. Basics. Proposition 0.1. Let f : A →B. (1) The map f is injective if and only if f has a left inverse. (2) The map f is surjective if and only if f has a right inverse. (3) The map f is a bijection if and only if there exists g : B →A such that f ◦ g is the identity map on B and g ◦ f is the identity map on A. (4) If A and B are finite sets with the same number of elements ([A[ = [B[), then f : A →B is bijective if and only if f is injective if and only if f is surjective. Proof. (a) Suppose f is injective so that f −1 (b) contains a single element for b ∈ f(A). Thus g(b) = f −1 (b) is well-defined for b ∈ f(A). Hence g(f(a)) = a for all a ∈ A. Now suppose that f has a left inverse g so that g(f(a)) = a for all a ∈ A. If f(a 1 ) = f(a 2 ) = b, then g(b) = a 1 and g(b) = a 2 . But g is a well-defined map so a 1 = a 2 and thus f is injective. (b) Suppose f is surjective. Take some b ∈ B so we can choose a ∈ A such that f(a) = b. Thus we can define g(b) = a such that f(a) = b. Hence f(g(b)) = f(a) = b. Now suppose that f has a right inverse g so that f(g(b)) = b for all b ∈ B. So f(g(B)) = B and thus f is surjective. (c) Suppose that f is bijective so there is a left inverse, g l , and a right inverse, g r , of f. Let f(a) = b so that a = g l (f(a)) = g l (b) and b = f(g r (b)). Since f is injective, g r (b) = a and thus g l = g r . Thus the right and left inverses of f are the same map. Now suppose that there exists g : B → A such that f ◦ g is the identity map on B and g ◦ f is the identity map on A. Then f is surjective because f(g(B)) = B. Now let f(a 1 ) = f(a 2 ) so that a 1 = g(f(a 1 )) = g(f(a 2 )) = a 2 . Thus f is injective. (d) Suppose that f is injective. Then f takes elements of A to unique elements of B. Since [A[ = [B[, f is surjective. Now suppose that f is surjective. Then since [A[ = [B[, f takes elements of A to unique elements of B so f is injective. Therefore f is either bijective or neither injective or surjective. Proposition 0.2. Let A be a nonepty set. (1) If ∼ defines an equivalence relation on A then the set of equivalence classes of ∼ form a partition of A. (2) If ¦A i [ i ∈ I¦ is a partition of A then there is an equivalence relation on A whose equivalence classes are preciselly the sets A i , i ∈ I. In Exercises 1 to 4 let / be the set of 2 2 matrices with real number entries. Recall that matrix multiplication is defined by a b c d p q r s = ap +br aq +bs cp +dr cq +ds Let M = 1 1 0 1 and let B = ¦X ∈ / [ MX = XM¦ . 0.1.1. The following elements of / lie in B: 1 1 0 1 , 0 0 0 0 , 1 0 0 1 . 0.1.2. Prove that if P, Q ∈ B, then P +Q ∈ B (where + denotes the usual sum of two matrices). Proof. (P +Q)M = PM +QM = MP +MQ = M(P +Q). 0.1.3. Prove that if P, Q ∈ B, then P Q ∈ B (where denotes the usual product of two matrices). Proof. (P Q)M = P M Q = M P Q = M(P Q). 4 DAVID S. DUMMIT AND RICHARD M. FOOTE 0.1.4. Find conditions on p, q, r, s which determine precisely when p q r s ∈ B. r = 0, p = s. 0.1.5. Determine whether the following functions f are well defined: (1) f : Q →Z defined by f(a/b) = a. (2) f : Q →Q defined by f(a/b) = a 2 /b 2 . Proof. (a) f is not a well defined map because 1/2 = 2/4 and 1 = 2. (b) f is a well defined map because if a 1 /b 1 = a 2 /b 2 , then a 2 1 /b 2 1 = a 2 2 /b 2 2 . 0.1.6. Determine whether the function f : 1 + → Z defined by mapping a real number r to the first digit to the right of the decimal point in a decimal expansion of r is well defined. f is well defined because the decimal expansion of a real number is unique. Thus there is one and only one possible first number to the right of the decimal for each real number. 0.1.7. Let f : A →B be a surjective map of sets. Prove that the relation a ∼ b if and only if f(a) = f(b) is an equivalence relation whose equivalence classes are the fibers of f. Proof. f(a) = f(a) so ∼ is reflexive. If f(a) = f(b), then f(b) = f(a) so ∼ is symmetric. If f(a) = f(b) and f(b) = f(c), then f(a) = f(c) so ∼ is transitive. If a 1 , a 2 ∈ f −1 (b), then f(a 1 ) = f(a 2 ) so a 1 ∼ a 2 . If a 1 ∼ a 2 then f(a 1 ) = f(a 2 ) so a 1 and a 2 are in the same fiber of f. 0.2. Properties of the Integers. (1) (Well Ordering of Z) If A is any nonempty subset of Z + , there is some element m ∈ A such that m ≤ a, for all a ∈ A. (2) If a, b ∈ Z with a = 0, we say a divides b if there is an element c ∈ Z such that b = ac. In this case we write a [ b; if a does not divide b we write a [ b. (3) If a, b ∈ Z` ¦0¦, there is a unique positive integer d, called the greatest common divisor of a and b, satisfying: (a) d [ a and d [ b, (b) if e [ a and e [ b, then e [ d. (4) If a, b ∈ Z` ¦0¦, there is a unique positive integer l, called the least common multiple of a and B, satisfying: (a) a [ l and b [ l, (b) if a [ m and b [ m, then l [ m. The connection between the greatest common divisor d and the least common multiple l of the two integers a and b is given by dl = ab. (5) The Division Algorithm: if a, b ∈ Z and b = 0, then there exist unique q, r ∈ Z such taht a = qb +r and 0 ≤ r < [b[ , where q is the quotient and r the remainder. (6) The Euclidean Algorithm is an important procedure which produces a greatest common divisor of two integers a and b by iterating the Division Algorithm: if a, b ∈ Z` ¦0¦, then we obtain a sequence of quotients and remainders a = q 0 b +r 0 b = q 1 r 0 +r 1 r 0 = q 2 r 1 +r 2 r 1 = q 3 r 2 +r 3 . . . r n−2 = q n r n−1 +r n r n−1 = q n+1 r n where r n is the last nonzero remainder. Such an r n exists since [b[ > [r 0 [ > [r 1 [ > > [r n [ is a decreasing sequence of strictly positive integers if the remainders are nonzero and such a sequence cannot continue indefinitely. Then r n = (a, b). 5 (7) One consequence of the Euclidean Algorithm which we chall use regularly is the following: if a, b ∈ Z` ¦0¦, then there exist x, y ∈ Z such that (a, b) = ax +by that is, the gcd of a and b is a Z-linear combination of a and b. This follows by recursively writing the element r n in the Euclidean Algorithm in terms of the previous remainders. (8) An element p of Z + is calle a prime if p > 1 and the only posotive divisors of p are 1 and p. An integer n > 1 which is not prime is called composite. An important property of primes is if p is a prime and p [ ab, for some a, b ∈ Z then either p [ a or p [ b. (9) The fundamental theorem of arithmetic says: if n ∈ Z, n > 1, then n can be factored uniquely into the product of primes, i.e., there are distinct primes p 1 , p 2 , . . . , p s and positive integers α 1 , α 2 , . . . , α s such that n = p α1 1 p α2 2 p αs s . This factorization is unique in the sense that if q 1 , q 2 , . . . , q t are any distinct primes and β 1 , β 2 , . . . , β t , are positive integers such that n = q β1 1 q β2 2 q βt t , then s = t and if we arrange the two sets of primes in increasing order, then q i = p i and α i = β i , 1 ≤ i ≤ s. Suppose the positive integers a and b are expressed as products of prime powers: a = p α1 1 p α2 2 p αs s , b = p β1 1 p β2 2 p βs s where p 1 , p 2 , . . . , p s are distinct and the exponents are ≥ 0. Then the gcd of a and b is (a, b) = p min(α1,β1) 1 p min(α2,β2) 2 p min(αs,βs) s and the lcm is obtained by taking the maximum of the α i and β i instead of the minimum. (10) The Euler ϕ-function is defined as follows: for n ∈ Z + let ϕ(n) be the number of positive integers a ≤ n with a relatively prime to n, i.e., (a, n) = 1. For primes p, ϕ(p) = p −1, and, more generally, for all a ≥ 1 we have the formula ϕ(p a ) = p a −p a−1 = p a−1 (p −1). The function ϕ is multiplicative in the sense that ϕ(ab) = ϕ(a)ϕ(b) if (a, b) = 1. Together with the formula above this gives a general formula for the values of ϕ: if n = p α1 1 p α2 2 p αs s , then ϕ(n) = ϕ(p α1 1 )ϕ(p α2 2 ) ϕ(p αs s ) = p α1−1 1 (p 1 −1)p α2−1 2 (p 2 −1) p αs−1 s (p s −1). 0.2.1. For each of the following paris of integers a and b, determine their gcd, their lcm, and write their gcd in the form ax +by for some integers x and y. (a) a = 20, b = 13. (b) a = 69, b = 372. (c) a = 792, b = 275. (d) a = 11391, b = 5673. (e) a = 1761, b = 1567. (f) a = 507885, b = 60808. Solution. 0.2.2. Prove that if the integer k divides the integers a and b then k divides as +bt for every pair of integers s and t. Proof. 0.2.3. Prove that if n is composite then there are integers a and b such that n divides ab but n does not divide either a or b. Proof. 6 DAVID S. DUMMIT AND RICHARD M. FOOTE 0.2.4. Let a, b and N be fixed integers with a and b nonzero and let d = (a, b) be the gcd of a and b. Suppose x 0 and y 0 are particular solutions to ax + by = N. Prove for any integer t that the integers x = x 0 + b d t and y = y 0 − a d t are also solutions to ax +by = N (this is in fact the general solution). Proof. 0.2.5. Determine the value ϕ(n) for each integer n ≤ 30 where ϕ denotes the Euler ϕ-function. Solution. 0.2.6. Prove the Well Ordering Property of Z by induction and prove the minimal element is unique. Proof. 0.2.7. If p is a prime prove that there do not exist nonzero integers a and b such that a 2 = pb 2 . Proof. 0.2.8. Let p be a prime, n ∈ Z + . Find a formula for the largest power of p which divides n! = n(n−1)(n−2) 2 1. Solution. 0.2.9. Write a computer program to determine the greatest common divisor (a, b) of two integers a and b and to express (a, b) in the form ax +by for some integers x and y. 0.2.10. Prove for any given positive integer N there exist only finitely many integers n with ϕ(n) = N where ϕ denotes Euler’s ϕ-function. Conclude in particular that ϕ(n) tends to infinity as n tends to infinity. Proof. 0.2.11. Prove that if d divides n then ϕ(d) divides ϕ(n) where ϕ denotes Euler’s ϕ-function. Proof. 0.3. Z/nZ: The Integers Modulo n. Theorem 0.3. The operations of addition and multiplication on Z/nZ are well defined, that is, they do not depend on the choices of representatives for the classes involved. More precisely, if a 1 , a 2 ∈ Z and b 1 , b 2 ∈ Z with a 1 = b 1 and a 2 = b 2 , then a 1 +a 2 = b 1 +b 2 and a 1 a 2 = b 1 b 2 , i.e., if a 1 ≡ b 1 mod n and a 2 ≡ b 2 mod n then a 1 +a 2 ≡ b 1 +b 2 mod n and a 1 a 2 ≡ b 1 b 2 mod n. Proposition 0.4. Z/nZ) × = ¦a ∈ Z/nZ [ (a, n) = 1¦. 0.3.1. Write down explicitly all the elements in the residue classes of Z/18Z. Solution. 0.3.2. Prove that the distinct equivalence classes in Z/nZ are precisely 0, 1, 2, . . . , n −1. Proof. 0.3.3. Prove that if a = a n 10 n +a n−1 n−1 + +a 1 10 +a 0 is any positive integer then a ≡ a n +a n−1 + +a 1 +a 0 mod 9. Proof. 0.3.4. Compute the remainder when 37 100 is divided by 29. Solution. 0.3.5. Compute the last two digits of 9 1500 . Solution. 0.3.6. Prove that the squares of the elements in Z/4Z are just 0 and 1. 7 Proof. 0.3.7. Prove for any integers a and b that a 2 +b 2 never leaves a remainder of 3 when divided by 4. Proof. 0.3.8. Prove that for any integers a and b that a 2 +b 2 = 3c 2 has no solutions in nonzero integers a, b, and c. Proof. 0.3.9. Prove that the square of any odd integer always leaves a remainder of 1 when divided by 8. Proof. 0.3.10. Prove that the number of elements of (Z/nZ) × is ϕ(n) where ϕ denotes the Euler ϕ-function. Proof. 0.3.11. Prove that if a, b ∈ (Z/nZ) × , then a b ∈ (Z/nZ) × . Proof. 0.3.12. Let n ∈ Z, n > 1, and let z ∈ Z with 1 ≤ a ≤ n. Prove if a and n are not relatively prime then there exists an itneger b with 1 ≤ b < n such that ab ≡ 0 mod n and deduce that there cannot be an itneger c such that ac ≡ 1 mod n. Proof. 0.3.13. Let n ∈ Z, n > 1, and let a ∈ Z with 1 ≤ a ≤ n. Prove that if a and n are relatively prime then there is an integer c such that ac ≡ 1 mod n. Proof. 0.3.14. Conclude from the previous two exercises that (Z/nZ) × is the set of elements a of Z/nZ with (a, n) = 1 and hence prove proposition 0.4. Verify this directly in the case n = 12. Solution. 0.3.15. For each of the following pairs of integers a and n, show that a is relatively prime to n and determine the multiplicative inverse of a in Z/nZ. (a) a = 13, n = 20. (b) a = 69, n = 89. (c) a = 1891, n = 3797. (d) a = 6003722857, n = 77695236973. Proof. 0.3.16. Write a computer program to add and multiply mod n, for any n given as input. The output of these operations should be the least residues of the usms and products of two integers. Also include the feature that if (a, n) = 1, and integer c between 1 and n −1 such taht a c = 1 may be printed on request. 8 DAVID S. DUMMIT AND RICHARD M. FOOTE Part I – Group Theory 1. Introduction to Groups 1.1. Basic Axioms and Examples. Proposition 1.1. If G is a group under the operation , then (1) the identity of G is unique (2) for each a ∈ G, a −1 is uniquely determined (3) (a −1 ) −1 = a for all a ∈ G (4) (a b) −1 = (b −1 ) (a −1 ) (5) for any a 1 , a 2 , . . . , a n ∈ G the value of a 1 a 2 a n is independent of how the expression is bracketed. Proposition 1.2. Let G be a group and let a, b ∈ G. The equations ax = b and ya = b have unique solutions for x, y ∈ G. In particular, the left and right cancellation laws hold in G, i.e., (1) if au = av, then u = v, (2) if ub = vb, then u = v. Let G be a group. 1.1.1. Determine which of the following binary operations are associative: (a) the operation on Z defined by a b = a −b (b) the operation on 1 defined by a b = a +b +ab (d) the operation on Z Z defined by (a, b) (c, d) = (ad +bc, bd). Note that (1 − 2) − 3 = −4 and 1 − (2 − 3) = 2, so (a) is not associative. We calculate that (b) and (d) are associative below. (a b) c = (a +b +ab) +c + (a +b +ab)c = a +b +c +ab +ac +bc +abc = a +b +c +bc +a(b +c +bc) = a (b c) (a, b) (c, d) (e, f) = (ad +bc)f +bde, bdf = adf +b(cf +de), bdf = (a, b) (c, d) (e, f) 1.1.2. Decide which of the following binary operations are commutative: (a) the operation on Z defined by a b = a −b (b) the operation on 1 defined by a b = a +b +ab (d) the operation on Z Z defined by (a, b) (c, d) = (ad +bc, bd). Note that 1 −2 = −1 and 2 −1 = 1 so (a) is not commutative. We calculate that (b) and (d) are commutative below. a b = a +b +ab = b +a +ba = b a (a, b) (c, d) = (ad +bc, bd) = (cb +da, db) = (c, d) (a, b) 1.1.3. Prove that addition of residue classes in Z/nZ is associative. Proof. 1.1.4. Prove that multiplication of residue classes in Z/nZ is associative. Proof. 9 1.1.5. Prove for all n > 1 that Z/nZ is not a group under multiplication of residue classes. Proof. 1.1.6. Determine which of the following sets are groups under addition: (b) the set of rational numbers in lowest terms whose denominators are even together with 0 (d) the set of rational numbers of absolute value ≥ 1 together with 0 (e) the set of rational numbers with denominators equal to 1 or 2. Note that 1/6 +1/6 = 1/3 so addition is not a binary relation on (b) and thus (b) is not a group. Also note that −3/2 + 1 = −1/2 so addition is not a binary relation on (d) and thus (d) is note a group. We notice that addition is a well defined binary relation on (e), addition is associative on the rational numbers so this subset must also have the associative property, the identity is 0, and the inverse of a is −a. Therefore (e) is a group. 1.1.7. Let G = ¦x ∈ 1 [ 0 ≤ x < 1¦ and for x, y ∈ G let xy be the fractional part of x+y (i.e., xy = x+y−[x+y] where [a] is the greatest integer less than or equal to a). Prove that is a well defined binary operation on G and that G is an abelian group under (called the real numbers mod 1). Proof. Note that the sum of two real numbers has a unique fractional part, x, such that 0 ≤ x < 1. Thus is a well defined operation that takes x y into G. We show that is associative on G. (x y) z = (x +y −[x +y]) +z −[(x +y −[x +y]) +z] = x +y +z −[x +y] + [x +y] −[x +y +z] = x +y +z −[y +z] + [y +z] −[x +y +z] = x + (y +z −[y +z]) −[x + (y +z −[y +z])] = x (y z) The identity element of G is clearly 0. The inverse of x ∈ G would be given by 1 −x for x = 0 because x (1 −x) = x + 1 −x −[x + 1 −x] = 1 −1 = 0. Note that 1 − x ∈ G for x ∈ G and x = 0. The inverse of 0 is clearly 0. We show that G is abelian by letting x, y ∈ G and computing that is commutative on G. x y = x +y −[x +y] = y +x −[y +x] = y x 1.1.8. Let G = ¦z ∈ C [ z n = 1 for some n ∈ Z + ¦. (a) Prove that G is a gorup under multiplication. (b) Prove that G is not a group under addition. Proof (a). Proof (b). 1.1.9. Let G = ¸ a +b √ 2 ∈ 1 [ a, b ∈ Q ¸ . (a) Prove that G is a gorup under addition. (b) Prove that the nonzero elements of G are a group under multiplication. Proof (a). Proof (b). 1.1.10. Prove that a finite gorup is abelian if and only if its group table is a symmetric matrix. Proof. 1.1.14. Find the orders of the following elements of the multiplicative group (Z/36Z) × : 1, −1, 5, 13, −13, 17. 1 = 1, −1 = 2, 5 = 6, 13 = 3, −13 = 6, 17 = 2. 1.1.25. Prove that if x 2 = 1 for all x ∈ G then G is abelian. 10 DAVID S. DUMMIT AND RICHARD M. FOOTE Proof. Since x 2 = 1 we know that x = x −1 for all x ∈ G. Thus xy = (xy) −1 = y −1 x −1 = yx. 1.2. Dihedral Groups. In these exercises, D 2n has the usual presentation D 2n = r, s [ r n = s 2 = 1, rs = sr −1 . 1.2.1. Compute the order of each of the elements in the following groups: (a) D 6 The order of elements from (a) are: [1[ = 1, [r[ = 3, r 2 = 3, [s[ = 2, [sr[ = 2, sr 2 = 2. 1.2.2. Use the generators and relations above to show that if x is any element of D 2n which is not a power of r, then rx = xr −1 . Proof. If x is not a power of r, then x = sr k for some 0 ≤ k < n. So rsr k = sr −1 r k = sr k−1 = sr k r −1 . 1.2.3. Use the generators and relations above to show that every element of D 2n which is not a power of r has order 2. Deduce that D 2n is generated by the two elements s and sr, both of which have order 2. Proof. If x ∈ D 2n is not a power of r, then x = sr k for some 0 ≤ k < n. Notice that sr 0 sr 0 = 1 and suppose that sr i sr i = 1 for 0 ≤ i < k. Then sr k sr k = sr k−1 sr −1 r k = sr k−1 sr k−1 = 1, by induction. 1.2.7. Show that a, b [ a 2 = b 2 = (ab) n = 1 gives a presentation for D 2n in terms of the two generators a = s and b = sr of order 2 computed in Exercise 1.2.3 above. [Show that the relations for r and s follow from the relations for a and b and, conversely, the relations for a and b follow from those for r and s.] Proof. First we start with the relations r n = s 2 = 1 and rs = sr −1 . Clearly a 2 = 1, Exercise 1.2.3 shows b 2 = 1, and (ab) n = (ssr) n = r n = 1. Hence the original relations imply the new relations. Now suppose that a 2 = b 2 = (ab) n = 1. Then clearly s 2 = 1, r n = (ssr) n = (ab) n = 1, and rs = ssrsrr −1 = sr −1 . In Exercise 1.2.9 you can find the order of the group of rigid motions in 1 3 (also called the group of rotations) of the given Platonic solid by following the proof for the order of D 2n : find the number of positions to which an adjacent pair of vertices can be sent. Alternatively, you can find the number of places to which a given face may be sent and, once a face is fixed, the number of positions to which a vertex on that face may be sent. 1.2.9. Let G be the group of rigid motions in 1 3 of a tetrahedron. Show that [G[ = 12. Proof. We use the second method described above to show that [G[ = 12. Since there are 4 faces of a tetrahedron, each face having 3 distinct rotations, there are 4 3 = 12 positions to which a vertex on each fixed face may be sent. 1.2.17. Let X 2n = x, y [ x n = y 2 = 1, xy = yx 2 . (a) Show that if n = 3k, then X 2k has order 6, and it has the same generators and relations as D 6 when x is replaced by r and y by s. (b) Show that if (3, n) = 1, then x satisfies the additional relation: x = 1. In this case deduce that X 2n has order 2. [Use the facts that x n = 1 and x 3 = 1.] 11 Proof (a). Suppose we have the relations given by D 6 . Then clearly x 3k = (x 3 ) k = y 2 = 1. We also have xy = yx −1 = yx 3−1 = yx 2 . Now suppose we have the relations given by X 2n . Then clearly x 3 = y 2 = 1. The relations xy = yx 2 and x 3 = 1 show us xy = yx 2 = yx −1 . Therefore D 6 = X 6 and [X 6 [ = 6. Proof (b). Since (3, n) = 1, there exist s, t ∈ Z such that 3s +nt = 1. So x = x 3s+nt = (x 3 ) s (x n ) t = 1. Thus X 2n has order 2, namely X 2n = ¦1, y¦. 1.3. Symmetric Groups. 1.3.4. Compute the order of each of the elments in the following groups: (a) S 3 (b) S 4 For elements in S 3 we calculate that [1[ = 1, [(12)[ = 2, [(13)[ = 2, [(23)[ = 2, [(123)[ = 3, [(132)[ = 3. For elements in S 4 we calculate that [1[ = 1, [(12)[ = 2, [(13)[ = 2, [(14)[ = 2, [(23)[ = 2, [(24)[ = 2, [(34)[ = 2, [(123)[ = 3, [(132)[ = 3, [(124)[ = 3, [(142)[ = 3, [(134)[ = 3, [(143)[ = 3, [(234)[ = 3, [(243)[ = 3, [(1234)[ = 4, [(1243)[ = 4, [(1324)[ = 4, [(1342)[ = 4, [(1423)[ = 4, [(1432)[ = 4, [(12)(34)[ = 2, [(13)(24)[ = 2, [(14)(23)[ = 2. 1.3.7. Write out the cycle decomposition of each element of order 2 in S 4 . (12), (13), (14), (23), (24), (34), (12)(34), (13)(24), (14)(23). 1.3.11. Let σ be the m-cycle (1 2 . . . m). Show that σ i is also an m-cycle if and only if i is relatively prime to m. Proof. First suppose that σ i is an m-cycle so that σ i = (a 1 a 2 . . . a m ). Then assume that α[i and α[m so there are s, t ∈ Z such that αs = i, αt = m. So σ i = σ αs = [(a 1 a α+1 . . . α m−α+1 ) (a α a 2α . . . a m )] s which can only be an m-cycle if α = 1. Thus i and m are relatively prime. Now suppose that i is relatively prime to m so that there are s, t ∈ Z such that si +tm = 1. But then σ = σ si+tm = σ si σ tm = (σ i ) s . Since a power of σ i is an m-cycle, then σ i must be an m-cycle as well. 1.3.20. Find a set of generators and relations for S 3 . S 3 = x 2 , x 3 [ x 2 2 = x 3 3 = 1, x 2 x 3 = x −1 3 x 2 1.4. Matrix Groups. 1.4.7. Let p be a prime. Prove that the order of GL 2 (F p ) is p 4 −p 3 −p 2 +p (do not just quote the order formula in this section). [Subtract the number of 2 2 matrices which are not invertible from the total number of 2 2 matrices over F p . You may use the fact that a 2 2 matrix is not invertible if and only if one row is a multiple of the other.] Proof. Notice that the total number of 22 matrices over F p is p 4 . We count the number of non-invertible matrices for a matrix a b c d in Figure 1.1 to be p 3 +p 2 −p. Thus the order of GL 2 (F p ) is p 4 −p 3 −p 2 +p. 1.4.8. Show that GL n (F) is non-abelian for any n ≥ 2 and any F. 12 DAVID S. DUMMIT AND RICHARD M. FOOTE a = 0 b = 0 b = 0 c = 0 c = 0 c = 0 c = 0 d = 0 d = 0 d = 0 d = 0 d = 0 d = 0 d = 0 d = 0 1 p −1 p −1 (p −1) 2 p −1 (p −1) 2 0 0 a = 0 b = 0 b = 0 c = 0 c = 0 c = 0 c = 0 d = 0 d = 0 d = 0 d = 0 d = 0 d = 0 d = 0 d = 0 p −1 0 (p −1) 2 0 (p −1) 2 (p −1) 3 Figure 1.1. Number of non-invertible matrices. Proof. First notice 1 1 0 1 0 1 1 0 = 1 1 1 0 0 1 1 0 1 1 0 1 = 0 1 1 1 so that GL 2 (F) is non-abelian for any F. Now notice that extending the matrices with zeros shows the result for n > 2. 1.4.10. Let G = a b 0 c [ a, b, c ∈ 1, a = 0, c = 0 . (a) Compute the product of a 1 b 1 0 c 1 and a 2 b 2 0 c 2 to show that G is closed under matrix multiplication. (b) Find the matrix inverse of a b 0 c and deduce that G is closed under inverses. (c) Deduce that G is a subgroup of GL 2 (1) (cf. Exercise 26, Section 1). (d) Prove that the set of elements of G whose two diagonal entries are equal (i.e., a = c) is also a subgroup of GL 2 (1). 13 Proof (a). Computing the product of two matrices in G, we find a 1 b 1 0 c 1 a 2 b 2 0 c 2 = a 1 a 2 a 1 b 2 +b 1 c 2 0 c 1 c 2 . Since a 1 , a 2 = 0 and c 1 , c 2 = 0, this matrix is also in G. (b). Notice that a b 0 c 1/a −b/ac 0 1/c = 1 −b/c +b/c 0 1 = 1 0 0 1 1/a −b/ac 0 1/c a b 0 c = 1 b/a −b/a 0 1 = 1 0 0 1 . Since 1/a −b/ac 0 1/c is in G, then G is closed under inverses. (c). First notice that G = ∅ because 1 0 0 1 ∈ G. To show that G is a subgroup of GL 2 (1), we must first show that elements of G have nonzero determinant. So let M = a b 0 c . det M = ac = 0. Thus G ⊆ GL 2 (1). Note matrix multiplication is a binary operation on G, the identity is in G, and G is closed under inverse. So G is a subgroup of GL 2 (1). Proof (d). We first show that matrix multiplication is a binary operation on G = ¦G [ a = c¦. So a 1 b 1 0 a 1 a 2 b 2 0 a 2 = a 1 a 2 a 1 b 2 +b 1 a 2 0 a 1 a 2 . Since a 1 a 2 = a 1 a 2 , matrix multipliaction is indeed a binary operation G . Notice that the identity element is in G . The inverse is given by a b 0 a −1 = 1/a −b/a 2 0 1/a due to the calculation of inverse in G. Thus G is closed under inverse and since G ⊆ G ⊆ GL 2 (1), we must have G as a subgroup of GL 2 (1). 1.5. Quaternion Groups. 1.5.1. Compute the order of each of the elements in Q 8 . Let Q 8 = ¦1, −1, i, −i, j, −j, k, −k¦ as given in the textbook. Then [1[ = 1, [−1[ = 2, [i[ = 4, [−i[ = 4, [j[ = 4, [−j[ = 4, [k[ = 4, [−k[ = 4. 1.5.2. Write out the group tables for S 3 , D 8 , and Q 8 . 1.5.3. Find a set of generators and relations for Q 8 . Q 8 = −1, i, j, k [ (−1) 2 = 1, i 2 = j 2 = k 2 = ijk = −1 14 DAVID S. DUMMIT AND RICHARD M. FOOTE 1.6. Homomorphisms and Isomorphisms. 1.6.2. If φ: G → H is an isomorphism, prove that [φ(x)[ = [x[ for all x ∈ G. Deduce that any two isomorphic groups have the same number of elements of order n for each n ∈ Z + . Is the result true if φ is only assumed to be a homomorphism? Proof. Let [φ(x)[ = n so that φ(x) n = 1 and φ(x) l = 1 for 0 < l < n. But 1 = φ(x) n = φ(x n ) and thus x n = 1. Now suppose that x l = 1. Then 1 = φ(x l ) which is a contradiction. Thus [x[ = n. Hence any two isomorphic groups have the same number of elements of order n for each n ∈ Z + because of the bijectivity of φ. The result that [φ(x)[ = [x[ is not true for φ any homomorphism. For example the trivial map φ(x) = 1 for all x ∈ G for G non-trivial. 1.6.7. Prove that D 8 and Q 8 are not isomorphic. Proof. Note that [s[ = r 2 = 2. But Exercise 1.5.1 shows there are not two elements with order two in Q 8 . Thus Exercise 1.5.2 shows that D 8 and Q 8 are not isomorphic. 1.6.13. Let G and H be groups and let φ: G → H be a homomorphism. Prove that the image of φ, φ(G), is a subgroup of H (cf. Exercise 26 of Section 1). Prove that if φ is injective then G ∼ = φ(G). Proof. Let h 1 , h 2 ∈ H so that there are g 1 , g 2 ∈ G such that φ(g 1 ) = h 1 and φ(g 2 ) = h 2 . Thus φ(g 1 g 2 ) = φ(g 1 )φ(g 2 ) = h 1 h 2 so h 1 h 2 ∈ H. Hence ∗ H×H : H H → H. Notice that φ(1) = 1 so H is nonempty. Let h ∈ H and let φ(g) = h. Then φ(g −1 ) = φ(g) −1 = h −1 , so H is closed under inverse. Therefore H ⊆ G is a subgroup. 1.6.14. Let Gand H be groups and let φ: G →H be a homomorphism. Define the kernel of φ to be ¦g ∈ G [ φ(g) = 1 H ¦ (so the kernel is the set of elements in G which map to the identity of H, i.e., is the fiber over the identity of H). Prove that the kernel of φ is a subgroup (cf. Exercise 26 of Section 1) of G. Prove that φ is injective if and only if the kernel of φ is the identity subgroup of G. Proof. First note that 1 ∈ ker φ so that ker φ = ∅. Now let g 1 , g 2 ∈ ker φ so that φ(g 1 ) = φ(g 2 ) = 1. Then φ(g 1 g 2 ) = φ(g 1 )φ(g 2 ) = 1 so that g 1 g 2 ∈ ker φ. Suppose that g ∈ ker φ so that φ(g) = 1. Then φ(g −1 ) = φ(g) −1 = 1 so g −1 ∈ ker φ and ker φ is closed under inverse. Thus ker φ ⊆ H is a subgroup. Suppose that φ is injective. Thus there is only one element that maps to 1 H . Since φ(1 G ) = 1 H , ker φ = ¦1 G ¦. Now suppose that ker φ = ¦1 G ¦ and let φ(g 1 ) = φ(g 2 ). Then φ(g 1 )φ(g 2 ) −1 = 1. So φ(g 1 g −1 2 ) = 1 and since ker φ = ¦1 G ¦, we must have g 1 g −1 2 = 1 G and thus g 1 = g 2 . Therefore φ is injective. 1.6.18. Let G be any group. Prove that the map from G to itself defined by g → g 2 is a homomorphism if and only if G is abelian. Proof. First suppose that g → g 2 is a homomorphism. Let g 1 , g 2 ∈ G so that g 1 g 2 g 1 g 2 = (g 1 g 2 ) 2 = g 2 1 g 2 2 . Thus g 2 g 1 = g 1 g 2 showing that G is abelian. Now suppose that G is abelian. Let g 1 , g 2 ∈ G so that (g 1 g 2 ) 2 = g 1 g 2 g 1 g 2 = g 2 1 g 2 2 and thus g → g 2 is a homomorphism. 1.7. Group Actions. 1.7.4. Let G be a group acting on a set A and fix some a ∈ A. Show that the following sets are subgroups of G (cf. Exercise 26 of Section 1): (a) the kernel of the action. (b) ¦g ∈ G [ ga = a¦ –this subgroup is called the stabilizer of a in G. Proof (a). Notice that 1 a = a for all a ∈ A so the kernel of the action is nonempty. Let g 1 , g 2 be two elements in the kernel of the action. Then (g 1 g 2 )(a) = g 1 (g 2 a) = g 1 a = a so the group operation on G is a binary operation on the kernel of the action. Now let g be in the kernel of the action. Then g −1 a = g −1 (ga) = (g −1 g)a = a so the kernel of the action is closed under inverse. Therefore the kernel of the action is a subgroup of G. Proof (b). Notice that 1 a = a for all a ∈ A so the stabilizer of a in G is nonempty. Let g 1 , g 2 ∈ ¦g ∈ G [ ga = a¦ so that (g 1 g 2 )a = g 1 (g 2 a) = g 1 a = a and hence the group operation on G is a binary operation on ¦g ∈ G [ ga = a¦. Now let g ∈ ¦g ∈ G [ ga = a¦ so that g −1 a = g −1 (ga) = (g −1 g)a = a and hence ¦g ∈ G [ ga = a¦ is closed under inverse. Therefore ¦g ∈ G [ ga = a¦ is a subgroup of G. 1.7.5. Prove that the kernel of an action of the group G on the set A is the same as the kernel of the corresponding permutation representation G →S A (cf. Exercise 14 in Section 6). 15 Proof. First suppose that ga = a for all a ∈ A. Then σ g (a) = a for all a ∈ A hence σ g = 1 S A . Now suppose that σ g (a) = a for all a ∈ A. Then ga = σ g (a) = a. Therefore these kernels are equivalent. 1.7.10. With reference to the preceding two exercises determine: (a) for which values of k the action of S n on k-element subsets is faithful. Let S n act on subsets of order k for some 1 ≤ k ≤ n. Note if n = 1, then the group action is trivially faithful. Now let σ ∈ S n be a non-identity element for n > 1. Since σ is not the identity, let σ(a i ) = a j for a i = a j . If k < n, then we can choose ¦a 1 , . . . , a k ¦ such that a i ∈ ¦a 1 , . . . , a k ¦ and a j / ∈ ¦a 1 , . . . , a k ¦. Thus σ ¦a 1 , . . . , a k ¦ = ¦a 1 , . . . , a k ¦ . Hence the homomorphism representing the group action is injective for k < n. If k = n, then σ is just a bijection of n-element sets so the homomorphism representing the group action is trivial. Therefore the values of k giving rise to a faithful group action are all values of k < n. 1.7.13. Find the kernel of the left regular action. If g is in the kernel of the left regular action, then ga = a for all a ∈ G. Thus g = 1 so the kernel of the left regular action is ¦1¦. 1.7.14. Let G be a group and let A = G. Show that if G is non-abelian then the maps defined by g a = ag for all g, a ∈ G do not satisfy the axioms of a (left) group action of G on itself. Proof. Since G is non-abelian we can find g 1 , g 2 ∈ G such that g 1 g 2 = g 2 g 1 . We assume the axioms of a group action hold. But g 1 g 2 = a −1 ag 1 g 2 = a −1 ((g 1 g 2 ) a) = a −1 (g 1 (g 2 a)) = a −1 (g 1 (ag 2 )) = a −1 (ag 2 g 1 ) = g 2 g 1 , a contradiction. Therefore this is not a group action. 2. Subgroups 2.1. Definition and Examples. Proposition 2.1 (The Subgroup Criterion). A subset H of a group G is a subgroup if and only if (1) H = ∅ (2) for all x, y ∈ H, xy −1 ∈ H Furthermore, if H is finite, then it suffices to check that H is nonempty and closed under multiplication. 2.2. Centralizers and Normalizers, Stabilizers and Kernels. 2.2.2. Prove that C G (Z(G)) = G and deduce that N G (Z(G)) = G. Proof. Notice C G (Z(G)) = ¸ g ∈ G [ gag −1 = a, ∀a ∈ Z(G) ¸ by definition. But a ∈ Z(G) if and only if ag = ga for all g ∈ G, if and only if a = gag −1 for all g ∈ G. Thus C G (Z(G)) = G. Since C G (Z(G)) ≤ N G (Z(G)) and N G (Z(G)) ⊆ G, we have N G (Z(G)) = G as well. 2.2.4. For each of S 3 , D 8 , and Q 8 compute the centralizers of each element and find the center of each group. Does Lagrange’s Theorem (Exercise 19 in Section 1.7) simplify your work? S 3 . C S3 (1) = S 3 , C S3 ((1 2)) = ¦1, (1 2)¦, C S3 ((1 3)) = ¦1, (1 3)¦, C S3 ((2 3)) = ¦1, (2 3)¦, C S3 ((1 2 3)) = ¦1, (1 2 3), (1 3 2)¦, C S3 ((1 3 2)) = ¦1, (1 2 3), (1 3 2)¦. D 8 . C D8 (1) = D 8 , C D8 (r) = 'r`, C D8 (r 2 ) = D 8 , C D8 (r 3 ) = 'r`, C D8 (s) = ¸ 1, r 2 , s, sr 2 ¸ , C D8 (sr) = ¸ 1, r 2 , sr, sr 3 ¸ , C D8 (sr 2 ) = ¸ 1, r 2 , s, sr 2 ¸ , C D8 (sr 3 ) = ¸ 1, r 2 , sr, sr 3 ¸ . Q 8 . C Q8 (1) = Q 8 , C Q8 (−1) = ¦1, −1¦, C Q8 (i) = ¦1, i¦, C Q8 (−i) = ¦1, −i¦, C Q8 (j) = ¦1, j¦, C Q8 (−j) = ¦1, −j¦, C Q8 (k) = ¦1, k¦, C Q8 (−k) = ¦1, −k¦. Note that Lagrange’s theorem lets us check our answers because the centralizer is a subgroup and thus must divide the order of the group. 2.2.5. In each of parts (a) to (c) show that for the specified group G and subgroup A of G, C G (A) = A and N G (A) = G. (a) G = S 3 and A = ¦1, (1 2 3), (1 3 2)¦. (b) G = D 8 and A = ¸ 1, s, r 2 , sr 2 ¸ . (c) G = D 10 and A = ¸ 1, r, r 2 , r 3 , r 4 ¸ . 16 DAVID S. DUMMIT AND RICHARD M. FOOTE Proof (a). Note that C G (A) = ∩ a∈A C G (a) so by Exercise 2.2.4, C S3 (A) = S 3 ∩¦1, (1 2 3), (1 3 2)¦∩¦1, (1 2 3), (1 3 2)¦ = A. Proof (b). Similarly to part (a) by Exercise 2.2.4, C D8 (A) = D 8 ∩ ¸ 1, r 2 , s, sr 2 ¸ ∩ D 8 ∩ ¸ 1, r 2 , s, sr 2 ¸ = A. Proof (c). Note that any power of r commutes with any power of r so A ⊆ C D10 (A). Since C D10 (r) = 'r`, we also must have C D10 (A) ⊆ A. 2.2.9. For any subgroup H of G and any nonempty subset A of G define N H (A) to be the set ¸ h ∈ H [ hAh −1 = A ¸ . Show that N H (A) = N G (A) ∩ H and deduce that N H (A) is a subgroup of H (note that A need not be a subset of H). Proof. Let x ∈ N H (A) so x ∈ H ≤ G and xAx −1 = A. Thus x ∈ N G (A) ∩ H. Now let x ∈ N G (A) ∩ H so that x ∈ HandxAx −1 = A. Thus x ∈ G H (A) and hence N H (A) = N G (A) ∩ H. Note that N G (A) and H are both subgroups of G and the intersection of two subgroups is again a subgroup. Therefore N H (A) ≤ G. 2.2.10. Let H be a subgroup of order 2 in G. Show that N G (H) = C G (H). Deduce that if N G (H) = G then H ≤ Z(G). Proof. Let H = ¦1, h¦, x ∈ N G (H) so that x ∈ G and xHx −1 = H. Since x1x −1 = 1, we must have xhx −1 = h so that x¦1, h¦ x −1 = ¦1, h¦. Thus x ∈ C G (H). Now if x ∈ C G (H), then x1x −1 = 1 and xhx −1 = h so xHx −1 = H. Thus x ∈ N G (H) and hence N G (H) = C G (H). Note if N G (H) = G, then G = C G (H). Thus any element of G commutes with any element in H so H ⊆ Z(G). Since H and Z(G) are subgroups we get H ≤ Z(G). 2.3. Cyclic Groups and Cyclic Subgroups. Proposition 2.2. If H = 'x`, then [H[ = [x[ (where if one side of this equality is infinite, so is the other). More specifically (1) if [H[ = n < ∞, then x n = 1 and 1, x, x 2 , . . . , x n−1 are all the distinct elements of H (2) if [H[ = ∞, then x n = 1 for all n = 0 and x a = x b for all a = b in Z. Proposition 2.3. Let G be an arbitrary group, x ∈ G and let m, n ∈ Z. If x n = 1 and x m = 1, then x d = 1, where d = (m, n). In particular, if x m = 1 for some m ∈ Z, then [x[ divides m. Theorem 2.4. Any two cyclic groups of the same order are isomorphic. More specifically, (1) if n ∈ Z + and 'x` and 'y` are both cyclic groups of order n, then the map ϕ: 'x` →'y` x k →y k is well defined and is an isomorphism (2) if 'x` is an infinite cyclic group, the map ϕ: Z →'x` k →x k is well defined and is an isomorphism. Proposition 2.5. Let G be a group, let x ∈ G and let z ∈ Z` ¦0¦. (1) If [x[ = ∞, then [x a [ = ∞. (2) If [x[ = n < ∞, then [x a [ = n (n,a) . (3) In particular, if [x[ = n < ∞ and a is a positive integer dividing n, then [x a [ = n a . Proposition 2.6. Let H = 'x`. (1) Assume [x[ = ∞. Then H = 'x a ` if and only if a = ±1. (2) Assume [x[ = n < ∞. Then H = 'x a ` if and only if (a, n) = 1. In particular, then number of generators of H is ϕ(n). Theorem 2.7. Let H = 'x` be a cyclic group. (1) Every subgroup of H is cyclic. More precisely, if K ≤ H, then either K = ¦1¦ or K = x d , where d is the smallest positive integer such that x d ∈ K. 17 (2) If [H[ = ∞, then for any distinct nonnegative integers a and b, 'x a ` = x b . Furthermore, for every integer m, 'x m ` = x |m| , where [m[ denotes the absolute value of m, so that the nontrivial subgroups of H correspond bijectively with the integers 1, 2, 3, . . .. (3) If [H[ = n < ∞, then for each positive integer a dividing n there is a unique subroups of H of order a. This subgroup is the cyclic group x d , where d = n a . Furthermore, for every integer m, 'x m ` = x (n,m) , so that the subgroups of H correspond bijectively with the positive divisors of n. 2.3.1. Find all subgroups of Z 45 = 'x`, giving a generator for each. Describe the containments between these subgroups. The subgroups of Z 45 are given by x 0 , 'x` , x 3 , x 5 , x 9 , x 15 . The containments between these subgroups are given by 'x` ⊇ x 3 ⊇ x 5 ⊇ x 9 ⊇ x 15 ⊇ x 0 . 2.3.5. Find the number of generators for Z/49000 Z. Since Z/49000 Z is cyclic of order 49000, we count the number of relatively prime integers to 49000. This is given by Euler’s ϕ function, so ϕ(49000) = ϕ(2 3 5 3 7 2 ) = ϕ(2 3 )ϕ(5 3 )ϕ(7 2 ) = 2 2 (1) 5 2 (4) 7(6) = 16800. 2.3.8. Let Z 48 = 'x`. For which integers a does the map ϕ a defined by ϕ a : ¯ 1 →x a extend to an isomorphism from Z/48Z onto Z 48 . All integers a such that (a, 48) = 1. So a ∈ ¦1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47¦. 2.3.12. Prove that the following groups are not cyclic: (b) Z 2 Z Proof. Suppose Z 2 Z is cyclic and let (x, y) be a generator. Note that x, y are not identity elements in their respective groups. Then there exists k ∈ Z such that (1, y) = (x, y) k = (x k , ky). Note k = 1 because x = 1. But then y = ky shows that y = 0. This is a contradiction so this group is not cyclic. 2.3.13. Prove that the following pairs of groups are not isomorphic: (b) QZ 2 and Q. Proof. Suppose that ϕ: (Q Z 2 ) → Q is an isomorphism. Let (0, x) ∈ Q Z 2 be a non-identity element. So ϕ((0, x)) = y = 0 but 0 = ϕ((0, 1)) = ϕ((0, x) 2 ) = 2y. Since y = 0 this is a contradiction, thus no such isomorphism exists. 2.3.18. Show that if H is any group and h is an element of H with h n = 1, then there is a unique homomorphism from Z n = 'x` to H such that x →h. Proof. Since h n = 1, there is an m ∈ N such that ['h`[ = m so m[n. Define a map ϕ(x k ) = h k . If x k = x l so that k ≡ l (mod n), then k − l = αn for some α ∈ Z. But then h k = h αn+l = h l so ϕ is well-defined. Now take x k , x l ∈ 'x` so that ϕ(x k )ϕ(x l ) = h k h l = h k+l = ϕ(x k+l ) shows that ϕ is such a homomorphism. Now suppose that ϕ and ψ are two homomorphisms such that ϕ(x) = h = ψ(x). Take some x k ∈ 'x` so that ϕ(x k ) = ϕ(x) k = h k = ψ(x) k = ψ(x k ). Therefore ϕ = ψ shows there is a unique map with these properties. 2.4. Subgroups Generated by Subsets of a Group. Proposition 2.8. If / is any nonempty collection of subgroups of G, then the intersection of all members of / is also a subgroup of G. Proposition 2.9. A = 'A`. 2.4.8. Prove that S 4 = '(1 2 3 4), (1 2 4 3)`. 18 DAVID S. DUMMIT AND RICHARD M. FOOTE Proof. First note that [S 4 [ = 4! = 24. Since '(1 2 3 4), (1 2 4 3)` ≤ S 4 , Lagrange’s theorem states we need to find 13 elements generated by this set so that '(1 2 3 4), (1 2 4 3)` = S 4 . So we have (1 2 3 4), (1 2 3 4) 2 = (1 3)(2 4), (1 2 3 4) 3 = (1 4 3 2), (1 2 4 3), (1 2 4 3) 2 = (1 4)(2 3), (1 2 4 3) 3 = (1 3 4 2), (1 2 3 4)(1 2 4 3) = (1 3 2), (1 2 3 4)(1 4)(2 3) = (2 4), (1 2 3 4)(1 3 4 2) = (1 4 3), (1 3)(2 4)(1 2 4 3) = (1 4), (1 4 3 2)(1 2 4 3) = (2 3 4), (1 4)(2 4) = (1 4 2), (1 3)(2 4)(2 4) = (1 3). 2.4.9. Prove that SL 2 (F 3 ) is the subgroup of GL 2 (F 3 ) generated by 1 1 0 1 and 1 0 1 1 . Proof. First note that [SL 2 (F 3 )[ = 24 so we use a method similar to exercise 2.4.8 to show this subgroup is generated by the two matrices. So we have 1 1 0 1 , 1 1 0 1 2 = 1 2 0 1 , 1 1 0 1 3 = 1 0 0 1 , 1 0 1 1 , 1 0 1 1 2 = 1 0 2 1 , 1 1 0 1 1 0 1 1 = 2 1 1 1 , 1 2 0 1 1 0 1 1 = 0 2 1 1 , 1 1 0 1 1 0 2 1 = 1 1 2 1 , 1 0 1 1 1 1 0 1 = 1 1 1 2 , 1 0 1 1 1 2 0 1 = 1 2 1 0 , 1 0 2 1 1 1 0 1 = 1 1 2 0 , 1 0 2 1 2 1 1 1 = 2 1 2 0 , 1 1 1 2 1 2 1 0 = 2 2 0 2 . 2.4.10. Prove that the subgroup of SL 2 (F 3 ) generated by 0 −1 1 0 and 1 1 1 −1 is isomorphic to the quaternion group of order 8. [Use a presentation for Q 8 .] Proof. Let Q 8 be given by the presentation Q 8 = i, j [ i 4 = 1 = j 4 , ij = −ji . Notice that 0 −1 1 0 4 = I = 1 1 1 −1 4 , and 0 −1 1 0 1 1 1 −1 = −1 1 1 1 = − 1 −1 −1 −1 = − 1 1 1 −1 0 −1 1 0 . Thus the relations of the subgroup of SL 2 (F 3 ) generated by the matrices are contained in the presentation of Q 8 . So the order of the matrix subgroup is less than or equal to 8. Since we have found 5 matrices in this subgroup and [SL 2 (F 3 )[ = 24, Lagrange’s theorem shows that the order of this subgroup is indeed 8. Therefore these groups are isomorphic. 2.4.11. Prove that SL 2 (F 3 ) and S 4 are two nonisomorphic groups of order 24. Proof. First notice that there is more than one element of order two in S 4 , namely (1 2) and (1 3). Now let a b c d ∈ SL 2 (F 3 ) be an element of order 2. So a b c d 2 = a 2 +bc ab +bd ac +cd bc +d 2 = 1 0 0 1 and ad − bc = 1. Since bc = ad − 1 we see ad − 1 + d 2 = 1. So d(a + d) = 2 shows that cd(a + d) = 2c = 0 hence c = 0. So ad = a 2 = d 2 = 1 together with b(a + d) = 0 shows that a 2 b + abd = 0 hence b = 0. Therefore the only element of order 2 is −1 0 0 −1 and hence these groups are not isomorphic. 2.4.12. Prove that the subgroup of upper triangular matrices in GL 3 (F 2 ) is isomorphic to the dihedral group of order 8. [First find the order of this subgroup.] Proof. Since the diagonal must be have nonzero entries, the diagonal is a string of 1’s. Thus the order of this subgroup is 2 3 = 8. We will show that the subgroup of upper triangular matrices in GL 3 (F 2 ) is equal to A = ¸ 1 1 1 0 1 0 0 0 1 ¸ , B = ¸ 1 1 0 0 1 1 0 0 1 ¸ ¸ 19 where AB = B −1 A and B 4 = I = A 2 . Calculating we find ¸ 1 1 1 0 1 0 0 0 1 ¸ 2 = ¸ 1 0 0 0 1 0 0 0 1 ¸ ¸ 1 1 0 0 1 1 0 0 1 ¸ 4 = ¸ 1 1 0 0 1 1 0 0 1 ¸ 2 ¸ 1 0 1 0 1 0 0 0 1 ¸ = ¸ 1 1 0 0 1 1 0 0 1 ¸ ¸ 1 1 1 0 1 1 0 0 1 ¸ = ¸ 1 0 0 0 1 0 0 0 1 ¸ . Thus we found five matrices generated by A and B to show that it generates the whole group by Lagrange’s theorem. Therefore these two groups are isomorphic. 2.5. Definitions and Examples. Let G and H be groups. 2.5.1. Let ϕ: G → H be a homomorphism and let E be a subgroup of H. Prove that ϕ −1 (E) ≤ G (ie., the preimage or pullback of a subgroup under a homormorphism is a subgroup). If E H prove that ϕ −1 (E) G. Deduce that ker ϕ G. Proof. Let g 1 , g 2 ∈ ϕ −1 (E) so ϕ(g 1 ) = h 1 and ϕ(g 2 ) = h 2 where h 1 , h 2 ∈ E. But ϕ(1 G ) = ϕ(1 H ) and ϕ(g 1 g −1 2 ) = h 1 h −1 2 ∈ E so ϕ −1 (E) ≤ G. If E H then hEh −1 ⊆ E for all h ∈ H. So take ϕ g ϕ −1 (E) g −1 ⊆ ϕ(g) E ϕ(g) −1 ⊆ E. Thus g ϕ −1 (E) g −1 ⊆ ϕ −1 (E) shows ϕ −1 (E) G. Since 1 H is normal in H then ker ϕ G. 2.5.5. Use the preceding exercise to prove that the order of the element gN in G/N is n, where n is the smallest positive integer such that g n ∈ N (and gN has infinite order if no such positive integer exists). Give an example to show that the order of gN in G/N may be strictly smaller than the order of g in G. Proof. First note if 1 < m < n then (gN) m = g m N _ N for g / ∈ N so the order of gN ≥ n. But (gN) n = g n N = N so the order of gN is n. An example with the desired property would be D 8 / r 2 . Note that r 2 = 2 in D 8 but r 2 r 2 = 1 in D 8 / r 2 . 2.5.6. Define ϕ: 1 × →¦±1¦ by letting ϕ(x) be x divided by the absolute value of x. Describe the fibers of ϕ and prove that ϕ is a homomorphism. Proof. The fibers of ϕ are given by ϕ −1 (1) = ¦α ∈ 1 × [ α > 0¦ and ϕ −1 (−1) = ¦α ∈ 1 × [ α < 0¦. To show ϕ is a homomorphism we calculate ϕ(αβ) = 1 αβ > 0, −1 αβ < 0, ϕ(α)ϕ(β) = 1 αβ > 0, −1 αβ < 0. 2.5.9. Define ϕ: C × → 1 × by ϕ(a + bi) = a 2 + b 2 . Prove that ϕ is a homomorphism and find the image of ϕ. Describe the kernel and the fibers of ϕ geometrically (as subsets of the plane). Proof. We calculate, ϕ((a +bi)(c +di)) = ϕ(ac +adi +bci −bc) = (ac −bd) 2 + (ad +bd) 2 = a 2 c 2 −2abcd +b 2 d 2 +a 2 d 2 + 2abcd +b 2 d 2 ϕ(a +bi)ϕ(c +di) = (a 2 +b 2 )(c 2 +d 2 ) = a 2 c 2 +a 2 d 2 +b 2 c 2 +b 2 d 2 20 DAVID S. DUMMIT AND RICHARD M. FOOTE so ϕ is a homomorphism. 2.5.11. Let F be a field and let G = a b 0 c [ a, b, c ∈ F, ac = 0 ≤ GL 2 (F). (a) Prove that the map ϕ: a b 0 c →a is a surjective homomorphism from G onto F × (recall that F × is the multiplicative group of nonzero elements in F). Describe the fibers and kernel of ϕ. (b) Prove that the map ψ: a b 0 c → (a, c) is a surjective homomorphism from G onto F × F × . Describe the fibers and kernel of ψ. (c) Let H = 1 b 0 1 [ b ∈ F . Prove that H is isomorphic to the additive group F. Proof (a). ϕ is clearly surjective so we show it is a homomorphism by a 1 b 1 0 c 1 a 2 b 2 0 c 2 = a 1 a 2 a 1 b 2 +b 1 c 2 0 c 1 c 2 . The fibers of ϕ are given by ϕ −1 (a) = a b 0 c [ b ∈ F, c = 0 . The kernel of ϕ is elements of G with a = 1. Proof (b). ψ is clearly surjective and the calculation from proof (a) shows it is a homomorphism. The fibers of ψ are given by ψ −1 ((a, c)) = a b 0 c [ b ∈ F . The kernel of ψ is elements of G with a = c = 1. Proof (c). The calculation 1 b 1 0 1 1 b 2 0 1 = 1 b 1 +b 2 0 1 makes it clear that H is isomorphic to F + . 2.5.12. Let G be the additive group of real numbers, let H be the multiplicative group of complex numbers of absolute value 1 (the unit circle S 1 in the complex plane) and let ϕ: G → H be the homomorphism ϕ: r → e 2πir . Draw the points on a real line which lie in the kernel of ϕ. Describe similarly the elements in the fibers of ϕ above the points −1, i, and e 4πi/3 of H. The points of the real line in the kernel of ϕ are the integers. The fibers are given by ϕ −1 (−1) = (1/2)(1 + /Z), ϕ −1 (i) = (1/4)(1 + /Z), ϕ(e 4πi/3 ) = (2/3)(1 + /Z). 2.6. The Lattice of Subgroups of a Group. 3. Quotient Groups and Homomorphisms 3.1. Definitions and Examples. Proposition 3.1. Let G and H be groups and let ϕ: G →H be a homomorphism. (1) ϕ(1 G ) = 1 H (2) ϕ(g −1 ) = ϕ(g) −1 for all g ∈ G. (3) ϕ(g n ) = ϕ(g) n for all n ∈ Z. (4) ker ϕ is a subgroup of G. (5) im(ϕ) is a subgroup of H. Proposition 3.2. Let ϕ: G → H be a homomorphism of groups with kernel K. Let X ∈ G/K be the fiber above a, i.e., X = ϕ −1 (a). Then (1) For any u ∈ X, X = ¦uk [ k ∈ K¦ (2) For any u ∈ X, X = ¦ku [ k ∈ K¦. Theorem 3.3. Let G be a group and let K be the kernel of some homomorphism from G to another group. Then the set whose eelements are the left cosets of K in G with operation defined by uK ◦ vK = (uv)K forms a group, G/K. In particular, this operation is well defined in the sense that if u 1 is any element in uK and v 1 is any element in vK, then u 1 v 1 ∈ uvK, e.e., u 1 v 1 K = uvK so that the multiplication does not depend on the choice of representatives for the cosets. The same statement is true with “right coset” in place of “left coset.” 21 Proposition 3.4. Let N be any subgroup of the group G. The set of left cosets of N in G form a partition of G. Furthermore, for all u, v ∈ G, uN = vN if and only if v −1 u ∈ N and in particular, uN = vN if and only if u and v are representatives of the same coset. Proposition 3.5. Let G be a group and let N be a subgroup of G. (1) The operation on the set of left cosets of N in G described by uN vN = (uv)N is well defined if and only if gng −1 ∈ N for all g ∈ G and all n ∈ N. (2) If the above operation is well defined, then it makes the set of left cosets of N in G into a group. In particular the identity of this group is the coset 1N and the inverse of gN is the coset g −1 N i.e., (gN) −1 = g −1 N. Theorem 3.6. Let N be a subgroup of the group G. The following are equivalent: (1) N G (2) N G (N) = G (3) gN = Ng for all g ∈ G (4) the operation on left cosets of N in G described in proposition 3.5 makes the set of left cosets into a group (5) gNg −1 ⊆ N for all g ∈ G. Proposition 3.7. A subgroup N of the group G is normal if and only if it is the kernel of some homomorphism. 3.1.36. Prove that if G/Z(G) is cyclic then G is abelian. Proof. If G/Z(G) is cyclic with generator xZ(G), then if gZ(G) ∈ G/Z(G) we have gZ(G) = (xZ(G)) n = x n Z(G) for some n ∈ N. Thus there is some z ∈ Z(G) such that g = x n z. Now take g 1 , g 2 ∈ G such that g 1 = x m z 1 and g 2 = x n z 2 . Then g 1 g 2 = x m z 1 x n z 2 = x m x n z 2 z 1 = x n x m z 2 z 1 = x n z 2 x m z 1 = g 2 g 1 . 3.2. More on Cosets and Lagrange’s Theorem. Theorem 3.8 (Lagrange’s Theorem). If G is a finite gorup and H is a subgroup of G, then the order of H divides the order of G (i.e., [H[ [ [G[) and the number of left cosets of H in G equals |G| |H| . Corollary 3.9. If G is a finite group and x ∈ G, then the roder of x divides the order of G. In particular x |G| = 1 for all x in G. Corollary 3.10. If G is a group of prime order p, then G is cyclic, hence G ∼ = Z p . Theorem 3.11 (Cauchy’s Theorem). If G is a finite group and p is a prime dividing [G[, then G has an element of order p. Theorem 3.12 (Sylow). If G is a finite group of order p α m, where p is a prime and p does not divide m, then G has a subgroup of order p α . Proposition 3.13. If H and K are finite subgroups of a group then [HK[ = [H[ [K[ [H ∩ K[ . Proposition 3.14. If H and K are subgroups of a group, HK is a subgroup if and only if HK = KH. Corollary 3.15. If H and K are subgroups of G and H and H ≤ N G (K), then HK is a subgroup of G. In particular, if K G then HK ≤ G for any H ≤ G. 3.2.8. Prove that if H and K are finite subgroups of G whose orders are relatively prime then H ∩ K = 1. Proof. Let [H[ = m and [K[ = n with (m, n) = 1. Now take x ∈ H ∩ K so [x[ [m and [x[ [n so x = 1. 22 DAVID S. DUMMIT AND RICHARD M. FOOTE 3.2.9. This exercise outlines a proof of Cauchy’s Theorem due to James McKay. Let G be a finite group and let p be a prime dividing [G[. Let o denote the set of p-tuples of elements of G the product of whose coordinates is 1: o = ¦(x 1 , x 2 , . . . , x p ) [ x i ∈ G and x 1 x 2 x p = 1¦ . Define the relation ∼ on o by letting α ∼ β if β is a cyclic permutation of α. (a) Show that o has [G[ p−1 elements, hence has order divisible by p. (b) Show that a cyclic permutation of an element of o is again an element of S. (c) Prove that ∼ is an equivalence relation on o. (d) Prove that an equivalence class contains a single element if and only if it is of the form (x, x, . . . , x) with x p = 1. (e) Prove that every equivalence class has order 1 or p (this uses the fact that p is a prime). Deduce that [G[ p−1 = k +pd, where k is the number of classes of size 1 and d is the number of classes of size p. (f) Since ¦(1, 1, . . . , 1)¦ is an equivalence class of size 1, conclude from (e) that there must be a nonidentity element x in G with x p = 1, i.e., G contains an element of order p. [Show p[k and so k > 1.] Proof (a). The first p − 1 coordinates have [G[ choices and the pth coordinate must be the inverse of the product of the first p −1 coordinates. Thus [o[ = [G[ p−1 . Proof (b). Let (x 1 , x 2 , . . . , x p ) ∈ o and notice x p x 1 x 2 x p−1 = 1 just by multiplication on the right by (x p ) −1 , then the left by x p . Thus (x p , x 1 , . . . , x p−1 ) ∈ o and any cyclic permutation can be realized by repeating this process. Proof (c). Reflexive is obvious. If α is a cyclic permutation of β, then β is a cyclic permutation of α so ∼ is symmetric. If β is a cyclic permutation of α and γ is a cyclic permutation of β, then γ is a cyclic permutation of α so ∼ is transitive. Thus ∼ is an equivalence relation. Proof (d). If some x is changed to a y = x, then a cyclic permutation would yield a different element of the equivalence class. If all the x are the same, then any cyclic permutation is indeed the same element of o. Proof (e). It is clear that every equivalence class has order less than or equal to p. Suppose the equivalence class is not of the form from part (d) and let σ ∈ S p be given by σ(i) ≡ i + j mod p for 0 ≤ j < p and suppose that (x 1 , . . . , x p ) ∼ (x σ(1) , . . . , x σ(p) ). Then x i = x σ(i) = x σ 2 (i) = = x σ p−1 (i) for all i. Since not every coordinate is allowed to be equal, j = 0 because [σ[ = p (for p prime, σ = 1). Thus the number of equivalence classes is the sum of size 1 and size p classes. Since equivalence classes are disjoint, [G[ p−1 = k +pd. Proof (f ). Since p divides [G[, then p divides [G[ p−1 and specifically p divides k. Thus k > 1 and hence there is an element x in G with order p. 3.2.10. Suppose H and K are subgroups of finite index in the (possibly infinity) group G with [G : H[ = m and [G : K[ = n. Prove that lcm(m, n) ≤ [G : H ∩ K[ ≤ mn. Deduce that if m and n are relatively prime then [G : H ∩ K[ = [G : H[ [G : K[. Proof by Dr. Schulze. By Problem 11, you have [G : H ∩ K] = [G : K][K : H ∩ K]. So n divides [G : H ∩ K], and also m divides by the same argument. Therefore lcm(m, n) ⇐[G : H ∩ K]. Now note that the first isomorphism theorem works also for non-normal subgroups, but you get a statement on maps of sets (not groups). Apply this to the map G →GG →G/H G/K sending g →(g, g) →(gH, gK). Its kernel is H ∩ K, so the (set version of the) first isomorphism theorem gives an injective map G/(H ∩ K) → G/H G/K. Thus, [G : H ∩ K] = [G/(H ∩ K)[ ⇐[G/H G/K[ = [G/H[[G/K[ = n m. 3.2.22. Use Lagrange’s Theorem in the multiplicative group (Z/nZ) × to prove Euler’s Theorem: a ϕ(n) ≡ 1 mod n for every integer a relatively prime to n, where ϕ denotes Euler’s ϕ-function. Proof. Since a is relatively prime to n, a ∈ (Z`nZ) × . Corollary 9 states that a ϕ(n) = a [(Z\nZ) × [ = 1 |(Z\nZ) × | ≡ 1 mod n. 23 3.3. The Isomorphism Theorems. Theorem 3.16 (The First Isomorphism Theorem). If ϕ: G → H is a homomorphism of groups, then ker ϕ G and G/ ker ϕ ∼ = ϕ(G). Corollary 3.17. Let ϕ: G →H be a homomorphism of groups. (1) ϕ is injective if and only if ker ϕ = 1. (2) [G: ker ϕ[ = [ϕ(G)[. Theorem 3.18 (The Second Isomorphism Theorem). Let G be a group, let A and B be subgroups of G and assume A ≤ N G (B). Then AB is a subgroup of G, B AB, A∩ B A and AB/B ∼ = A/A∩ B. Theorem 3.19 (The Third Isomorphism Theorem). Let G be a group and let H and K be normal subgroups of G with H ≤ K. Then K/H G/H and (G/H)/(K/H) ∼ = G/K. If we denote the quotient by H with a bar, this can be written G/K ∼ = G/K. Theorem 3.20 (The Fourth Isomorphism Theorem). Let G be a group and let N be a normal subgroup of G. Then there is a bijection from the set of subgroups A of G which contain N onto the set of subgroups A = A/N of G/N. In particular, every subgroup of G is of the form A/N for some subgroup A of G containing N (namely, its preimage in G under the natural projection homomorphism from G to G/N). This bijection has the following properties: for all A, B ≤ G with N ≤ A and N ≤ B, (1) A ≤ B if and only if A ≤ B, (2) if A ≤ B, then [B: A[ = B: A , (3) 'A, B` = A, B , (4) A∩ B = A∩ B, and (5) A G if and only if A G. 3.3.3. Prove that if H is a normal subgroup of G of prime index p then for all K ≤ G either (i) K ≤ H or (ii) G = HK and [K: K ∩ H[ = p. Proof. Suppose that K < K so since K ≤ N G (H), HK = KH ≤ G. Since [K[ = 1, [HK[ > H so HK = G because [G : H[ = p and Lagrange’s theorem. The fact that [K : H ∩ H[ = p is immediate by the diamond isomorphism theorem. 3.3.4. Let C be a normal subgroup of the group A and let D be a normal subgroup of the group B. Prove that (C D) (AB) and (AB)/(C D) ∼ = (A/C) (B/D). Proof. Let π 1 : A → A/C and π 2 : B → B/D which exist by the normality of C and D. Now let Φ: (A B) → (A/C B/D) be defined by π 1 and π 2 . Thus ker Φ = C D and thus (C D) (A B). Since Φ is surjective, the first isomorphism theorem shows that (AB)/(C D) ∼ = (A/C) (B/D). 3.3.7. Let M and N be normal subgroups of G such that G = MN. Prove that G/(M ∩ N) ∼ = (G/M) (G/N). [Draw the lattice.] Proof. Define a homomorphism ϕ: G → (G/M) (G/N) by ϕ(g) = (g, g). Take some (m, n) ∈ (G/M) (G/N). Then nm n −1 = m so take nm = g = mn so ϕ(g) = (g, g) = (m, n) so ϕ is surjective. The kernel of ϕ is clearly M∩N because g ∈ M and g ∈ N if and only if (g, g) = 1 G/M×G/N . Therefore the first isomorphism theorem states that G/(M ∩ N) ∼ = (G/M) (G/N). 3.3.9. Let p be a prime and let G be a group of order p a m, where p does not divide m. Assume P is a subgroup of G of order p a and N is a normal subgroup of G of order p b n, where p does not divide n. Prove that [P ∩ N[ = p b and [PN/N[ = p a−b . (The subgroup P of G is called a Sylow p-subgroup of G. This exercise shows that the intersection of any Sylow p-subgroup of G with a normal subgroup N is a Sylow p-subgroup of N.) Proof. First note that [PN[ = ([P[ [N[)/ [P ∩ N[ = (p a+b n)/ [P ∩ N[. But this divides p a m so [P ∩ N[ = p b α. But then p b α divides p b n so α divides n shows α does not divide p. But now p b α divides p a so α = 1 and hence [P ∩ N[ = p b . The diamond isomorphism theorem shows that [PN/N[ = p a−b . 24 DAVID S. DUMMIT AND RICHARD M. FOOTE 3.4. Composition Series and the Holder Program. Proposition 3.21. If G is a finite abelian group and P is a prime dividing [G[, then G contains an element of order p. Theorem 3.22 (Jordan-Holder). Let G be a finite group with G = 1. Then (1) G has a composition series and (2) The composition factors in a composition sereis are unique, namely, if 1 = N 0 ≤ N 1 ≤ ≤ N r = G and 1 = M 0 ≤ M 1 ≤ ≤ M s = G are two composition series for G, then r = s and there is some permutation, π, of ¦1, 2, . . . , r¦ such taht M π(i) /M π(i)−1 ∼ = N i /N i−1 , 1 ≤ i ≤ r. 3.5. Transpositions and the Alternating Group. Proposition 3.23. The map : S n → ¦±1¦ is a homomorphism (where ¦±1¦ is a multiplicative version of the cyclic group of order 2). Proposition 3.24. Transpositions are all odd permutations and is a surjective homomorphism. Proposition 3.25. The permutation σ is odd if and only if the number of cycles of even length in its cycle decomposition is odd. 3.5.3. Prove that S n = '¦(i i + 1) [ 1 ≤ i ≤ n −1¦`. Proof. Let σ ∈ S n and decompose σ into a product of transpositions. But if (i j) is any transposition, then (i j) = (j −1 j) (i i + 1) (j −1 j). Therefore this set generates S n . 3.5.5. Show that if p is prime, S p = 'σ, τ` where σ is any transposition and τ is any p-cycle. Proof. Without loss of generality, suppose σ = (1 2) and τ = (1 p). If 1 ≤ i ≤ n −1, then (i i + 1) = τ i−1 στ −i+1 . Exercise 3.5.3 shows this generates S p . 3.5.6. Show that '(1 3), (1 2 3 4)` is a proper subgroup of S 4 . What is the isomorphism type of this subgroup? Proof. Notice that (1 3)(1 2 3 4) 2 (1 3) = (1 2 3 4) 2 shows that (1 2 3 4) 2 ∈ Z('(1 3), (1 2 3 4)`). But (1 2 3 4) 2 (1 4) = (1 2 4 3) = (1 3 4 2) = (1 4)(1 2 3 4) 2 . Therefore '(1 3), (1 2 3 4)` < S 4 . The isomorphism type of this subgroup is D 8 because we can construct 8 elements and we have (1 3) 2 = (1 2 3 4) 4 = (1 3)(1 2 3 4)(1 3)(1 2 3 4) = 1. 3.5.7. Prove that the group of rigid motions of a tetrahedron is isomorphic to A 4 . Proof. First notice that both groups are order 12. Now we represent the group of rigid motions of a tetrahedron as a subgroup of S n . The elements in this subgroup are 1, (1 2 3), (1 3 2), (1 3 4), (1 4 3), (1 2 4), (1 4 2), (2 3 4), (2 4 3), (1 3)(2 4), (1 2)(3 4), (1 4)(2 3). Since all 12 of these elements are even, this is A 4 . 3.5.10. Find a composition series for A 4 . Deduce that A 4 is solvable. Solution. Let N 1 = ¦1, (1 3)(2 4)¦ and N 2 = ¦1, (1 3)(2 4), (1 2)(3 4), (1 4)(2 3)¦. Simple calculations show that 1 = N 0 N 1 N 2 N 3 = A 4 , while [N i+1 /N i [ = p shows that each composition factor is simple. Since the order of each composition factor is prime, they are also abelian and hence A 4 is solvable. 3.5.17. If x and y are 3-cycles in S n , prove that 'x, y` is isomorphic to Z 3 , A 4 , A 5 , or Z 3 Z 3 . 25 Proof. If x = y, then 'x, y` = 'x` ∼ = Z 3 by order argument. Now let x = (i j k) and y = (j k l) and notice these are both even permutations hence generate even permutations. Since 1, (i j k), (i k j), (j k l), (j l k), (i k l) and (i j)(k l) are all generated by x and y, then 'x, y` ∼ = A 4 by Lagrange’s theorem. Now let x = (i j k), y = (k l m) and we will show that 'x, y` A 5 , so they are equal because A 5 is simple. If we can show conjugating (i j k) by any transposition remains in 'x, y`, then we are done because any permutation can be represented by a transposition, and the argument for (k l m) will be similar. But (i j k) (i j) = (i j k) (j k) = (i j k) (i k) = (i j k) −1 , (i j k) (l m) = (i j k), (i j k) (i l) = (j k l) = [(i j k) (k l m) ] (i j k) , (i j k) (j l) = [(j i k) (k l m) ] (j i k) , (i j k) (i m) = [(i j k) (k m l) ] (i j k) , (i j k) (j m) = [(j i k) (k m l) ] (j i k) , (i j k) (k l) = (i j k) (k l m) , and (i j k) (k m) = (i j k) (k m l) . If x and y are disjoint 3-cycles, then it is clear that 'x, y` ∼ = Z 3 Z 3 . 4. Group Actions 4.1. Group Actions and Permutation Representations. 4.2. Groups Acting on Themselves by Left Multiplication – Cayley’s Theorem. 4.2.2. List the elements of S 3 as 1, (1 2), (2 3), (1 3), (1 2 3), (1 3 2) and label these with the integers 1,2,3,4,5,6 respectively. Exhibit the image of each element of S 3 under the left regular representation of S 3 into S 6 . Solution. 4.2.4. Use the left regular representation of Q 8 to produce two elements of S 8 which generate a subgroup of S 8 isomorphic to the quaternion group Q 8 . Solution. 4.2.8. Prove that if H has finite index n then there is a normal subgroup K of G with K ≤ H and [G : K[ ≤ n!. Proof. 4.2.10. Prove that every non-abelian group of order 6 has a nonnormal subgroup of order 2. Use this to classify groups of order 6. [Produce an injective homomorphism into S 3 .] Proof. 4.3. Groups Acting on Themselves by Conjugation – The Class Equation. 4.3.3. Find all conjugacy classes and their sizes in the following groups: (c) A 4 . Solution (c). 4.3.6. Assume G is a non-abelian group of order 15. Prove that Z(G) = 1. Use the fact that 'g` ≤ C G (g) for all g ∈ G to show that there is at most one possible class equation for G. [Use exercise 36, section 3.1.] Proof. 4.3.13. Find all finite groups which have exactly two conjugacy classes. Solution. 4.3.27. Let g 1 , g 2 , . . . , g r be representatives of the conjugacy classes of the finite group G and assume these elements pairwise commute. Prove that G is abelian. Proof. 4.3.29. Let p be a prime and let G be a group of order p α . Prove that G has a subgroup of order p β , for every β with 0 ≤ β ≤ α. [Use theorem 8 and induction on α.] Proof. 26 DAVID S. DUMMIT AND RICHARD M. FOOTE 4.4. Automorphisms. Let G be a group. 4.4.1. If σ ∈ Aut(G) and ϕ g is conjugation by g prove σϕ g σ −1 = ϕ σ(g) . Deduce that Inn(G) Aut(G). Proof. Let x ∈ G, so σϕ g σ −1 (x) = σ(gσ −1 (x)g −1 ) = σ(g)xσ(g) −1 = ϕ σ(g) (x) shows the desired equation. Since conjugating an element of Inn(G) with an element of Aut(G) is contained in Inn(G), normality must hold. 4.4.2. Prove that if G is an abelian group of order pq, where p and q are distinct primes, then G is cyclic. Proof. We use Cauchy’s theorem to pick x, y ∈ G with [x[ = p and [y[ = q. But if 1 = (xy) k = x k y k for k ≥ 1, then k [ p and k [ q. Since x = y −1 , then k ≥ pq. Thus 'xy` = G and therefore G is cyclic. 4.4.3. Prove that under any automorphism of D 8 , r has at most 2 possible images and s has at most 4 possible images (r and s are the usual generators). Deduce that [Aut(D 8 )[ ≤ 8. Proof. Since there is only one subgroup of order 4, 'r` is characteristic so if σ ∈ Aut(D 8 ), then σ('r`) = 'r`. Since [r[ = 4, then σ(r) ∈ ¸ r, r 3 ¸ . Since σ('r`) = 'r`, we must have σ(s) ∈ ¸ s, sr, sr 2 , sr 3 ¸ . If x ∈ D 8 , then x = sr k for some 0 ≤ k ≤ 3. So σ(x) = σ(sr k ) = σ(s)σ(r) k and thus the automorphism is uniquely determined by the value on s and r. So [Aut(D 8 )[ ≤ 2 4 = 8. 4.4.11. If p is a prime and P is a subgroup of S p of order p, prove N Sp (P)/C Sp (P) ∼ = Aut(P). Proof. We know from a previous exercise that N Sp (P) = p(p −1) and from Corollary 15 that N Sp (P)/C Sp (P) ∼ = H ≤ Aut(P). Since C Sp (P) = P and [Aut(P)[ = p −1, we have the desired isomorphism. 4.4.18. This exercise shows that for n = 6 every automorhpism of S n is inner. Fix an integer n ≥ 2 with n = 6. (a) Prove that the automorphism group of a group G permutes the conjugacy classes of G, i.e., for each σ ∈ Aut(G) and each conjugacy class / of G the set σ(/) is also a conjugacy class of G. (b) Let / be the conjugacy class of transpositions in S n and let / be the conjugacy class of any element of order 2 in S n that is not a transposition. Prove that [/[ = [/ [. Deduce that any automorphism of S n sends transpositions to transpositions. (c) Prove that for each σ ∈ Aut(S n ) σ: (1 2) →(a b 2 ), σ: (1 3) →(a b 3 ), . . . , σ: (1 n) →(a b n ) for some distinct integers a, b 2 , b 3 , . . . , b n ∈ ¦1, 2, . . . , n¦. (d) Show that (1 2), (1 3), . . ., (1 n) generate S n and deduce that any automorphism of S n is uniquely determined by its action on these elements. Use (c) to show that S n has at most n! automorphisms and conclude that Aut(S n ) = Inn(S n ) for n = 6. Proof (a). 4.5. The Sylow Theorems. Let G be a finite group and let p be a prime. 4.5.1. Prove that if P ∈ Syl p (G) and H is a subgroup of G containing P then P ∈ Syl p (H). Give an example to show that, in general, a Sylow p-subgroup of a subgroup of G need not be a Sylow p-subgroup of G. Proof. Let [P[ = p k . Then p k [H[ [G[, so P ∈ Syl p (H). Example. Let 1 = H _ G and p [G[. Then 1 ∈ Syl p (H) and 1 / ∈ Syl p (G). 4.5.3. Use Sylow’s Theorem to prove Cauchy’s Theorem. (Note that we only used Cauchy’s Theorem for abelian groups – Proposition 3.21 – in the proof of Sylow’s Theorem so this line of reasoning is not circular.) Proof. Suppose that p [G[ and let P ∈ Syl p (G) with [P[ = p k . Now take x ∈ P`1 so [x[ p k and let [x[ = p l . Then x l = p. 4.5.9. Exhibit all Sylow 3-subgroups of SL 2 (F 3 ). 27 Solution. Note [SL 2 (F 3 )[ = 24 so if P ∈ Syl 3 (SL 2 (F 3 )) then [P[ = 3. Since n 3 ≡ 1 mod 3, n 3 ∈ ¦1, 4¦. But 1 1 0 1 , 1 0 1 1 , 0 1 2 2 , 0 2 1 2 , all generate different subgroups of order 3. 4.5.13. Prove that a group, G, of order 56 has a normal Sylow p-subgroup for some prime p dividing its order. Proof. Note 56 = 2 3 7. Suppose n 2 , n 7 > 1. Then ( ¸ Syl 2 (G))`1 ≥ 14 and ( ¸ Syl 7 (G))`1 ≥ 48. These are clearly all distinct elements, so n 2 or n 7 is equal to 1. 4.5.18. Prove that a group of order 200 has a normal Sylow 5-subgroup. Proof. Note 200 = 2 3 5 2 . But n 5 ∈ ¦1, 6¦ and since 6 [ 8 we must have n 5 = 1. 4.5.19. Prove that if [G[ = 6545 then G is not simple. Proof. Note 6545 = 5 7 11 17. But since n 11 ≡ 1 mod 11 and n 11 [ 595, we must have n 11 = 1. 4.5.30. How many elements of order 7 must there be in a simple group of order 168? Solution. Note 168 = 2 3 3 7. Since the group is simple, n 7 > 1. Since n 7 ≡ 1 mod 7 and n 7 [ 24, we must have n 7 = 8. Therefore there are 6 8 = 48 elements of order 7. 4.5.32. Let P be a Sylow p-subgroup of H and let H be a subgroup of K. If P H and H K, prove that P is normal in K. Deduce that if P ∈ Syl p (G) and H = N G (P), then N G (H) = H. Proof. If P H, then P is characteristic in H. Since P is characteristic in H and H K, then P K. Now if P ∈ Syl p (G) and H = N G (P), then P ∈ Syl p (H) and P is characteristic in H. Thus if H g = H, then P g = P so g ∈ N G (P) = H. Therefore N G (H) = H. 4.5.40. Prove that the number of Sylow p-subgroups of G = GL 2 (F p ) is p + 1. Proof. Note [GL 2 (F p )[ = (p −1) 2 p(p + 1) and n p ≡ 1 mod p. Since P 1 = 1 γ 0 1 , P 2 = 1 0 γ 1 ∈ Syl p (G), for γ ∈ F p ` ¦0, 1¦, we must have n p ≥ p + 1. Now we show that the upper triangular matrices are a subgroup of N G (P 1 ) with order (p − 1) 2 p. Simple calculations show that for a, b, c ∈ F p , we have a b 0 c −1 = a −1 −a −1 bc −1 0 c −1 . Now if γ ∈ F p ` ¦0, 1¦ we get a b 0 c −1 1 γ 0 1 a b 0 c = 1 a −1 γb −a −1 bc −1 b 0 1 . Since the upper diagonal matrices have nonzero diagonal elements and any upper right entry, the order must be (p −1) 2 p. Thus (p −1) 2 p [ N G (P 1 ). Sylow’s theorem states that n p = [G: N G (P 1 )], so n p [ p + 1. Therefore n p = p + 1. 4.6. The Simplicity of A n . 28 DAVID S. DUMMIT AND RICHARD M. FOOTE 5. Direct and Semidirect Products and Abelian Groups 5.1. Direct Products. 5.1.10. Let p be a prime. Let A and B be two cyclic groups of order p with generators x and y, respectively. Set E = AB so that E is the elementary abelian groups of order p 2 : E p 2. Prove that the distinct subgroups of E of order p are 'x` , 'xy` , xy 2 , . . . , xy p−1 , 'y` (note that there are p + 1 of them). Proof. Example 3 in this section shows that there are p + 1 subgroups of order p. Note that these p + 1 subgroups are all distinct and nontrivial, so if the pth power of each generator is the identity element, then we are done. But x p = 1, (xy) p = x p y p = 1, (xy k ) p = x p (y k ) p = 1, y p = 1. 5.1.11. Let p be a prime and let n ∈ Z + . Find a formula for the number of subgroups of order p in the elementary abelian group E p n. Solution. Every nonidentity element has order p, so there are p n − 1 many elements of order p. By Lagrange’s theorem, each distinct subgroup intersects trivially. So there are (p n − 1)/(p − 1) = p n−1 + p n−2 + + 1 many distinct subgroups of order p. 5.1.12. Let A and B be groups. Assume Z(A) contains a subgroup Z 1 and Z(B) contains a subgroup Z 2 with Z 1 ∼ = Z 2 . Let this isomorphism be given by the map x i → y i for all x i ∈ Z 1 . A central product of A and B is a quotient (AB)/Z where Z = ¸ (x i , y −1 i ) [ x i ∈ Z 1 ¸ and is denoted by A∗ B – it is not unique since it depends on Z 1 , Z 2 and the isomorphism between them. (Think of A ∗ B as the direct product of A and B “collapsed” by identifying each element x i ∈ Z 1 with its corresponding element y i ∈ Z 2 .) (a) Prove that the images of A and B in the quotient group A ∗ B are isomrophic to A and B, respectively, and that these images intersect in a central subgroup isomorphic to Z 1 . Find [A∗ B[. (b) Let Z 4 = 'x`. Let D 8 = 'r, s` and Q 8 = 'i, j` be given by their usual generators and realtions. Let Z 4 ∗ D 8 be the central product of Z 4 and D 8 which identifies x 2 and r 2 (i.e., Z 1 = x 2 , Z 2 = r 2 and the isomorphism is x 2 → r 2 ) and let Z 4 ∗ Q 8 be the central product of Z 4 and Q 8 which identifies x 2 and −1. Prove that Z 4 ∗ D 8 ∼ = Z 4 ∗ Q 8 . Proof (a). Proof (b). 5.1.18. In each of (a) to (e) give an example of a group with the specified properties: (a) an infinite group in which every element has order 1 or 2 (b) an infinite group in which every element has finite order but for each positive integer n there is an element of order n (c) a group with an element of infinite order and an element of order 2 (d) a group G such that every finite group is isomorphic to some subgroup of G (e) a nontrivial group G such that G ∼ = GG. Solution (a). The direct sum Z 2 Z 2 . Solution (b). The direct sum Z 2 Z 3 . Solution (c). Z Z 2 . Solution (d). S 2 S 3 . Solution (e). G = Z 2 Z 2 . 29 5.2. The Fundamental Theorem of Finitely Generated Abelian Groups. 5.2.2. In each of parts (a) to (e) give the lists of invariant factors for all abelian groups of the specified order: (a) order 270 Solution (a). Since 270 = 2 3 3 5, the list of invariant factors is given by (30, 3, 3), (90, 3). 5.2.3. In each of parts (a) to (e) give the lists of elementary divisors for all abelian groups of the specified order and then match each list with the corresponding list of invariant factors found in the preceding exercise: (a) order 270 Solution (a). The elementary divisors of an abelian group of order 270 are either (2, 3, 3, 5) or (2, 9, 5) with lists corresponding to respective orderings. 5.2.7. Let p be a prime and let A = 'x 1 ` 'x 2 ` 'x n ` be an abelian p-group, where [x i [ = p αi > 1 for all i. Define the pth-power map ϕ: A →A by ϕ: x →x p . (a) Prove that ϕ is a homomorphism. (b) Describe the image and kernel of ϕ in terms of the given generators. (c) Prove both ker ϕ and A/imϕ have rank n (i.e., have the same rank as A) and prove these groups are both isomorphic to the elementary abelian group, E p n, of order p n . Proof (a). This map is clearly well-defined, so we calculate ϕ(xy) = (xy) p = x p y p = ϕ(x)ϕ(y). Solution (b). The image of ϕ is given by ϕ(A) = ¦(β 1 , . . . , β n ) [ β i = (x γ i ) p , γ ∈ Z¦ . The kernel of ϕ is given by ¦(k 1 , . . . , k n )¦ where k i ∈ ¦x γ i [ (x γ i ) p = 1, γ ∈ Z¦ . Proof (c). The kernel of ϕ is of type (k 1 , . . . , k n ) so it has rank n, while A/imϕ is of type (¯ a 1 , . . . , ¯ a n ) so also has rank n. If (x γ i ) p = 1, then [x γ i [ p so x p α i −1 i = p, with each element to the power of p clearly one. Thus ker ϕ ∼ = n ¸ i=1 x p α i −1 so it is isomorphic to E p n. Since ϕ: A →A and ker ϕ[ xi = p, then imϕ[ xi = α p−1 i . Thus 'x i ` /imϕ[ xi ∼ = Z p . Therefore A/imϕ ∼ = E pn . 5.2.5. Prove that A n is the commutator subgroup of S n for all n ≥ 5. Proof. Since A n S n and S n /A n ∼ = Z 2 is abelian, proposition 7 part (4) states S n ≤ A n . Also by proposition 7 part (3), S n is characteristic in S n . Since A n is simple, S n A n , and S n is non abelian, we must have A n = S n . 5.2.11. Prove that if G = HK where H and K are characteristic subgroups of G with H ∩ K = 1 then Aut(G) ∼ = Aut(H) Aut(K). Deduce that if G is an abelian group of finite order then Aut(G) is isomorphic to the direct product of the automorphism groups of its Sylow subgroups. 30 DAVID S. DUMMIT AND RICHARD M. FOOTE Proof. Let φ, ψ ∈ Aut(G) and notice that φ[ H ∈ Aut(H) and φ[ K ∈ Aut(K) since H, K are characteristic in G. Now let f : Aut(G) →Aut(H) Aut(K) be given by f(φ) = (φ[ H , φ[ K ). If φ = ψ, then (φ[ H , φ[ K ) = (ψ[ H , ψ[ K ) so f is well-defined. Now we calculate f(φ ◦ ψ) = ((φ ◦ ψ)[ H , (φ ◦ ψ)[ K ) = (φ[ H ◦ ψ[ H , φ[ K ◦ ψ[ K ) = (φ[ H , φ[ K )(ψ[ H , ψ[ K ) = f(φ)f(ψ), so f is a homomorphism. Now if φ H ∈ Aut(H) and φ K ∈ Aut(K), then let φ ∈ Aut(G) be given by φ[ H = φ H and φ[ K = φ K . So f(φ) = (φ[ H , φ[ K ) = (φ H , φ K ) shows f is surjective. Now if (φ[ H , φ[ K ) = (ψ[ H , ψ[ K ), then for g = hk ∈ G, φ(hk) = φ(h)φ(k) = φ[ H (h)φ[ K (k) = ψ[ H (h)ψ[ K (k) = ψ(h)ψ(k) = ψ(hk), so φ = ψ. Therefore f shows Aut(G) ∼ = Aut(H) Aut(K). If G is finite abelian, then it is the direct product of its (characteristic) Sylow subgroups. Therefore Aut(G) is isomorphic to the direct product of the automorphism groups of the Sylow subgroups. 5.2.14. Let G = ¦(a ij ) ∈ GL n (F) [ a ij = 0 if i > j, and a 11 = a 22 = = a nn ¦, where F is a field, be the group of upper triangular matrices all of whose diagonal entries are equal. Prove that G ∼ = D U, where D is the group of all nonzero multiples of the identity matrix and U is the group of upper triangular matrices with 1’s down the diagonal. Proof. It is clear that D ∩ U = 1. Now let A, B ∈ G such that A = ¸ ¸ ¸ ¸ a a 12 a 1n a a 2n . . . . . . a ¸ , B = ¸ ¸ ¸ ¸ b b 12 b 1n b b 2n . . . . . . b ¸ and note if AB = I, then ab = 1. If λI ∈ D for λ ∈ F, then AλIB = λAIB = λAB = λI. If | ∈ U, then note that there exist | A , | B ∈ U such that A = a | A and B = b | B and A|B = (a | A ) |(b | B ) = ab (| A | | B ) ∈ U. Thus D and U are normal in G and G ∼ = D U. 5.3. Table of Groups of Small Order. 5.4. Recognizing Direct Products. 5.5. Semidirect Products. 5.5.7. This exercise describes thirteen isomorphism types of groups of order 56. (a) Prove that there are three abelian groups of order 56. (b) Prove that every group of order 56 has either a normal Sylow 2-subgroup or a normal Sylow 7-subgroup. (c) Construct the following non-abelian groups of order 56 which have a normal Sylow 7-subgroup and whose Sylow 2-subgroup S is as specified: one group when S ∼ = Z 2 Z 2 Z 2 two non-isomorphic groups when S ∼ = Z 4 Z 2 one group when S ∼ = Z 8 two non-isomorphic groups when S ∼ = Q 8 three non-isomorphic groups when S ∼ = D 8 . 31 (d) Let G be a group of order 56 with a non-normal Sylow 7-subgroup. Prove that if S is the Sylow 2-subgroup of G then S ∼ = Z 2 Z 2 Z 2 . (e) Prove that there is a unique group of order 56 with a non-normal Sylow 7-subgroup. Proof (a). Suppose that G is an abelian group of order 56 = 2 3 7. The invariant factors are given by (2 3 7), (2 2 7, 2), (2 7, 2, 2). Thus there are three abelian groups of order 56 given by Z 56 , Z 28 Z 2 , Z 14 Z 2 Z 2 . Proof (b). Suppose that n 2 , n 7 > 1. Since n p ≡ 1 mod p, we have n 2 ≥ 3 and n 7 ≥ 8. Counting [ ¸ Syl 2 (G)`1[ + [ ¸ Syl 7 (G)`1[ = 3 7 + 8 6 > 56 reaches a contradiction. Therefore every group of order 56 has either a normal Sylow 2-subgroup or a normal Sylow 7-subgroup. Solution (c). Note for all of the Sylow 2-subgroup isomorphism types, if P ∈ Syl 7 (G), we have P ∩ S = 1 and P G so G is a semidirect product. If S ∼ = Z 2 Z 2 Z 2 , then we consider the non-trivial homomorphisms ϕ: S → Aut(P), where Aut(P) ∼ = Z 6 . Thus [ker ϕ[ = 4 and [ϕ(S)[ = 2 for any such ϕ. Therefore P ϕ S is the unique isomorphism type. If S ∼ = Z 4 Z 2 , then ker ϕ ∈ ¸ Z 4 , Z 2 2 ¸ . Thus there are two isomorphism types. If S ∼ = Z 8 , then ker ϕ = Z 4 giving rise to one isomorphism type. If S ∼ = Q 8 , then the trivial map gives rise to the non-abelian Z 7 Q 8 . The non-trivial maps ϕ: S →Aut(P) give rise to a second isomorphism type. If S ∼ = D 8 , then the trivial map gives rise to the non-abelian Z 7 D 8 . Two possible kernels of non-trivial maps are 'r` ∼ = Z 4 and r 2 , s ∼ = Z 2 Z 2 . Since these kernels are not isomorphic, the comment in part (c) states there are three isomorphism types. Proof (d). If n 7 > 1, then the map ϕ: P → Aut(S) is non-trivial so that P is not normal. Thus ker ϕ = 1 and ϕ(P) ∼ = Z 7 . Let ϕ(P) = 'x` so if s ∈ S`1, then s x k shows all elements in S`1 have the same order. Since every non-identity element in S must have order 2, it is given by the elementary abelian group, S ∼ = Z 3 2 . Proof (e). If S ∼ = Z 3 2 , then Aut(S) ∼ = GL 3 (F 2 ) with [Aut(S)[ = 7 6 4 = 168. Let P 1 , P 2 ∈ Syl 7 (G) so that for any non-trivial ϕ: P i → Aut(S), ker ϕ = 1. Thus ϕ(P i ) ∈ Syl 7 (Aut(S)), so P 1 = P σ 2 for some σ ∈ Aut(S). Therefore by exercise 6, S ϕ P 1 ∼ = S ϕ P 2 . 5.5.20. Let p be an odd prime. Prove that if P is a non-cyclic p-group then P contains a normal subgroup U with U ∼ = Z p Z p . Deduce that for odd primes p a p-group that contains a unique subgroup of order p is cyclic. Proof. Let [P[ = p n with P non-abelian. If n = 2, then P ∼ = Z p Z p since P is not cyclic. Now suppose there exists U P with U ∼ = Z p Z p for all k < n. Since p [ [Z(P)[, let Z ≤ Z(P) with [Z[ = p. Thus the non-cyclic quotient group, P/Z, has a subgroup U P/Z with U ∼ = Z p Z p . Now we have the non-cyclic U P with [U[ = p 3 , so there is a U P with U ∼ = Z p Z p . 6. Further Topics in Group Theory 6.1. p-groups, Nilpotent Groups, and Solvable Groups. 6.1.3. If G is finite prove that G is nilpotent if and only if it has a normal subgroup of each order dividing [G[, and is cyclic if and only if it has a unique subgroup of each order dividing [G[. Proof. Suppose G is nilpotent so theorem 3 states G ∼ = P 1 P s for P i ∈ Syl pi . But if n [ [G[, then n = p k1 1 p ks s . Theorem 1 states each P i has a normal subgroup, P i , with order p ki i . Thus [P 1 P s [ = n with the desired normality property. Now suppose G has a normal subgroup of each order dividing [G[. Then each Sylow subgroup is normal in G, so theorem 3 states G is nilpotent. If G is cyclic, then theorem 2.7 states it has a unique subgroup of each order dividing [G[. If G has a unique subgroup of each order dividing [G[, then proposition 5 shows G is cyclic. 6.1.4. Prove that a maximal subgroup of a finite nilpotent group has prime index. Proof. Let M < G be a maximal subgroup of G. Since G is nilpotent, MG. Since G/M is nilpotent, exercise 6.1.3 shows it has a normal subgroup of each order dividing [G/M[. If P G/M with P = p for p prime, then M ≤ P ≤ G so P = G by maximality of M. Therefore [G : M] = p. 6.1.8. Prove that if p is a prime and P is a non-abelian group of order p 3 then [Z(P)[ = p and P/Z(P) ∼ = Z p Z p . 32 DAVID S. DUMMIT AND RICHARD M. FOOTE Proof. Since [P[ = p 3 we know P is nilpotent. Since P is a p-group Z(P) = 1, since P is non-abelian, Z(P) = P. Thus [Z(P)[ ∈ ¸ p, p 2 ¸ . If [Z(P)[ = p 2 , then P/Z(P) is cyclic contradicting P being non-abelian. Thus [Z(P)[ = p. Since P/Z(P) is not cyclic, we must have P/Z(P) ∼ = Z p Z p . 6.1.12. Find the upper and lower central series for A 4 and S 4 . Solution. The upper central series of A 4 is Z i (A 4 ) = 1 and of S 4 is Z i (S 4 ) = 1 for all i since the center of S 4 is trivial. Since s −1 a −1 sa is an even permutation, S 1 4 ≤ A 4 . But if α ∈ A 4 , we can choose s ∈ S 4 such that s −1 a −1 s = αa so that α ∈ S 1 4 . The same argument shows S 2 4 = A 4 . Since S 1 4 = A 4 and S 2 4 = A 4 , we have found the lower central series for S 4 . Now N = ¦1, (12)(34), (13)(24), (14)(23)¦ is the only proper normal subgroup of A 4 . Thus A i 4 = N for all i so this is the lower central series for A 4 . 33 Part II – Ring Theory 7. Introduction to Rings 7.1. Basic Definitions and Examples. Proposition 7.1. Let R be a ring. Then (1) 0a = a0 = 0 for all a ∈ R. (2) (−a)b = a(−b) = −(ab) for all a, b ∈ R. (3) (−a)(−b) = ab for all a, b ∈ R. (4) if R has an identity 1, then the identity is unique and −a = (−1)a. Proposition 7.2. Assume a, b and c are element of any ring with A not a zero divisor. If ab = ac, then either a = 0 or b = c. In particular, if a, b, c are any elements in an integral domain and ab = ac, then either a = 0 or b = c. Corollary 7.3. Any finite integral domain is a field. 7.1.3. Let R be a ring with identity and let S be subring of R containing the identity. Prove that if u is a unit in S then u is a unit in R. Show by example that the converse if false. Proof. If u is a unit in S, then there is a v ∈ S such that uv = 1 = vu. But then v ∈ R and hence u is a unit in R. Now consider 2 ∈ Q and notice that it is a unit because 1/2 ∈ Q. But Z ⊆ Q is a subring of Q and 2 is not a unit. Thus the converse is false. 7.1.5. Decide which of the following (a)-(f) are subrings of Q: (a) the set of all rational numbers with odd denominators (when written in lowest terms) (b) the set of all rational numbers with even denominators (when written in lowest terms) (c) the set of nonnegative rational numbers (d) the set of squares of rational numbers (e) the set of all rational numbers with odd numerators (when written in lowest terms) (f) the set of all rational numbers with even numerators (when written in lowest terms). Solution. (a), (f). 7.1.6. Decide which of the following are subrings of the ring of all functions from the closed interval [0, 1] to 1. (a) the set of all functions f(x) such that f(q) = 0 for all q ∈ Q∩ [0, 1] (b) the set of all polynomial functions (c) the set of all functions which have only a finite number of zeros, together with the zero function (d) the set of all functions which have an infinite number of zeros (e) the set of all functions f such that lim x→1 − f(x) = 0 (f) the set of all rational linear combinations of the functions sin nx and cos mx, where m, n ∈ Z + . Solution. (a), (b), (c), (e). 7.1.7. The center of a ring R is ¦z ∈ R [ zr = rz, ∀r ∈ R¦ (i.e., is the set of all elements which commute with every element of R). Prove that the center of a division ring is a field. Proof. 7.1.12. Prove that any subring of a field which contains the identity is an integral domain. Proof. Let F be a field and let R ⊆ F be a subring of F with 1 ∈ R. Then since R ⊆ F, every element of R is a unit. Therefore there are no zero divisors in R. 7.1.13. An element x in R is called nilpotent if x m = 0 for some m ∈ Z + . (a) Show that if n = a k b for some integers a and b then ab is a nilpotent element of Z/nZ. (b) If a ∈ Z is an integer, show that the element a ∈ Z/nZ is nilpotent if and only if every prime divisor of n is also a divisor of a. In particular, determine the nilpotent elements of Z/72Z explicitly. (c) Let R be the ring of functions from a nonempty set X to a field F. Prove that R contains no nonzero nilpotent elements. 7.1.14. Let x be a nilpotent element of the commutative ring R (cf. Exercise 7.1.13). (a) Prove that x is either zero or a zero divisor. 34 DAVID S. DUMMIT AND RICHARD M. FOOTE (b) Prove that rx is nilpotent for all r ∈ R. (c) Prove that 1 +x is a unit in R. (d) Deduce that the sum of a nilpotent element and a unit is a unit. Proof (a). Suppose x = 0 and since x m = 0, let n be the smallest positive integer such that x n = 0. Then xx n−1 = 0 where x n−1 = 0 so x is a zero divisor. Proof (b). Since R is commutative, we have (rx) m = r m x m = 0. Thus rx is nilpotent for all r ∈ R. Proof (c). Observing the calculation (1 +x)(1 −x +x 2 −x 3 + x m−1 ) = 1 shows that 1 +x is a unit. Solution (d). Let u be a unit so vu = 1 and let x be our nilpotent element. Then part (b) shows vx is nilpotent and part (c) shows 1 +vx is a unit. Thus v(u +x) is a unit and hence u +x is a unit. 7.1.23. Let D be a squarefree integer, and let O be the ring of integers in the quadratic field Q( √ D). For any positive integer f prove that the set O f = Z[fω] = ¦a +bfω [ a, b ∈ Z¦ is a subring of O containing the identity. Prove that [O: O f ] = f (index as additive abelian groups). Prove conversely that a subring of O containing the identity and having finite index f in O (as additive abelian group) is equal to O f . The ring O f is called the order of conductor f in the field Q( √ D). The ring of integers O is called the maximal order in Q( √ D). Proof. Let a + bfω ∈ O f . Since bf ∈ Z, then a + bfω ∈ O so O f ⊆ O. Now (a + bfω)(c + dfω) = ac + (ad + bd)fω +bdf 2 ω 2 ∈ O f and (a +bfω) −(c +dfω) = (a −c) + (b −d)fω ∈ O f . Thus O f is a subring of O. Let a +bω +O f = a +b ω +O f . Then (a +bω) −(a +b ω) ∈ O f and (a −a ) +(b −b )ω ∈ O f . So b −b = b f thus b ≡ b mod f. Now let b ≡ b + α for 1 ≤ α < f and suppose ∃a, a such that a + bω + O f = a + b ω + O f . But this is a contradiction because f would have to divide b −b . Thus a +bω +O f and a +b ω +O f are the same coset if and only if b ≡ b mod f hence [O: O f ] = f. Suppose that a subring, S, of O contains the identity and has finite index f. Since 1 ∈ S then a ∈ S for all a ∈ Z because S is closed under addition. Since [O/S[ = f, [ω +S[ divides f. But then S = f(ω +S) = fω +S so fω ∈ S. Since S is closed under addition then bfω ∈ S for all b ∈ Z. Now we have a + bfω ∈ S for all a, b ∈ Z so O f ⊆ S. Since [O f [ = [S[, we must have O f = S. 7.1.26. Let K be a field. A discrete valuation on K is a function ν : K × →Z satisfying (i) ν(ab) = ν(a) +ν(b) (ii) ν is surjective, and (iii) ν(x +y) ≥ min ¦ν(x), ν(y)¦ for all x, y ∈ K × with x +y = 0. The set R = ¦x ∈ K × [ ν(x) ≥ 0¦ ∪ ¦0¦ is called the valuation ring of ν. (a) Prove that R is a subring of K which contains the identity. In general, a ring R is called a discrete valuation ring if there is some field K and some discrete valuation ν on K such that R is the valuation ring of ν. (b) Prove that for each nonzero element x ∈ K either x or x −1 is in R. (c) Prove that an element x is a unit of R if and only if ν(x) = 0. Proof (a). Proof (b). Proof (c). 7.2. Examples: Polynomials Rings, Matrix Rings, and Group Rings. Proposition 7.4. Let R be an integral domain and let p(x), q(x) be nonzero elements of R[x]. Then (1) degree p(x)q(x) = degree p(x) + degree q(x), (2) the units of R[x] are just the units of R, (3) R[x] is an integral domain. Let R be a commutative ring with 1. 7.2.1. Let p(x) = 2x 3 − 3x 2 + 4x − 5 and let q(x) = 7x 3 + 33x − 4. In each of parts (a), (b), and (c) compute p(x) + q(x) and p(x)q(x) under the assumption that the coefficients of the two given polynomials are taken from the specified ring (where the integer coefficients are taken mod n in parts (b) and (c)): 35 (a) R = Z, (b) R = Z/2Z, (c) R = Z/3Z. Solution (a). p(x) +q(x) = 9x 3 −3x 2 + 37x −9 p(x)q(x) = 14x 6 −21x 5 + 94x 4 −142x 3 + 144x 2 −181x + 20 Solution (b). p(x) +q(x) = x 3 −x 2 +x −1 p(x)q(x) = x 5 −x Solution (c). p(x) +q(x) = x p(x)q(x) = 2x 6 +x 4 +x 3 + 2x 7.2.6. Let S be a ring with identity 1 = 0. Let n ∈ Z + and let A be an n n matrix with entries from S whose i, j entry is a ij . Let E ij be the element of M n (S) whose i, j entry is 1 and whose other entries are all 0. (a) Prove that E ij A is the matrix whose i th row equals the j th row of A and all other rows are zero. (b) Prove that AE ij is the matrix whose j th column equals the i th column of A and all other columns are zero. (c) Deduce that E pq AE rs is the matrix whose p, s entry is a qr and all other entries are zero. Proof (a). Proof (b). Solution (c). 7.2.7. Prove that the center of the ring M n (R) is the set of scalar matrices (cf. Exercise 7.1.7). Use Exercise 7.2.6. Proof. Let [z ij ] = Z ∈ Z(M n (R)) so ZE ij = E ij Z for all E ij . If i = j, then the (j, j) th element of E ji Z is z ij and the (j, j) th element of ZE ji is 0, so z ij = 0. Now we observe that the (j, i) th element of E ji Z is z ii and the (j, i) th element of ZE ji is z jj , so z ii = z jj . Thus Z must be a scalar matrix. If Z = rI, then AZ = ArI = rAI = rIA = ZA, for all A ∈ M n (R). 7.2.10. Consider the following elements of the integral group ring ZS 3 : α = 3(1 2) −5(2 3) + 14(1 2 3) and β = 6(1) + 2(2 3) −7(1 3 2) where (1) is the identity of S 3 . Compute the following elements: (a) α +β, (b) 2α −3β, (c) αβ, (d) βα, (e) α 2 . Solution (a). α +β = 6(1) + 3(1 2) −3(2 3) + 14(1 2 3) −7(1 3 2) 36 DAVID S. DUMMIT AND RICHARD M. FOOTE Solution (b). 2α −3β = −18(1) + 6(1 2) −16(2 3) + 28(1 2 3) + 21(1 3 2) Solution (c). αβ = −108(1) + 81(1 2) −21(1 3) −30(2 3) + 90(1 2 3) Solution (d). βα = −108(1) + 18(1 2) + 63(1 3) −51(2 3) + 84(1 2 3) + 6(1 3 2) Solution (e). α 2 = 34(1) −70(1 2) −28(1 3) + 42(2 3) −15(1 2 3) + 196(1 3 2) 7.2.12. Let G = ¦g 1 , . . . , g n ¦ be a finite group. Prove that the element N = g 1 + g 2 + + g n is in the center of the group ring RG (cf. Exercise 7.1.7). Proof. Let X = x 1 g 1 + +x n g n for x i ∈ R. Then NX = n ¸ k=1 ¸ ¸ gigj=g k x j ¸ g k = n ¸ k=1 ¸ n ¸ j=1 x j ¸ g k = n ¸ k=1 ¸ ¸ gigj=g k x i ¸ g k = XN. 7.3. Ring Homomorphisms and Quotient Rings. Proposition 7.5. Let R and S be rings and let ϕ: R →S be a homomorphism. (1) The image of ϕ is a subring of S. (2) The kernel of ϕ is a subring of R. Furthermore, if α ∈ ker ϕ then rα and αr ∈ ker ϕ for every r ∈ R, i.e., ker ϕ is closed under multiplication by elements from R. Proposition 7.6. Let R be a ring and let I be an ideal of R. Then the (additive) quotient group R/I is a ring under the binary operations: (r +I) + (s +I) = (r +s) +I and (r +I) (s +I) = (rs) +I for all r, s ∈ R. Conversely, if I is any subgroup such that the above operations are well defined, then I is an ideal of R. Theorem 7.7. (1) (The First Isomorphism Theorem for Rings) If ϕ: R → S is a homomorphism of rings, then the kernel of ϕ is an ideal of R, the image of ϕ is a subring of S and R/ ker ϕ is isomorphic as a ring to ϕ(R). (2) If I is any ideal of R, then the map R →R/I defined by r →r +I is a surjective ring homomorphism with kernel I (this homomorphism is called the natural projection of R onto R/I). Thus every ideal is the kernel of a ring homomorphism and vice versa. Theorem 7.8. Let R be a ring. (1) (The Second Isomorphism Theorem for Rings) Let A b ea subring and let B be an ideal of R. Then A+B = ¦a +b [ a ∈ A, b ∈ B¦ is a subring of R, A∩ B is an ideal of A and A+B/B ∼ = A/(A∩ B). 37 (2) (The Third Isomorphism Theorem for Rings) Let I and J be ideal of R with I ⊆ J. Then J/I is an ideal of R/I and (R/I)/(J/I) ∼ = R/J. (3) (The Fourth Isomorphism Theorem for Rings) Let I be an ideal of R. The correspondence A ↔ A/I is an inclusion preserving bijection between the wset of subrings A of R that contain I and the set of subrings of R/I. Furthermore, A (a subring containing I) is an ideal of R if and only if A/I is an ideal of R/I. 7.3.12. Let D be an integer that is not a perfect square in Z and let S = a b Db a [ a, b ∈ Z . (a) Prove that S is a subring of M 2 (Z). (b) If D is not a perfect square in Z prove that the map ϕ: Z[ √ D] →S defined by ϕ(a +b √ D) = a b Db a is a ring isomorphism. (c) If D ≡ 1 mod 4 is squarefree, prove that the set a b (D −1)b/4 a +b [ a, b ∈ Z is a subring of M 2 (Z) and is isomorphic to the quadratic integer ring O. Proof (a). S is clearly a group under addition, so we show that S is closed under multiplication. But a 1 b 1 Db 1 a 1 a 2 b 2 Db 2 a 2 = a 1 a 2 +Db 1 b 2 a 1 b 2 +a 2 b 1 D(a 2 b 1 +a 1 b 2 ) Db 1 b 2 +a 1 a 2 so S is a subring of M 2 (Z). Proof (b). We calculate ϕ((a 1 +b 1 √ D) + (a 2 +b 2 √ D)) = ϕ((a 1 +a 2 ) + (b 1 +b 2 ) √ D) = a 1 +a 2 b 1 +b 2 D(b 1 +b 2 ) a 1 +a 2 = a 1 b 1 Db 1 a 1 + a 2 b 2 Db 2 a 2 ϕ((a 1 +b 1 √ D)(a 2 +b 2 √ D)) = ϕ((a 1 a 2 +b 1 b 2 D) + (a 1 b 2 +a 2 b 1 ) √ D) = a 1 a 2 +b 1 b 2 D a 1 b 2 +a 2 b 1 D(a 1 b 2 +a 2 b 1 ) a 1 a 2 +b 1 b 2 D = a 1 b 1 Db 1 a 1 a 2 b 2 Db 2 a 2 so ϕ is a homomorphism of rings. Now let a 1 b 1 Db 1 a 1 = a 2 b 2 Db 2 a 2 , so a 1 = a 1 , b 1 = b 2 and thus a 1 +b 1 √ D = a 2 +b 2 √ D. Now if a b Db a ∈ S, then ϕ maps a +b √ D to this element. Therefore ϕ is an isomorphism. Proof (c). This is clearly a group under addition, so we show it is closed under multiplication. But a 1 b 1 (D −1)b 1 /4 a 1 +b 1 a 2 b 2 (D −1)b 2 /4 a 2 +b 2 = a 1 a 2 + (D −1)a 1 b 2 /4 a 1 b 2 +a 2 b 1 +b 1 b 2 (D −1)(a 1 b 2 +b 1 b 2 )/4 (D −1)b 1 b 2 /4 +a 1 a 2 +a 1 b 2 +a 2 b 1 +b 1 b 2 so this is a subring of M 2 (Z). Similarly to the proof in part (b), this is isomorphic to the quadratic integer ring O. 7.3.15. Let X be a nonempty set and let {(X) be the Boolean ring of all subsets of X defined in exercise 21 of section 1. Let R be the ring of all functions from X into Z/2Z. For each A ∈ {(X) define the function χ A : X →Z/2Z by χ A (x) = 1, x ∈ A, 0, x / ∈ A, χ A is called the characteristic function of A with values in Z/2Z. Prove that the map {(X) → R defined by A →χ A is a ring isomorphism. 38 DAVID S. DUMMIT AND RICHARD M. FOOTE Proof. Let A = A so that ϕ(A) = χ A = χ A = ϕ(A ), so ϕ is well defined. Now let A, B ⊆ X so that ϕ(A+B) = ϕ((A`B) ∪ (B`A)) = χ (A\B)∪(B\A) . But χ (A\B)∪(B\A) = 1, x ∈ A`B, 1, x ∈ B`A, 0 x ∈ A∩ B, 0, x ∈ X`(A∩ B) But simple calculations show this is ϕ(A) +ϕ(B). Similarly, ϕ(AB) = ϕ(A∩ B) = χ A∩B = 0, x ∈ A`B, 0, x ∈ B`A, 1, x ∈ A∩ B, 0, x ∈ X`(A∩ B), = ϕ(A)ϕ(B). Now let f ∈ R so let A ⊆ X be the set such that f(A) = 1 and f(X`A) = 0. So ϕ(A) = χ A = f and ϕ is surjective. Now let f = g ∈ R so there exist A, B such that ϕ(A) = f = g = ϕ(B). So χ A = χ B , thus A = B and ϕ is injective. Therefore ϕ is a ring isomorphism. 7.3.17. Let R and S be nonzero rings with identity and denote their respective identities by 1 R and 1 S . Let ϕ: R →S be a nonzero homomorphism of rings. (a) Prove that if ϕ(1 R ) = 1 S then ϕ(1 R ) is a zero divisor in S. Deduce that if S is an integral domain then every ring homomorphism from R to S sends the identity of R to the identity of S. (b) Prove that if ϕ(1 R ) = 1 S then ϕ(u) is a unit in S and that ϕ(u −1 ) = ϕ(u) −1 for reach unit u of R. Proof (a). If ϕ(1 R ) = 1 S , then (ϕ(1 R ) −1 S )ϕ(1 R ) = ϕ(1 R )ϕ(1 R ) −1 S ϕ(1 R ) = ϕ(1 R ) −ϕ(1 R ) = 0. So ϕ(1 R ) is a zero divisor in S. If S is an integral domain, then there are no zero divisors. Since an integral domain has a 1 = 0 and ϕ(1 R ) = 0 if ϕ(R) = 0, then ϕ(1 R ) −1 S = 0. Therefore ϕ(1 R ) = 1 S . Proof (b). If u ∈ R is a unit, there exists v ∈ R such that uv = 1 R . So ϕ(u)ϕ(v) = ϕ(uv) = ϕ(1 R ) = 1 S . Thus ϕ(u) is a unit, with ϕ(u) −1 = ϕ(v) = ϕ(u −1 ). 7.3.27. Prove that a nonzero Boolean ring has characteristic 2 (cf. exercise 15, section 1). Proof. Let a = −1 so that 0 = (−1) 2 + 1 = 1 + 1. Since 1 = 0, then R has characteristic 2. 7.3.29. Let R be a commutative ring. Recall (cf. exercise 7.1.13) that an element x ∈ R is nilpotent if x n = 0 for some n ∈ Z + . Prove that the set of nilpotent element form an ideal – called the nilradical of R and denoted by A(R). Proof. 7.3.30. Prove that if R is a commutative ring and A(R) is its nilradical (cf. exercise 7.3.29) then zero is the only nilpotent element of R/A(R) i.e., prove that A(R/A(R)) = 0. Proof. Let x ∈ R/A(R) with x n = 0 for some n ∈ N. Then there exists r ∈ R such that x = r +A(R). But then 0 = x n = (r +A(R)) n = r n +A(R). This implies r n ∈ A(R) so r ∈ A(R) and thus x = 0. 7.3.33. Assume R is commutative. Let p(x) = a n x n +a n−1 x n−1 + +a 1 x +a 0 be an element of the polynomial ring R[x]. (a) Prove tha tp(x) is a unit in R[x] if and only if a 0 is a unit and a 1 , a 2 , . . . , a n are nilpotent in R. [See exercise 7.1.14.] 39 (b) Prove that p(x) is nilpotent in R[x] if and only if a 0 , a 1 , . . . , a n are nilpotenet elements of R. Proof (a). Proof (b). 7.4. Properties of Ideals. Proposition 7.9. Let I be an ideal of R. (1) I = R if and only if I contains a unit. (2) Assume R is commutative. Then R is a field if and only if its only ideals are 0 and R. Corollary 7.10. If R is a field then any nonzero ring homomorphism from R into another ring is an injection. Proposition 7.11. In a ring with identity every proper ideal is contained in a maximal ideal. Proposition 7.12. Assume R is commutative. The ideal M is a maximal ideal if and only if the quotient ring R/M is a field. Proposition 7.13. Assume R is commutative. Then the ideal P is a prime ideal in R if and only if the quotient ring R/P is an integral domain. Corollary 7.14. Assume R is commutative. Every maximal ideal of R is a prime ideal. Let R be a ring with identity 1 = 0. 7.4.2. Assume R is commutative. Prove that the augmentation ideal in the group ring RG is generated by ¦g −1 [ g ∈ G¦. Prove that if G = 'σ` is cyclic then the augmentation ideal is generated by σ −1. 7.4.3. (a) Let p be a prime and let G be an abelian group of order p n . Prove that the nilradical of the group ring F p G is the augmentation ideal (cf. exercise 7.3.29). (b) Let G = ¦g 1 , . . . , g n ¦ be a finite group and assume R is commutative. Prove that if r is any element of the augmentation ideal of RG then r(g 1 + +g n ) = 0. Proof (a). Suppose there exists m such that ( ¸ p n i=1 a i g i ) m = 0. If ϕ is the augmentation map, then 0 = ϕ( p n ¸ i=1 a i g i ) m = ( p n ¸ i=1 a i ) m . Since F p ` ¦0¦ is a multiplicative group, we must have ¸ p n i=1 a i = 0. Thus ¸ p n i=1 a i g i ∈ ker ϕ. Now let g −1 ∈ ker ϕ. Then (g −1) p n = g p n +p(. . .) −1 = 0 in F p G. Since exercise 7.4.2 states ker ϕ is generated by ¦g −1 [ g ∈ G¦, we must have ker ϕ ⊆ A(F p G). Therefore ker ϕ = A(F p G). Proof (b). By exercise 7.4.2 we may assume r = g i −1. But then (g i −1)(g 1 + +g n ) = g i g 1 + g i g n −g 1 − −g n = g 1 + +g n −g 1 − −g n = 0. 7.4.7. Let R be a commutative ring with 1. Prove that the principal ideal generated by x in the polynomial ring R[x] is a prime ideal if and only if R is an integral domain. Prove that (x) is a maximal ideal if and only if R is a field. Proof. Let ϕ: R[x] → R be given by ϕ(p(x)) = p(0) and notice ker ϕ = (x). Thus R[x]/(x) ∼ = R so by proposi- tion 7.13, (x) is prime if and only if R is an integral domain. Similarly, proposition 7.12 shows (x) is maximal if and only if R is a field. 7.4.10. Assume R is commutative. Prove that if P is a prime ideal of R and P contains no zero divisors then R is an integral domain. Proof. Since P is prime, we know that R = R/P is an integral domain. So if r, a ∈ R`P, then ra = r a = 0. Thus ra = 0 and R is an integral domain. 7.4.13. Let ϕ: R →S be a homomorphism of commutative rings. 40 DAVID S. DUMMIT AND RICHARD M. FOOTE (a) Prove that if P is a prime ideal of S then either ϕ −1 (P) = R or ϕ −1 (P) is a prime ideal of R. Apply this to the special case when R is a subring of S and ϕ is the inclusion homomorphism to deduce that if P is a prime ideal of S then P ∩ R is either R or a prime ideal of R. (b) Prove that if M is a maximal ideal of S and ϕ is surjective then ϕ −1 (M) is a maximal ideal of R. Give an example to show that this need not be the case if ϕ is surjective. Proof (a). Let r ∈ ϕ −1 (P), so ϕ(rs) = ϕ(r)ϕ(s) ∈ P because ϕ(r) ∈ P and P is an ideal. Since R is commutative, ϕ −1 (P) is an ideal. Now let rs ∈ ϕ −1 (P), so ϕ(r)ϕ(s) = ϕ(rs) ∈ P so ϕ(r) or ϕ(s) is in P because P is prime. So r or s is in ϕ −1 (P) so ϕ −1 (P) is prime or ϕ −1 (P) = R. Specifically if R ⊂ S and ϕ is the inclusion homomorphism, then ϕ −1 (P) is either R or a prime ideal of R. But ϕ −1 (P) = P ∩ R. Proof (b). Let π 1 : S → S/M and π 2 : R/ ker(π 1 ◦ ϕ) be the canonical projection maps. Since π 1 ◦ ϕ is surjective, R/ ker(π 1 ◦ ϕ) ∼ = S/M. But ker(π 1 ◦ ϕ) = ϕ −1 (M). So R/ϕ −1 (M) is a field and ϕ −1 (M) is maximal. 7.4.15. Let x 2 +x +1 be an element of the polynomial ring E = F 2 [x] and use the bar notation to denote passage to the quotient ring F 2 [x]/(x 2 +x + 1). (a) Prove that E has 4 elements: 0, 1, x and x + 1. (b) Write out the 4 4 addition table for E and deduce that the additive group E is isomorphic to the Klein 4-group. (c) Write out the 4 4 multiplication table for E and prove that E × is isomorphic to the cyclic group of order 3. Deduce that E is a field. Proof (a). Let ¸ n i=1 a i x i be an element of F 2 [x]. If n = 1 then ¸ 1 i=0 = a 0 + a 1 x. Suppose there exists c 0 , c 1 such that ¸ n−1 i=0 a i x i = c 0 +c 1 x. Then if n is even, n ¸ i=0 a i x i = n−1 ¸ i=0 a i x i +a n x n = n−1 ¸ i=0 a i x i +a n (x + 1) n/2 = n−1 ¸ i=0 c i x i = c 0 +c 1 x. Now if n > 2 is odd, n ¸ i=0 a i x i = n−1 ¸ i=0 a i x i +a n x n = n−1 ¸ i=0 a i x i +a n x(x + 1) (n−1)/2 = n−1 ¸ i=0 b i x i = c 0 +c 1 x. So there exist c 0 , c 1 such that ¸ n i=0 a i x i = c 0 +c 1 x for all n. Since c 0 , c 1 ∈ ¦0, 1¦, F 2 [x]/(x 2 +x+1) ⊆ ¸ 0, 1, x, x + 1 ¸ . Solution (b). The addition table in table 7.1 shows F 2 [x]/(x 2 +x + 1) ∼ = V 4 . Solution (c). The multiplication table in table 7.2 shows E × ∼ = Z 3 . Therefore E is a field. 7.4.30. Let I be an ideal of the commutative ring R and define rad I = ¸ r ∈ R [ r n ∈ I for some n ∈ Z + ¸ called the radical of I. Prove that rad I is an ideal containing I and that (rad I)/I = A(R/I) (cf. exercise 7.3.29). 41 Addition Table 0 1 x 1 +x 0 0 1 x 1 +x 1 1 0 1 +x x x x 1 +x 0 1 1 +x 1 +x x 1 0 Table 7.1. F 2 [x]/(x 2 +x + 1). Multiplication Table 0 1 x 1 +x 0 0 0 0 0 1 0 1 x 1 +x x 0 x 1 +x 1 1 +x 0 1 +x 1 x Table 7.2. F 2 [x]/(x 2 +x + 1). 7.4.31. An ideal I of the commutative ring R is called a radical ideal if rad I = I. (a) Prove that every prime ideal of R is a radical ideal. (b) Let n > 1 be an integer. Prove that 0 is a radical ideal in Z/nZ if and only if n is a product of distinct primes to the first power (i.e., n is square free). Deduce that (n) is a radical of Z if and only if n is a product of distinct primes in Z. Proof (a). Note P ⊆ rad P. Let r ∈ R so if r n ∈ P, either r or r n−1 is in P. Induction shows r ∈ P, so rad P = P. Proof (b). Suppose rad 0 = 0 and n = mp 2 . Then mp = 0 mod n and (mp) 2 = 0 mod n which is a contradiction. Thus n is a product of distinct primes to the first power. Now suppose n is a product of distinct primes to the first power. If α m = 0 mod n, then α m contains all of the prime factors of n. Thus α must contain all prime factors of n so α = 0 mod n. Therefore rad 0 = 0. 7.4.39. Following the notation of exercise 7.1.26, let K be a field, let ν be a discrete valuation on K and let R be the valuation ring of ν. For each integer k ≥ 0 define A k = ¦r ∈ R [ ν(r) ≥ k¦ ∪ ¦0¦. (a) Prove that A k is a principal ideal and that A 0 ⊇ A 1 ⊇ A 2 ⊇ . (b) Prove that if I is any nonzero ideal of R, then I = A k for some k ≥ 0. Deduce that R is a local ring with unique maximal ideal A 1 . Proof (a). Since ν is surjective, A k is nonempty. Since −1 ∈ R × , ν(−1) = 0. So ν(a − b) ≥ min ¦ν(a), ν(−b)¦ = min ¦ν(a), ν(b)¦ ≥ k. Also ν(ab) = ν(a) + ν(b) ≥ k so A k is a subring. If r ∈ R and a ∈ A k then ν(ra) = ν(r) +ν(a) ≥ k and similarly ν(ar) ≥ k. So A k is an ideal. Now let ν(a) = k by surjectivity of ν. Notice that 0 = ν(1) = ν(aa −1 ) = ν(a) + ν(a −1 ) so ν(a) = −ν(a −1 ). If b ∈ A k then ν(ba −1 ) = ν(b) −ν(a) ≥ 0 so ba −1 ∈ R. Thus b = ba −1 a ∈ (a), so A k ⊆ (a). Therefore A k is a principal ideal and it is clear that A 0 ⊇ A 1 ⊇ A 2 ⊃ . Proof (b). Let k = min ¦ν(I)¦ with ν(r) = k. Then A k = (r) = I. From part (a) any proper ideal is contained in A 1 so A 1 is the unique maximal ideal of R. 7.4.41. A proper ideal Q of the commutative ring R is called primary if whenever ab ∈ Q and a / ∈ Q then b n ∈ Q for some positive integer n. Establish the following facts about primary ideals. (a) The primary ideal of Z are 0 and (p n ), where p is a prime and n is a positive integer. (b) Every prime ideal of R is a primary ideal. (c) An ideal Q of R is primary if and only if every zero divisor in R/Q is a nilpotent element of R/Q. (d) If Q is a primary ideal then rad (Q) is a prime ideal (cf. exercise 7.4.30). 42 DAVID S. DUMMIT AND RICHARD M. FOOTE Proof (a). Let I ⊂ Z be a primary ideal so I = (n) for some n ∈ Z since every ideal of Z is principal. Let n = pa for p some prime, so p / ∈ (n), a / ∈ (n) with (n) a primary ideal implies that p i ∈ (n). So n[p i shows that n = p k for some k ≥ 1. Thus (n) = (p k ). Now consider (p k ) ⊆ Z. Let ab ∈ (p k ) with a / ∈ (p k ). Then p k [ab and p k [ a, so p[b. Thus p k [b k , so b k ∈ (p k ) and (p k ) is a primary ideal. Proof (b). Let P be a prime ideal of R with ab ∈ P. If a / ∈ P then b ∈ P so P is a primary ideal. Proof (c). Let a b = 0 in R = R/Q. Then ab ∈ Q with a k ∈ Q and b l ∈ Q for k, l > 1. So a k = 0 thus every zero divisor is nilpotent in R. Now suppose every zero divisor in R is a nilpotent element. Let ab ∈ Q with a / ∈ Q. Then b is a zero divisor in R so b k = 0. Thus b k ∈ Q and Q is a primary ideal of R. Proof (d). Let ab ∈ rad (Q) and a / ∈ rad Q, so (ab) n ∈ Q and a m / ∈ Q for all m ∈ Z + . Since Q is primary then b nk ∈ Q so b ∈ rad (Q). Therefore rad (Q) is a prime ideal. 7.5. Rings of Fractions. Theorem 7.15 (cf. theorem 15.36). Let R be a commutative ring. Let D be any nonempty subset of R that does not contain 0, does not contain any zero divisors and is closed under multiplication (i.e., ab ∈ D for all a, b ∈ D). Then there is a commutative ring Q with 1 such that Q contains R as a subring and every element of D is a unit in Q. The ring Q has the following additional properties. (1) every element of Q is of the form rd −1 for some r ∈ R and d ∈ D. In particular, if D = R` ¦0¦ then Q is a field. (2) (uniqueness of Q) The ring Q is the “smallest” ring containing R in which all elements of D become units, in the following sense. Let S be any commutative ring with idenitty and let ϕ: R →S be any injective ring homomorhpism such that ϕ(d) is a unit in S for every d ∈ D. Then there is an injective ring homomorphism Φ: Q → S such that Φ[ R = ϕ. In other words, any ring containing an isomorphic copy of R in which all the elements of D become units must also contain an isomorphic copy of Q. Corollary 7.16. let R be an integral domain and let Q be the field of fractions of R. If a field F contains a subring R isomorphic to R then the subfield of F generated by R is isomorphic to Q. Let R be a commutative ring with identity 1 = 0. 7.5.1. Prove theorem 7.15. Proof. Let T = ¦(r, d) [ r ∈ R, d ∈ D¦ and define the relation ∼ on T by (r, d) ∼ (s, e) if and only if re = sd. It is immediate that this relation is reflexive and symmetric. Suppose (r, d) ∼ (s, e) and (s, e) ∼ (t, f). Then re − sd = 0 and sf − te = 0. Multiplying the first of these equations by f and the second by d and adding them gives rf − td)e = 0. Since e ∈ D is neither zero nor a zero divisor we must have rf − td = 0, i.e., (r, d) ∼ (t, f). This proves ∼ is transitive, hence an equivalence relation. Done the equivalence class of (r, d) by r d : r d = ¦(a, b) [ a ∈ R, b ∈ D and rb = ad¦ . Let Q be the set of equivalence classes under ∼. Note that r d = re de in Q for all e ∈ D, since D is closed under multiplication. We now define an additive and multiplicative structure on Q: a b + c d = ad +bc bd and a b c d = ac bd . In order to prove that Q is a commutative ring with identity there are a number of things to check: (1) these operations are well defined (2) Q is an abelian group under addition, where the additive identity is 0 d for any d ∈ D and the additive inverse of a d is −a d , (3) multiplication is associative, distributive and commutative, and (4) Q has an identity. 43 To check that addition is well defined assume a b = a b (i.e., ab = a b) and c d = c d . We must show that ad+bc bd = a d +b c b d , i.e., (ad +bc)(b d ) = (a d +b c )(bd). The left hand side of this equation is ab dd + cd bb substituting a b for ab and c d for cd gives a bdd + c dbb , which is the right hand side. Hence addition of fractions is well defined. To check that multiplication is well defined assume a b = a b and c d = c d . We must show (ac)(b d ) = (a c )(bd). The left hand side of this equation is (ab )(cd ) substituting a b for ab and c d for cd gives (a b)(c d) which is the right hand side. Hence multiplication of fractions is well defined. To show Q is an abelian group under addition note that 0 d ∈ Q with 0 d1 + r d2 = d1r d1d2 = r d2 so there exists an additive identity in Q. Now if r1 d1 , r2 d2 , r3 d3 ∈ Q, then ( r 1 d 1 + r 2 d 2 ) + r 3 d 3 = r 1 d 2 +r 2 d 1 d 1 d 2 + r 3 d 3 = r 1 d 2 d 3 +r 2 d 1 d 3 +r 3 d 1 d 2 d 1 d 2 d 3 = r 1 d 1 + r 2 d 3 +d 1 d 2 d 2 d 3 = r 1 d 1 + ( r 2 d 2 + r 3 d 3 ). Thus addition is associative. Now if r d ∈ Q, then −r d ∈ Q with r d + −r d = rd −rd d 2 = 0 d . So Q is closed under inverse. Since addition is clearly commutative in Q, we see Q is an abelian group under addition. To show multiplication is associative, let r1 d1 , r2 d2 , r3 d3 ∈ Q so ( r 1 d 1 r 2 d 2 ) r 3 d 3 = r 1 r 2 d 1 d 2 r 3 d 3 = r 1 r 2 r 3 d 1 d 2 d 3 = r 1 d 1 r 2 r 3 d 2 d 3 = r 1 d 1 ( r 2 d 2 r 3 d 3 ). To show multiplication is distributive, let r1 d1 , r2 d2 , r3 d3 ∈ Q so ( r 1 d 1 + r 2 d 2 ) r 3 d 3 = r 1 d 2 +r 2 d 1 d 1 d 2 r 3 d 3 = r 1 r 3 d 2 +r 2 r 3 d 1 d 1 d 2 d 3 = r 1 d 1 r 3 d 2 d 2 d 3 + r 2 d 2 r 3 d 1 d 1 d 3 = r 1 d 1 r 3 d 3 + r 2 d 2 r 3 d 3 r 1 d 1 ( r 2 d 2 + r 3 d 3 ) = r 1 d 1 r 2 d 3 +r 3 d 2 d 2 d 3 = r 1 r 2 d 3 +r 1 r 3 d 2 d 1 d 2 d 3 = r 1 d 1 r 2 d 3 d 2 d 3 + r 1 d 1 r 3 d 2 d 2 d 3 = r 1 d 1 r 2 d 2 + r 1 d 1 r 3 d 3 . 44 DAVID S. DUMMIT AND RICHARD M. FOOTE To show multiplication is commutative, let r1 d1 , r2 d2 ∈ Q so r 1 d 1 r 2 d 2 = r 1 r 2 d 1 d 2 = r 2 r 1 d 2 d 1 = r 2 d 2 r 1 d 1 . To show Q has an identity, notice that d d ∈ Q for any d ∈ D. Now let r d1 ∈ Q so r d 1 d d = rd d 1 d = r d 1 . Next we embed R into Q by defining ι : R →Q by ι : r → rd d where d is any element of D. Since rd d = re e for all d, e ∈ D, ι(r) does not depend on the choice of d ∈ D. Now let r 1 , r 2 ∈ R so ι(r 1 +r 2 ) = (r 1 +r 2 )d d = r 1 d d + r 2 d d = ι(r 1 ) +ι(r 2 ) ι(r 1 r 2 ) = (r 1 r 2 )d d = r 1 d d r 2 d d = ι(r 1 )ι(r 2 ), so ι is a ring homomorphism. Furthermore, ι is injective because ι(r) = 0 ⇔ rd d = 0 d ⇔rd 2 = 0 ⇔r = 0 because d is neither zero nor a zero divisor. The subring ι(R) of Q is therefore isomorphic to R. We henceforth identify each r ∈ R with ι(r) and so consider R as a subring of Q. Next note that each d ∈ D has a multiplicative inverse in Q: namely, if d is represented by the fraction de e then its multiplicative inverse is e de . One then sees that every element of Q may be written as r d −1 for some r ∈ R and some d ∈ D. In particular, if D = R` ¦0¦, every nonzero element of Q has a multiplicative inverse and Q is a field. It remains to establish the uniqueness property of Q. Assume ϕ: R →S is an injective ring homomorphism such that ϕ(d) is a unit in S for all d ∈ D. Extend ϕ to a map Φ: Q → S by defining Φ(rd −1 ) = ϕ(r)ϕ(d) −1 for all r ∈ R, d ∈ D. This map is well defined, since rd −1 = se −1 implies re = sd, so ϕ(r)ϕ(e) = ϕ(s)ϕ(d), and then Φ(rd −1 ) = ϕ(r)ϕ(d) −1 = ϕ(s)ϕ(e) −1 = Φ(se −1 ). Now let r 1 d −1 1 , r 2 d −1 2 ∈ Q. Then Φ(r 1 d −1 1 +r 2 d −1 2 ) = Φ((r 1 d 2 +r 2 d 1 )(d 1 d 2 ) −1 ) = ϕ(r 1 d 2 +r 2 d 1 )ϕ(d 1 d 2 ) −1 = ϕ(r 1 )ϕ(d 2 )ϕ(d 1 ) −1 ϕ(d 2 ) −1 +ϕ(r 2 )ϕ(d 1 )ϕ(d 1 ) −1 ϕ(d 2 ) −1 = ϕ(r 1 )ϕ(d 1 ) −1 +ϕ(r 2 )ϕ(d 2 ) −1 = Φ(r 1 d −1 1 ) + Φ(r 2 d −1 2 ) Φ(r 1 d −1 1 r 2 d −1 2 ) = Φ((r 1 r 2 )(d 1 d 2 ) −1 ) = ϕ(r 1 r 2 )ϕ(d 1 d 2 ) −1 = ϕ(r 1 )ϕ(d 1 ) −1 ϕ(r 2 )ϕ(d 2 ) −1 = Φ(r 1 d −1 1 )Φ(r 2 d −1 2 ). So Φ is a ring homomorphism. Finally, Φ is injective because rd −1 ∈ ker Φ implies r ∈ ker Φ∩R = ker ϕ; since ϕ is injective this forces r and hence also rd −1 to be zero. 45 7.5.2. Let R be an integral domain and let D be a nonempty subset of R that is closed under multiplication. Prove that the ring of fractions D −1 R is isomorphic to a subring of the quotient field of R (hence is also an integral domain). Proof. Let Q be the quotient field of R. Since R is a domain, D` ¦0¦ is also a multiplicative set of R. By theorem 7.15 there exists an injective ring homomorhpism ι : R →Q with ι(R` ¦0¦) ⊆ Q × . Since D` ¦0¦ ⊆ R` ¦0¦, theorem 7.15 states there exists an injective ring homomorhpism ψ: D −1 R →Q with ψ[ R = ι. 8. Euclidean, Principal Ideal, and Unique Factorization Domains 8.1. Euclidean Domains. Proposition 8.1. Every ideal in a Euclidean Domain is principal. More precisely, if I is any nonzero ideal in the Euclidean Domain R then I = (d), where d is any nonzero element of I of minimum norm. Proposition 8.2. If a and B are nonzero element sin the commutative ring R such that the ideal generated by a and b is a principal ideal (d), then d is a greatest common divisor of a and b. Proposition 8.3. Let R be an integral domain. If two elements d and d of R generate the same principal ideal, i.e., (d) = (d ), then d = ud for some u ∈ R × . In particular, if d and d are both greatest common divisors of a and b, then d = ud for some unit u. Theorem 8.4. Let R be a Euclidean Domain and let a and b be nonzero elements of R. Let d = r n be the last nonzero remainder in the Euclidean Algorithm for a and b described at the beginning of this chapter. Then (1) d is a greatest common divisor of a and b, and (2) the principal ideal (d) is the ideal generated by a and b. In particular, d can be written as an R-linear combination of a and b, i.e., there are elements x and y in R such that d = ax +by. Proposition 8.5. Let R be an integral domain that is not a field. If R is a Euclidean Domain then there are universal side divisors in R. 8.1.3. Let R be a Euclidean Domain. Let m be the minimum integer in the set of norms of nonzero elements of R. Prove that every nonzero element of R of norm m is a unit. Deduce that a nonzero element of norm zero (if such an element exists) is a unit. Proof. Let a ∈ R with N(a) = m. Since R is a Euclidean Domain, there exists q, r ∈ R such that 1 = qa + r with N(r) < N(a) or r = 0. Since a is chosen to be minimum norm, r = 0. Thus a is a unit. 8.1.5. Determine all integer solutions of the following equations: (a) (b) 17x + 29y = 31 (c) Proof (b). Notice that 372 17 −217 29 = 31. So any solution to this equation has form x = 372 +m 899, y = −217 −m 527, for any m ∈ Z. 8.1.8. Let F = Q( √ D) be a quadratic field with associated quadratic integer ring O and field norm N as in section 7.1. (a) Suppose D is −1, −2, −3, −7 or −11. Prove that O is a Euclidean Domain with respect to N. (b) Suppose that D = −43, −67 or −163. Prove that O is not a Euclidean Domain with respect to any norm. 8.1.10. Prove that the quotient ring Z[i]/I is finite for any nonzero ideal I of Z[i]. Proof. By exercise 8.1.8(a), Z[i] is a Euclidean Domain. So I = (α) for some α ∈ Z[i]. Let x + I be a coset of I. Then there exists q, r such that x +α = qα +r. Since x / ∈ (α), then r = 0 so N(r) < N(α). But now x+I = r +I, so any coset of I has a representative with norm less than N(α). Let F = ¦f ∈ R [ N(f) < N(α)¦. Then if f = f 1 +f 2 i and α = a 1 +a 2 i we have f 2 1 +f 2 2 < a 2 1 +a 2 2 . Since there are only a finite number of (f 1 , f 2 ) ∈ Z Z with this property, Z[i]/I is finite for any nonzero ideal I of Z[i]. 46 DAVID S. DUMMIT AND RICHARD M. FOOTE 8.1.11. Let R be a commutative ring with 1 and let a and b be nonzero elements of R. A least common multiple of a and b is an element e of R such that (i) a [ e and b [ e, and (ii) if a [ e and b [ e then e [ e . (a) Prove that a least common multiple of a and b (if such exists) is a generator for the unique largest principal ideal contained in (a) ∩ (b). (b) Deduce that any two nonzero elements in a Euclidean Domain have a least common multiple which is unique up to multiplication by a unit. (c) Prove that in a Euclidean Domain the least common multiple of a and b is ab (a,b) , where (a, b) is the greatest common divisor of a and b. Proof (a). First suppose there exists a least common multiple e of a and b. Since a [ e and b [ e we must have (e) ⊆ (a) and (e) ⊆ (b). If (e ) is a principal ideal such that (e ) ⊆ (a) ∩ (b) then e [ a and e [ b. Thus e [ e and (e ) ⊆ (e) so (e) is the largest principal ideal contained in (a) ∩ (b). Now suppose there is a largest principal ideal (e) contained in (a) ∩ (b). Then e [ a and e [ b. If a [ e and b [ e , then (e ) ⊆ (a) ∩ (b). Since (e) is the largest such principal ideal, then (e ) ⊆ (e). But then e [ e . Proof (b). If a and b are nonzero elements of R then (a) ∩ (b) is the largest ideal contained in (a) ∩ (b). Since R is a Euclidean Domain, there exists e ∈ R with (e) = (a) ∩ (b). So part (a) shows e is a least common multiple of a and b. Now let e, e be two least common multiples of a and b. Then e [ e and e [ e. So e = se and et = e . Since a and b are nonzero then e is also nonzero so e = ste and e(1 −st) = 0 implies 1 = st because R is a domain. Thus least common multiples are unique up to multiplication by a unit. Proof (c). First note that ( ab (a,b) ) ⊆ (a) ∩ (b) so ( ab (a,b) ) ⊆ (e). Now let ax = e = by. Then abx = eb and aby = ea so ab [ eb and ab [ ea. Thus ab [ (ea, eb) and ab (a,b) [ e. So we have (e) ⊆ ( ab (a,b) ). Therefore ab (a,b) is a least common multiple of a and b. 8.1.12 (A Public Key Code). Let N be a positive integer. Let M be an integer relatively prime to N and let d be an integer relatively prime to ϕ(N), where ϕ denotes Euler’s ϕ-function. Prove that if M 1 ≡ M d mod N then M ≡ M d 1 mod N where d is the inverse of d mod ϕ(N): dd ≡ 1 mod ϕ(N). Proof. Since (M, N) = 1 we know that M ϕ(N) ≡ 1 mod N. Thus M d 1 ≡ M dd mod N ≡ M αϕ(N)+1 mod N ≡ M mod N. 9. Polynomial Rings 9.1. Definitions and Basic Properties. Proposition 9.1. Let R be an integral domain. Then (1) degree p(x)q(x) = degree p(x) + degree q(x) if p(x), q(x) are nonzero (2) the units of R[x] are just the units of R (3) R[x] is an integral domain. Proposition 9.2. Let I be an ideal of the ring R and let (I) = I[x] denote the ideal of R[x] generated by I (the set of polynomials with coefficients in I). Then R[x]/(I) ∼ = (R/I)[x]. In particular, if I is a prime ideal of R then (I) is a prime ideal of R[x]. 9.1.13. Prove that the rings F[x, y]/(y 2 −x) and F[x, y]/(y 2 −x 2 ) are not isomorphic for any field F. Proof. Since y 2 −x 2 = (y −x)(y +x) then F[x, y]/(y 2 −x 2 ) is not a domain. Not let ϕ: R[x, y] →R[y] be given by ϕ(p(x, y)) = p(y 2 , y). Let p(x, y)(y 2 −x) ∈ (y 2 −x) so that ϕ(p(x, y)(y 2 −x)) = p(y 2 , y)(y 2 −y 2 ) = 0. 47 Thus (y 2 −x) ⊆ ker ϕ. Now let p(x, y) ∈ ker ϕ so that 0 = ϕ(p(x, y)) = p(y 2 , y). Note that we can write p(x, y) = r(x)+ys(x)+q(x, y)(y 2 −x) by considering cosets of F[x, y]/(y 2 −x). But then 0 = ϕ(r(x)+ys(x)) = r(y 2 )+ys(y 2 ). So r = 0 and s = 0. Therefore p(x, y) = q(x, y)(y 2 −x) so ker ϕ ⊆ (y 2 −x). 9.1.14. Let R be an integral domain and let i, j be relatively prime integers. Prove that the ideal (x i − y j ) is a prime ideal in R[x, y]. Proof. 9.1.15. Let p(x 1 , x 2 , . . . , x n ) be a homogeneous polynomial of degree k in R[x 1 , . . . , x n ]. Prove that for all λ ∈ R we have p(λx 1 , λx 2 , . . . , λx n ) = λ k p(x 1 , x 2 , . . . , x n ). Proof. First we write p(x 1 , x 2 , . . . , x n ) = m ¸ i=0 a i n ¸ j=1 x kj j , where ¸ n j=1 k j = k. Then p(λx 1 , λx 2 , . . . , λx n ) = m ¸ i=0 a i n ¸ j=1 (λx j ) kj = λ k m ¸ i=0 a i n ¸ j=1 x kj j = λ k p(x 1 , x 2 , . . . , x n ). 9.1.17. An ideal I in R[x 1 , . . . , x n ] is called a homogeneous ideal if whenever p ∈ I then each homogeneous component of p is also in I. Prove that an ideal is a homogeneous ideal if and only if it may be generated by homogeneous polynomials. Proof. 9.2. Polynomial Rings Over Fields I. Theorem 9.3. Let F be a field. The polynomial ring F[x] is a Euclidean Domain. Specifically, if a(x) and b(x) are two polynomials in F[x] with b(x) nonzero, then there are unique q(x) and r(x) in F[x] such that a(x) = q(x)b(x) +r(x) with r(x) = 0 or deg r(x) < deg b(x). Corollary 9.4. If F is a field, then F[x] is a Principal Ideal Domain and a Unique Factorization Domain. 9.2.4. Let F be a finite field. Prove that F[x] contains infinitely many primes. Proof. Suppose on the other hand that p 1 (x), . . . , p n (x) are all of the primes in F[x]. Let p(x) = ¸ n i=1 p i (x) and notice 1 +p(x) ∈ F[x]. Since F[x] is a unique factorization domain we can write 1 +p(x) = m ¸ i=1 p αi (x) where each p αi (x) is irreducible. But now 1 = m ¸ i=1 p αi (x) −p(x) = p α1 (x) m ¸ i=2 p αi (x) − p(x) p α1 (x) . This is a contradiction because a prime is not a unit. 9.2.5. Exhibit all the ideals in the ring F[x]/(p(x)), where F is a field and p(x) is a polynomial in F[x]. 48 DAVID S. DUMMIT AND RICHARD M. FOOTE Proof. Since F is a field, F[x] is a Euclidean domain and in particular a unique factorization domain. So let p(x) = n ¸ i=1 p i (x), where each p i (x) is irreducible. Let I/(p(x)) ⊆ F[x]/(p(x)) be an ideal. By the lattice isomorphism theorem, (p(x)) ⊆ I, and I is an ideal of F[x]. Since F[x] is a Euclidean domain, it is also a principal ideal domain so let I = (q(x)). Since (p(x)) ⊆ (q(x)), p(x) = r(x)q(x) for some r(x) ∈ F[x]. Let r(x) = ¸ s i=1 r i (x) and q(x) = ¸ t i=1 q i (x) for r i (x), q i (x) irreducible. Then α ¸ s i=1 r i (x) ¸ t i=1 q i (x) = ¸ n i=1 p i (x), for some α ∈ F. So there exists σ ∈ S n such that α i q i (x) = p σ(i) (x) for all 1 ≤ i ≤ t, α i ∈ F. Therefore (q(x)) = ( ¸ t i=1 q i (x)) = ( ¸ t i=1 p σ(i) (x)). 9.2.6. Describe the ring structure of the following rings: (d) Z[x, y]/(x 2 , y 2 , 2). Show that α 2 = 0 or 1 for every α in the last ring and determine those elements with α 2 = 0. Determine the characteristics of each of these rings. Proof (d). The highest power of x and y is 1 along with the largest coefficient. Let α = α(x, y) + (x 2 , y 2 , 2) so α 2 = (α(x, y)) 2 + (x 2 , y 2 , 2) ∈ ¸ 0, 1 ¸ because if there is no constant term then every term has even degree and if there is a constant term, let α(x, y) = β(x, y) +1 where β(x, y) has no constant term. So α(x, y) 2 = (β(x, y) +1) 2 = β(x, y) 2 + 2β(x, y) + 1 so this is in 1. The elements with α = 0 are the elements with no constant term. 9.2.10. Determine the greatest common divisor of a(x) = x 3 + 4x 2 + x − 6 and b(x) = x 5 − 6x + 5 in Q[x] and write it as a linear combination in Q[x] of a(x) and b(x). Solution. Using division of polynomials we calculate x 5 −6x + 5 = (x 3 + 4x 2 −x −6)(x 2 −4x + 15) −(50x 2 + 45x −95) x 3 + 4x 2 +x −6 = (50x 2 + 45x −95)(x/50 + 31/500) + (31/10 x 2 + 1395/500 x −2945/500) 50x 2 + 45x −95 = (31/10 x 2 + 1395/500 x −2945/500)(500/31). So we write the greatest common divisor as 31/10 x 2 + 1395/500 x −2945/500 = = (x 3 + 4x +x −6)[(x 2 −4x + 15)(x/50 + 31/500) −1] −(x 5 −6x + 5)[x/50 + 31/500]. 9.3. Polynomial Rings That are Unique Factorization Domains. Proposition 9.5 (Gauss’ Lemma). Let R be a Unique Factorization Domain with field of fractions F and let p(x) ∈ R[x]. If p(x) is reducible in F[x] then p(x) is reducible in R[x]. More precisely, if p(x) = A(x)B(x) for some nonconstant polynomials A(x),B(x) ∈ F[x], then there are nonzero elements r, s ∈ F such that rA(x) = a(x) and sB(x) = b(x) both lie in R[x] and p(x) = a(x)b(x) is a factorization in R[x]. Corollary 9.6. Let R be a Unique Factorization Domain, let F be its field of fractions and let p(x) ∈ R[x]. Suppose the greatest common divisor of the coefficients of p(x) is 1. Then p(x) is irreducible in R[x] if and only if it is irreducible in F[x]. In particular, if p(x) is a moni polynomial that is irreducible in R[x], then p(x) is irreducible in F[x]. Theorem 9.7. R is a Unique Factorization Domain if and only if R[x] is a Unique Factorization Domain. Corollary 9.8. If R is a Unique Factorization Domain, then a polynomial ring in an arbitrary number of variable with coefficients in R is also a Unique Factorization Domain. 9.4. Irreducibility Criteria. Proposition 9.9. Let F be a field and let p(x) ∈ F[x]. Then p(x) has a facto of degree one if and only if p(x) has a roote in F, i.e., there is an α ∈ F with p(α) = 0. Proposition 9.10. A polynomial of degree two or three over a field F is reducible if and only if it has a root in F. Proposition 9.11. Let p(x) = a n x n +a n−1 x n−1 + +a 0 be a polynomial of degree n with integer coefficients. If r/x ∈ Q is in lowest terms (i.e., r and s are relatively prime integers) and r/s is a root of p(x), then r divides the constant term and s divides the leading coefficient of p(x). In particular, if p(x) is a monic polynomial with integer coefficients and p(d) = 0 for all itnegers d dividing the constant tterm of p(x), then p(x) has no roots in Q. 49 Proposition 9.12. Let I be a proper ideal in the integral domain R and let p(x) be a nonconstant monic polynomial in R[x]. If the image of p(x) in (R/I)[x] cannot be factored in (R/I)[x] into two polynomials of smaller degree, then p(x) is irreducible in R[x]. Proposition 9.13 (Eisenstein’s Criterion). Let P be a prime ideal of the integral domain R and let f(x) = x n +a n−1 x n−1 + +a 1 x +a 0 be a polynomial in R[x] (here n ≥ 1). Suppose a n−1 , . . . , a 1 , a 0 are all elements of P and suppose a 0 is not an element of P 2 . Then f(x) is irreducible in R[x]. Corollary 9.14 (Eisenstein’s Criterion for Polynomial Ring over Z). Let p be a prime in Z and let f(x) = x n +a n−1 x n−1 + +a 1 x +a 0 ∈ Z[x], n ≥ 1. Suppose p divides a i for all i ∈ ¦0, 1, . . . , n −1¦ but that p 2 does not divide a 0 . Then f(x) is irreducible in both Z[x] and Q[x]. 9.4.1. Determine whether the following polynomials are irreducible in the rings indicated. For those that are reducible, determine their factorization into irreducibles. The notaion F p denotes the finite field Z/pZ, p a prime. (b) x 3 +x + 1 in F 3 [x]. (c) x 4 + 1 in F 5 [x]. Solution (b). Solution (c). 9.4.2. Prove that the following polynomials are irreducible in Z[x]: (c) x 4 + 4x 3 + 6x 2 + 2x + 1 (d) (x+2) p −2 p x , where p is an odd prime. Proof (c). Proof (d). 9.4.3. Show that the polynomial (x −1)(x −2) (x −n) −1 is irreducible over Z for all n ≥ 1. Proof. 9.4.6. Construct fields of each of the following orders: (a) 9, (b) 49, (c) 8, (d) 81. Solution (a). Solution (b). Solution (c). Solution (d). 9.4.8. Prove that K 1 = F 11 [x]/(x 2 + 1) and K 2 = F 11 [y]/(y 2 + 2y + 2) are both fields with 121 elements. Prove that the map which sends the element p(x) of K 1 to the element p(y + 1) of K 2 (where p is any polynomial with coefficients in F 11 ) is well defined and gives a ring isomorphism from K 1 to K 2 . Proof. 9.4.16. Let F be a field and let f(x) be a polynomial of degree n in F[x]. The polynomial g(x) = x n f(1/x) is called the reverse of f(x). (a) Describe the coefficients of g in terms of the coefficients of f. (b) If f(0) = 0 prove that f is irreducible if and only if g is irreducible. Solution (a). Proof (b). 9.4.19. Let F be a field and let f(x) = a n x + a n−1 x n−1 + + a 0 ∈ F[x]. The derivative, D x (f(x)), of f(x) is defined by D x (f(x)) = na n x n−1 + (n −1)a n−1 x n−2 + +a 1 where, as usual, na = a + +a (n times). Note that D x (f(x)) is again a polynomial with coefficients in F. The polynomial f(x) is said to have a multiple root if there is some field E containing F and some α ∈ E such that x−α) 2 divides f(x) in E[x]. For example, the polynomial f(x) = (x−1) 2 (x−2) ∈ Q[x] has α = 1 as a multiple root and the polynomial f(x) = x 4 + 2x 2 + 1 = (x 2 + 1) 2 ∈ 1[x] has α = ±i ∈ C as multiple roots. We shall prove in section 13.5 that a nonconstant polynomial f(x) has a multiple root if and only if f(x) is not relatively prime to 50 DAVID S. DUMMIT AND RICHARD M. FOOTE its derivative (which can be detected by the Euclidean Algorithm in F[x]). Use this criterion to determine whether the following polynomials have multiple roots: (a) x 3 −3x −2 ∈ Q[x] (b) x 3 + 3x + 2 ∈ Q[x] (c) x 6 −4x 4 + 6x 3 + 4x 2 −12x + 9 ∈ Q[x] (d) Show for any prime p and any a ∈ F p that the polynomial x p −a has a multiple root. 9.5. Polynomial Rings Over Fields II. Proposition 9.15. The maximal ideals in F[x] are the ideal (f(x)) generated by irreducible polynomials f(x). In particular, F[x]/(f(x)) is a field if and only if f(x) is irreducible. Proposition 9.16. Let g(x) be a nonconstant monic element of F[x] and let g(x) = f 1 (x) n1 f 2 (x) n2 f k (x) n k be its factorization into irreducibles, where the f i (x) are distinct. Then we have the following isomorphism of rings: F[x]/(g(x)) ∼ = F[x]/(f 1 (x) n1 ) F[x]/(f 2 (x) n2 ) F[x]/(f k (x) n k ). Proposition 9.17. If the polynomial f(x) has roots α 1 , α 2 , . . . , α k in F (not necessarily distinct), then f(x) has (x −α 1 ) (x −α k ) as a factor. In particular, a polynomial of degree n in one variable over a field F has at most n roots in F, even counted with multiplicity. Proposition 9.18. A finite subgroup of the multiplicative group of a field is cyclic. In particular, if F is a finite field, then the multiplicative group F × of nonzero elements of F is a cyclic group. Corollary 9.19. Let p be a prime. The multiplicative group (Z/pZ) × of nozero residue classes mod p is cyclic. Corollary 9.20. Let n ≥ 2 be an integer with factorization n = p α1 1 p αr r in Z, where p 1 , . . . , p r are distinct primes. We have the following isomorphisms of multiplicative groups: (1) (Z/nZ) × ∼ = (Z/p αi i Z) × (Z/p αr r Z) × (2) (Z/2 α Z) × is the direct product of a cyclic group of order 2 and a cyclic group of order 2 α−2 , for all α ≥ 2 (3) (Z/p α Z) × is a cyclic group of order p α−1 (p −1), for all odd primes p. 9.6. Polynomials in Several Variables Over a Field and Gr¨obner Bases. Theorem 9.21 (Hilbert’s Basis Theorem). If R is a Noetherian ring then so is the polynomial ring R[x]. Corollary 9.22. Every ideal in the polynomial ring F[x 1 , x 2 , . . . , x n ] with coefficients from a field F is finitely generated. Theorem 9.23. Fix a monomial ordering on R = F[x 1 , . . . , x n ] and suppose ¦g 1 , . . . , g m ¦ is a Gr¨obner basis for the nozero ideal I in R. Then (1) Every polynomial f ∈ R can be written uniquely in the form f = f I +r where f I ∈ I and no nonzero monomial term of the remainder r is divisible by any of the leading terms LT(g 1 ), . . . , LT(g m ). (2) Both f I and r can be computed by general polynomial division by g 1 , . . . , g m and are independent of the order in which these polynomials are used in the division. (3) The remainder r provides a unique representative for the coset of f in the quotient ring F[x 1 , . . . , x n ]/I. In particular, f ∈ I if and only if r = 0. Proposition 9.24. Fix a monomial ordering on R = F[x 1 , . . . , x n ] and let I be a nonzero ideal in R. (1) If g 1 , . . . , g m are any elements of I such that LT(I) = (LT(g 1 ), . . . , LT(g m )), then ¦g 1 , . . . , g m ¦ is a Gr¨obner basis for I. (2) The ideal I has a Gr¨ obner basis. Lemma 9.25. Suppose f 1 , . . . f m ∈ F[x 1 , . . . , x n ] are polynomials with the same multidegree α and that the linear combination h = a 1 f 1 + +a m f m with constants a i ∈ F has strictly smaller multidegree. Then h = m ¸ i=2 b i S(f i−1 , f i ), for some constants b i ∈ F. 51 Proposition 9.26 (Buchberger’s Criterion). Let R = F[x 1 , . . . , x n ] and fix a monomial ordering on R. If I = (g 1 , . . . , g m ) is a nonzero ideal in R, then G = ¦g 1 , . . . , g m ¦ is a Gr¨ obner basis for I if and only if S(g i , g j ) ≡ 0 mod G for 1 ≤ i < j ≤ m. Theorem 9.27. Fix a monomial ordering on R = F[x 1 , . . . , x n ]. Then there is a unique reduced Gr¨obner basis for every nonzero ideal I in R. Corollary 9.28. Let I and J be two ideal in F[x 1 , . . . , x n ]. Then I = J if and only if I and J have the same reduced Gr¨obner basis with respect to any fixed monomial ordering on F[x 1 , . . . , x n ]. Proposition 9.29 (Elimination). Suppose G = ¦g 1 , . . . , g m ¦ is a Gr¨obner basis for the nonzero ideal I in F[x 1 , . . . , x n ] with respect to the lexicographic monomial ordering x 1 > > x n . Then G ∩ F[x i+1 , . . . , x n ] is a Gr¨obner basis of the i th elimination ideal I i = I ∩ F[x i+1 , . . . , x n ] of I. In particular, I ∩ F[x i+1 , . . . x n ] = 0 if and only if G∩ F[x i+1 , . . . , x n ] = ∅. Proposition 9.30. If I and J are any two ideals in F[x 1 , . . . , x n ] then tI + (1 −t)J is an ideal in F[t, x 1 , . . . , x n ] and I ∩ J = (tI + (1 −t)J) ∩ F[x 1 , . . . , x n ]. In particular, I ∩ J is the first elimination ideal of tI + (1 −t)J with respect to the ordering t > x 1 > > x n . 52 DAVID S. DUMMIT AND RICHARD M. FOOTE 9.6.24. Use reduced Gr¨obner bases to show that the ideal I = (x 3 −yz, yz +y) and the ideal J = (x 3 z +x 3 , x 3 +y) in F[x, y] are equal. Proof. By corollary 9.28 two ideals in F[x, y, z] are equal if and only if I, J have the same reduced Gr¨obner basis for any monomial ordering. So fix the lexicographical ordering x > y > z for F[x, y, z]. To find a Gr¨obner basis for I, S(f 1 , f 2 ) = yzf 1 −x 3 f 2 = −y 2 z 2 −x 3 y ≡ 0 mod ¦f 1 , f 2 ¦ . Since x 3 −yz = −1(yz +y) + (x 3 +y), then our reduced Gr¨obner basis for I is G = ¸ x 3 +y, yz +y ¸ . Now to find the reduced Gr¨obner basis for J, S(g 1 , g 2 ) = g 1 −zg 2 = x 3 −yz ≡ −yz −y mod ¦g 1 , g 2 ¦ S(g 1 , g 3 ) = −yg 1 −x 3 g 3 = −x 3 y +x 3 y ≡ 0 mod ¦g 1 , g 2 , g 3 ¦ S(g 2 , g 3 ) = x 3 g 2 +yzg 3 = x 3 −y 2 z ≡ 0 mod ¦g 1 , g 2 , g 3 ¦ . Thus ¦g 1 , g 2 , g 3 ¦ is a Gr¨obner basis for J. Since LT(g 2 ) mod LT(g 1 ), G = ¸ x 3 +y, yz +y ¸ is the reduced Gr¨obner basis for J. 9.6.32. Use Gr¨obner bases to show that (x, z) ∩ (y 2 , x −yz) = (xy, x −yz) in F[x, y, z]. Proof. Let I = (x, z) and J = (y 2 −yz). Then proposition 9.30 states I ∩ J = (tI + (1 −t)J) ∩ F[x, y, z]. By calculating the S(f i , f j ) we find tI + (1 −t)J = (tx, tz, ty 2 −y 2 , tx −tyz −x +yz, xy 2 , tyz +x −yz, x 2 −xyz, y 2 z, xz −yz 2 , x −yz) is given by a Gr¨ obner basis. Thus tI + (1 −t)J = (tz, ty 2 −y 2 , tyz +x −yz, y 2 z, x −yz) and finally I ∩ J = (y 2 z, x −yz). Now to find a Gr¨obner basis for (xy, x −yz), we calculate S(f 1 , f 2 ) = f 1 −yf 2 = y 2 z S(f 1 , f 3 ) = yzf 1 −xf 3 = 0 S(f 2 , f 3 ) = y 2 zf 2 −xf 3 = −y 3 z 2 ≡ 0 mod ¦f 1 , f 2 , f 3 ¦ . Thus ¸ xy, x −yz, y 2 z ¸ is a Gr¨obner basis so ¸ x −yz, y 2 z ¸ is the reduced Gr¨obner basis because x [ xy. Since the reduced Gr¨obner basis is the same for both ideals, corollary 9.28 states the ideals are equal. 53 Part III – Modules and Vector Spaces 10. Introduction to Module Theory 10.1. Basic Definitions and Examples. Proposition 10.1 (The Submodule Criterion). Let R be a ring and let M be an R-module. A subset N of M is a submodule of M if and only if (1) N = ∅, and (2) x +ry ∈ N for all r ∈ R and for all x, y ∈ N. 10.1.8. An element m of the R-module M is called a torsion element if rm = 0 for some nonzero element r ∈ R. The set of torsion elements is denoted Tor(M) = ¦m ∈ M [ rm = 0 for some nonzero r ∈ R¦ . (a) Prove that if R is an integral domain then Tor(M) is a submodule of M called the torsion submodule of M. (b) Give an example of a ring R and an R-module M such that Tor(M) is not a submodule. (c) If R has zero divisors show that every nonzero R-module has nonzero torsion elements. Proof (a). Tor(M) = ∅ because r0 = r(m−m) = rm−rm = 0. If r 1 x = r 2 y = 0, r ∈ R, then r 1 r 2 (x +ry) = r 2 r 1 x +rr 1 r 2 y = 0 so Tor(M) is a submodule of M because r 1 r 2 = 0. Example (b). If R = 0, then this is such an example. Proof (c). Let x, y ∈ R` ¦0¦ with xy = 0. If ym = 0, then xym = 0 so either way there are nonzero torsion elements. 10.1.14. Let z be an element of the center of R, i.e., zr = rz for all r ∈ R. Prove that zM is a submodule of M, where zM = ¦zm [ m ∈ M¦. Show that if R is the ring of 2 2 matrices over a field and e is the matrix with a 1 in position 1, 1 and zeros elsewhere then eR is not a left R-submodule (where M = R is considered as a left R-module as in example 1) – in this case the matrix e is not in the center of R. Proof. zM is clearly nonempty. Let x = zm 1 , y = zm 2 , r ∈ R so x +ry = zm 1 +rzm 2 = zm 1 +zrm 2 = z(m 1 +rm 2 ) so zM is a submodule of M. If a ∈ eR then a = f 1 f 2 0 0 but 1 1 1 1 f 1 f 2 0 0 = f 1 f 2 f 1 f 2 so eR is not a left R-submodule. 10.1.16. Prove that the submodules U k described in the example of F[x]-modules are all of the F[x]-submodules for the shift operator. Proof. Note since F is a field, then if a component contains a nonzero element then that component of the submodule must contain all of F. But if a component is zero to the left of some nonzero component, then it would not be T-stable. Thus it is clear that the U k s are all of the F[x]-submodules by the bjiection of F[x]-submodules and subspaces invariant under T. 10.1.22. Suppose A is a ring with identity 1 A that is a (unital) left R-module satisfying r (ab) = (r a)b = a(r b) for all r ∈ R and a, b ∈ A. Prove that the map f : R →A defined by f(r) = r 1 A is a ring homomorphism mapping 1 R to 1 A and that f(R) is contained in the center of A. Conclude that A is an R-algebra and that the R-module structure on A induced by its algebra structure is precisely the original R-module structure. Proof. If f(x) = f(y) then x = x 1 A = y 1 A = y so f is well defined. Now f(x +y) = (x +y) 1 A = f(x) +f(y), f(xy) = (xy) 1 A = f(x)f(y) and f(1 R ) = 1 R 1 A = 1 A . Since (r 1 A ) a = (r a) 1 A = a (r 1 A ). Thus by definition A is an R-algebra and the R-module structure on A induced by its algebra structure is precisely the original R-module structure. 54 DAVID S. DUMMIT AND RICHARD M. FOOTE 13. Field Theory 13.1. Basic Theory of Field Extensions. Proposition 13.1. The characteristic of a field F, ch (F), is either 0 or a prime p. If ch (F) = p then for any α ∈ F, p α = α +α + +α = 0. Proposition 13.2. Let ϕ: F →F be a homomorphism of fields. Then ϕ is either identically 0 or is injective, so that the image of ϕ is either 0 or isomorphic to F. Theorem 13.3. Let F be a field and let p(x) ∈ F[x] be an irreducible polynomial. Then there exists a field K containing an isomorphic copy of F in which p(x) has a root. Identifying F with this isomorphic copy shows that there exists an extension of F which p(x) has a root. Theorem 13.4. Let p(x) ∈ F[x] be an irreducible polynomial of degree n over the field F and let K be the field F[x]/(p(x)). Let θ = x mod (p(x)) ∈ K. Then the elements 1, θ, θ 2 , . . . , θ n−1 are a basis for K as a vector space over F, so the degree of the extension is n, i.e., [K : F] = n. Hence K = ¸ a 0 +a 1 θ +a 2 θ 2 + +a n−1 θ n−1 [ a 0 , a 1 , . . . , a n−1 ∈ F ¸ consists of all polynomials of degree < n in θ. Corollary 13.5. Let K be as in theorem 13.4, and let a(θ), b(θ) ∈ K be two polynomials of degree < n in θ. Then addition in K is defined simply by usual polynomial addition and multiplication in K defined by a(θ)b(θ) = r(θ) where r(x) is the remainder (of degree < n) obtained after dividing the polynomial a(x)b(x) by p(x) in F[x]. Theorem 13.6. Let F be a field and let p(x) ∈ F[x] be an irreducible polynomial. Suppose K is an extension field of F containing a root α of p(x): p(α) = 0. Let F(α) denote the subfield of K generated over F by α. Then F(α) ∼ = F[x]/(p(x)). Corollary 13.7. Suppose in theorem 13.6 that p(x) is of degree n. Then F(α) = ¸ a 0 +a 1 α +a 2 α 2 + +a n−1 α n−1 [ a 0 , a 1 , . . . , a n−1 ∈ F ¸ ⊆ K. Theorem 13.8. Let ϕ: F ˜ →F be an isomorphism of fields. Let p(x) ∈ F[x] be an irreducible polynomial and let p (x) ∈ F (x) be the irreducible polynomial obtained by applying the map ϕ to the coefficients of p(x). Let α be a root of p(x) (in some extension of F) and let β be a root of p (x) (in some extension of F ). Then there is an isomorphism σ: F(α) ˜ →F (β) mapping α to β and extending ϕ, i.e., such that σ restricted to F is the isomorphism ϕ. 13.2. Basic Theory of Field Extensions. Proposition 13.9. Let α be algebraic over F. Then there is a unique monic irreducible polynomial m α,F (x) ∈ F[x] which has α as a root. A polynomial f(x) ∈ F[x] has α as a root if and only if m α,F (x) divides f(x) in F[x]. Corollary 13.10. If L ⊂ F is an extension of fields and α is algebraic over both F and L, then m α,L (x) divides m α,F (x) in L[x]. Proposition 13.11. Let α be algebraic over the field F and let F(α) be the field generated by α over F. Then F(α) ∼ = F[x]/(m α (x)) so that in particular [F(α): F] = deg m α (x) = deg α, i.e., the degree of α over F is the degree of the extension it generates over F. Proposition 13.12. The element α is algebraic over F if and only if the simple extension F ⊂ F(α) is finite. More precisely, if α is an element of an extension of degree n over F then α satisfies a polynomial of degree at most n over F and if α satisfies a polynomial of degree n over F then the degree of F(α) over F is at most n. Corollary 13.13. If the extension F ⊂ K is finite, then it is algebraic. 55 Theorem 13.14. Let F ⊆ K ⊆ L be fields. Then [L: F] = [L: K][K: F], i.e., extension degrees are multiplicative, where if one side of the equation is infinite, the other side is also infinite. Corollary 13.15. Suppose F ⊂ L is a finite extension and let K be any subfield of L containing F, F ⊆ K ⊆ L. Then [K: F] divides [L: F]. Lemma 13.16. F(α, β) = (F(α))(β), i.e., the field generated over F by α and β is the field generated by β over the field F(α) generated by α. Theorem 13.17. The extension F ⊂ K is finite if and only if K is generated by a finite number of algebraic elements over F. More precisely, a field generated over F by a finite4 number of algebraic elements of degrees n 1 , n 2 , . . . , n k is algebraic of degree ≤ n 1 n 2 n k . Corollary 13.18. Suppose α and β are algebraic over F. Then α ± β, αβ, α/β (for β = 0), (in particular α −1 for α = 0) are all algebraic. Corollary 13.19. Let F ⊂ L be an arbitrary extension. Then the collection of elements of L that are algebraic over F form a subfield K of L. Theorem 13.20. If K is algebraic over F and L is algebraic over K, then L is algebraic over F. Proposition 13.21. Let K 1 and K 2 be two finite extensions of a field F contained in K. Then [K 1 K 2 : F] ≤ [K 1 : F][K 2 : F] with equality if and only if an F-basis for one of the fields remains linearly independent over the other field. If α 1 , α 2 , . . . , α n and β 1 , β 2 , . . . , β m are bases for K 1 and K 2 over F, respectively, then the elements α i β j for i = 1, . . . , n and j = 1, . . . , m span K 1 K 2 over F. Corollary 13.22. Suppose that [K 1 : F] = n, [K 2 : F] = m in proposition 13.21, where n and m are relatively prime: (n, m) = 1. Then [K 1 K 2 : F] = [K 1 : F][K 2 : F] = nm. 13.2.1. Let F be a finite field of characteristic p. Prove that [F[ = p n for some positive integer n. Proof. Let ϕ: Z →F be the ring map given by ϕ(n) = n 1 F . Since pZ ⊆ ker ϕ then ¯ ϕ: F p →F is well defined. But ¯ ϕ is a ring map of fields so it is trivial or injective. Notice ¯ ϕ(1 Fp ) = 1 F so ¯ ϕ is injective. Thus we can consider F as a vector space over F p . Since F is finite then [F[ = [F p [ [F: Fp] = p n for some n ∈ Z + . 13.2.7. Prove that Q( √ 2+ √ 3) = Q( √ 2, √ 3). Conclude that [Q( √ 2+ √ 3): Q] = 4. Find an irreducible polynomial satisfied by √ 2 + √ 3. Proof. Since Q, √ 2, √ 3 are contained in Q( √ 2, √ 3), then Q, √ 2 + √ 3 are contained in Q( √ 2, √ 3). Hence by definition Q( √ 2 + √ 3) ⊆ Q( √ 2, √ 3). To prove the other inclusion notice that ( √ 2 + √ 3) 3 −9( √ 2 + √ 3) /2 = √ 2. Thus √ 2 ∈ Q( √ 2 + √ 3) thus √ 3 ∈ Q( √ 2 + √ 3). By definition we have Q( √ 2, √ 3) ⊆ Q( √ 2 + √ 3) and thus equality. Since Q( √ 2, √ 3) = Q( √ 2 + √ 3) then [Q( √ 2 + √ 3): Q] = [Q( √ 2, √ 3): Q] = 4. Notice that p( √ 2 + √ 3) = 0 where p(x) = x 4 −10x 2 + 1. If p were to be reducible, then there would be an irreducible polynomial with degree less than 4. This contradicts the minimality of the degree of m √ 2+ √ 3 = 4. Therefore p is irreducible in Q. 13.2.10. Determine the degree of the extension Q( 3 + 2 √ 2) over Q. Solution. First note that [Q( 3 + 2 √ 2): Q] = [Q( 3 + 2 √ 2: Q( √ 2)][Q( √ 2): Q]. But x 2 − 3 + 2 √ 2 = (x − 1 + 2 √ 2)(x + 1 + 2 √ 2) so [Q( 3 + 2 √ 2): Q( √ 2)] = 1. Thus [Q( 3 + 2 √ 2): Q] = 2. 13.2.13. Suppose F = Q(α 1 , . . . , α n ) where α 2 i ∈ Q for i = 1, . . . , n. Prove that 3 √ 2 / ∈ F. Proof. Since [Q(α 1 , . . . , α k ): Q(α 1 , . . . , α k−1 )] ∈ ¦1, 2¦, then [Q(α 1 , . . . , α n ): Q] = 2 l for some l ∈ Z + . Now if 3 √ 2 ∈ F then Q ⊂ Q( 3 √ 2) ⊆ F then [Q( 3 √ 2: Q)] [ [F : Q]. Since 3 does not divide 2 l for l ∈ Z + then 3 √ 2 / ∈ F. 56 DAVID S. DUMMIT AND RICHARD M. FOOTE 13.2.16. Let F ⊂ K be an algebraic extension and let R be a ring contained in K and containing F. Show that R is a subfield of K containing F. Proof. Let α ∈ R` ¦0¦ so since α ∈ F then there is some irreducible p ∈ R[x] such that p(α) = 0. Let p(x) be given by p(x) = a n x n + +a 0 , and notice since p is irreducible, a 0 = 0. Since p(α) = 0 we have α −1 = −a −1 0 (a n α n−1 + +a 1 ). Since a i ∈ F ⊆ R and α ∈ R then α −1 ∈ R and R is a field. 13.3. Classical Straightedge and Compass Constructions. Proposition 13.23. If the element α ∈ 1 is obtained from a field F ⊆ 1 by a series of compass and straightedge constructions then [F(α): F] = 2 k for some integer k ≥ 0. Theorem 13.24. None of the classical Greek problems: (I) Doubling the Cube, (II) Trisecting an Angle, and (III) Squaring the Circle, is possible. 13.4. Splitting Fields and Algebraic Closures. Theorem 13.25. For any field F, if f(x) ∈ F[x] then there exists an extension K of F which is a splitting field for f(x). Proposition 13.26. A splitting field of a polynomial of degree n over F is of degree at most n! over F. Theorem 13.27. Let ϕ: F ˜ →F be an isomorphism of fields. Let f(x) ∈ F[x] be a polynomial and let f (x) ∈ F [x] be the polynomial obtained by applying ϕ to the coefficients of f(x). Let E be a splitting field for f(x) over F and let E be a splitting field for f (x) over F . Then the isomorphism ϕ extends to an isomorphism σ: E ˜ →E , i.e., σ restricted to F is the isomorphism ϕ. Corollary 13.28 (Uniqueness of Splitting Fields). Any two splitting fields for a polynomial f(x) ∈ F[x] over a field F are isomorphic. Proposition 13.29. Let F be an algebraic closure of F. Then F is algebraically closed. Proposition 13.30. For any field F there exists an algebraically closed field K containing F. Proposition 13.31. Let K be an algebraically closed field and let F be a subfield of K. Then the collection of elements F of K that are algebraic over F is an algebraic closure of F. An algebraic closure of F is unique up to isomorphism. Theorem (Fundamental Theorem of Algebra). The field C is algebraically closed. Corollary 13.32. The field C contains an algebraic closure for any of its subfields. In particular, Q, the collection of complex numbers algebraic over Q, is an algebraic closure of Q. 13.5. Separable and Inseparable Extensions. Proposition 13.33. A polynomial f(x) has a multpile root α if and only if α is also a root of D x f(x), i.e., f(x) and D x f(x) are both divisible by the minimal polynomial for α. In particular, f(x) is separable if and only if it is relatively prime to its derivative: (f(x), D x f(x)) = 1. Corollary 13.34. Every irreducible polynomial over a field of characteristic 0 is separable. A polynomial over such a field is separable if and only if it is the product of distinct irreducible polynomials. Proposition 13.35. Let F be a field of characteristic p. Then for any a, b ∈ F, (a +p) p = a p +b p , and (ab) p = a p b p . Put another way, the p th -power map defined by ϕ(a) = a p is an injective field homomorphism from F to F. Corollary 13.36. Suppose that F is a finite field of characteristic p. Then every element of F is a p th power in F, i.e., F = F p . Proposition 13.37. Every irreducible polynomial over a finite field F is separable. A polynomial in F[x] is separable if and only if it is the product of distinct irreducible polynomials in F[x]. 57 Proposition 13.38. Let p(x) be an irreducible polynomial over a field F of characteristic p. Then there is a unique integer k ≥ 0 and a unique irreducible separable polynomial p sep (x) ∈ F[x] such that p(x) = p sep (x p k ). Corollary 13.39. Every finite extension of a perfect field is separable. In particular, every finite extension of either Q or a finite field is separable. 13.6. Cyclotomic Polynomials and Extensions. Lemma 13.40. The cyclotomic polynomial Φ n (x) is a monic polynomial in Z[x] of degree ϕ(n). Theorem 13.41. The cyclotomic polynomial Φ n (x) is an irreducible monic polynomial in Z[x] of degree ϕ(n). Corollary 13.42. The degree over Q of the cyclotomic field of n th roots of unity is ϕ(n): [Q(ζ n ): Q] = ϕ(n). 14. Galois Theory 14.1. Basic Definitions. Proposition 14.1. Aut(K) is a group under composition and Aut(K/F) is a subgroup. Proposition 14.2. Let K/F be a field extension and let α ∈ K be algebraic over F. Then for any σ ∈ Aut(K/F), σα is a root of the minimal polynomial for α over F i.e., Aut(K/F) permutes the roots of irreducible polynomials. Equivalently, any polynomial with coefficients in F having α as a root also has σα as a root. Proposition 14.3. Let H ≤ Aut(K) be a subgroup of the group of automorphisms of K. Then the collection F of elements of K fixed by all the elements of H is a subfield of K. Proposition 14.4. The association of groups to fields and fields to groups defined above is inclusion reversing, namely (1) if F 1 ⊆ F 2 ⊆ K are two subfields of K then Aut(K/F 2 ) ≤ Aut(K/F 1 ), and (2) if H 1 ≤ H 2 ≤ Aut(K) are two subgroups of automorphisms with associated fixed fields F 1 and F 2 , respec- tively, then F 2 ⊆ F 1 . Proposition 14.5. Let E be the splitting field over F of the polynomial f(x) ∈ F[x]. Then [Aut(E/F)[ ≤ [E: F] with equality if f(x) is separable over F. Corollary 14.6. If K is the splitting field over F of a separable polynomial f(x) then K/F is Galois. 14.2. The Fundamental Theorem of Galois Theory. Theorem 14.7 (Linear Independence of Characters). If χ 1 , . . . , χ n are distinct characters of G with values in L then they are linearly independent over L. Corollary 14.8. If σ 1 , . . . , σ n are distinct embeddings of a field K into a field L, then they are linearly independent as functions on K. In particular distinct automorphisms of a field K are linearly independent as functions on K. Theorem 14.9. Let G = ¦σ 1 = 1, σ 2 , . . . , σ n ¦ be a subgroup of automorphisms of a field K and let F be the fixed field. Then [K: F] = n = [G[ . Corollary 14.10. Let K/F be any finite extension. Then [Aut(K/F)[ ≤ [K: F] with equality if and only if F is the fixed field of Aut(K/F). Put another way, K/F is Galois if and only if F is the fixed field of Aut(K/F). Corollary 14.11. Let G be a finite subgroup of automorphisms of a field K and let F be the fixed field. Then every automorphism of K fixing F is contained in G, i.e., Aut(K/F) = G, so that K/F is Galois, with Galois group G. Corollary 14.12. If G 1 = G 2 are distinct finite subgroups of automorphisms of a field K then their fixed fields are also distinct. Theorem 14.13. The extension K/F is Galois if and only if K is the splitting field of some separable polynomial over F. Furthermore, if this is the case then every irreducible polynomial with coeffiecients in F which has a root in K is separable and has all its roots in K (so in particular K/F is a separable extension). Theorem 14.14 (Fundamental Theorem of Galois Theory). 58 DAVID S. DUMMIT AND RICHARD M. FOOTE 15. Commutative Rings and Algebraic Geometry 15.1. Noetherian Rings and Affine Algebraic Sets. 15.2. Radicals and Affine Varieties. 15.3. Integral Extensions and Hilbert’s Nullstellensatz. 15.4. Localization. Theorem 15.36 (cf. theorem 7.15). let R be a commutative ring with 1 = 0 and let D be a multiplicatively closed subset of R containing 1. Then there is a commutative ring D −1 R and a ring homomorphism π: R → D −1 R satisfying the following universal property: for any homomorphism ψ: R → S of commutative rings that sends 1 to 1 such that ψ(d) is a unit in S for every d ∈ D, there is a unique homomorphism Ψ: D −1 R → S such that Ψ◦ π = ψ. Corollary 15.37. In the notation of theorem 15.36, (1) ker π = ¦r ∈ R [ xr = 0 for some x ∈ D¦; in particular, π: R → D −1 R is an injetion if and only if D contains neither zero nor any zero divisors of R, and (2) D −1 R = 0 if and only if 0 ∈ D, hence if and only if D contains nilpotent elements. 15.4.18. Let R be any commutative ring with 1 and let f be any element of R. Let D be the multiplicative set ¦f n [ n ≥ 0¦ of nonnegative powers of f in R. Define R f = D −1 R. Note that R f = 0 if and only if f is nilpotent. If f is not nilpotent, then f becomes a unit in R f . Prove that R f ∼ = R[x]/(fx − 1) where R[x] is the polynomial ring in the variable x, if f is not nilpotent in R. Proof. 15.4.21. Suppose ϕ: R → S is a ring homomorphism with ϕ(1 R ) = 1 S and D is a multiplicatively closed subset of S. Let D = ϕ −1 (D ). Prove D is a multiplicatively closed subset of R and the map ϕ : D −1 R → D −1 S given by ϕ (r/d) = ϕ(r)/ϕ(d) is a ring homomorphism. Proof. Let d 1 , d 2 ∈ D with ϕ(d 1 d 2 ) = ϕ(d 1 )ϕ(d 2 ) so d 1 d 2 ∈ ϕ −1 (D ) because D is a multiplicative set. Let r1 d1 = r2 d2 . Then r 1 d 2 = r 2 d 1 so ϕ ( r 1 d 1 ) = ϕ ( r 1 r 2 d 2 r 2 d 1 d 2 ) = ϕ(r 1 r 2 d 2 ) ϕ(r 2 d 1 d 2 ) = ϕ(r 2 d 1 )ϕ(r 2 ) ϕ(r 2 d 1 )ϕ(d 2 ) = ϕ ( r 2 d 2 ) 59 So ϕ is well defined. Now let r1 d1 , r2 d2 ∈ D −1 R. Then ϕ ( r 1 d 1 + r 2 d 2 ) = ϕ ( r 1 d 2 +r 2 d 1 d 1 d 2 ) = ϕ(r 1 d 2 +r 2 d 1 ) ϕ(d 1 d 2 ) = ϕ(r 1 ) ϕ(d 1 ) + ϕ(r 2 ) ϕ(d 2 ) = ϕ ( r 1 d 1 ) +ϕ ( r 2 d 2 ) ϕ ( r 1 d 1 r 2 d 2 ) = ϕ ( r 1 r 2 d 1 d 2 ) = ϕ(r 1 r 2 ) ϕ(d 1 d 2 ) = ϕ(r 1 ) ϕ(d 1 ) ϕ(r 2 ) ϕ(d 2 ) = ϕ ( r 1 d 1 )ϕ ( r 2 d 2 ). 15.4.22. Suppose P ⊆ Q are prime ideals in R and let R Q be the localization of R at Q. Prove that the localization R P is isomorphic to the localization of R Q at the prime ideal P R Q (cf. exercise 15.4.21). Proof. 2 DAVID S. DUMMIT AND RICHARD M. FOOTE 4.4. Automorphisms 4.5. The Sylow Theorems 4.6. The Simplicity of An 5. Direct and Semidirect Products and Abelian Groups 5.1. Direct Products 5.2. The Fundamental Theorem of Finitely Generated Abelian Groups 5.3. Table of Groups of Small Order 5.4. Recognizing Direct Products 5.5. Semidirect Products 6. Further Topics in Group Theory 6.1. p-groups, Nilpotent Groups, and Solvable Groups Part II – Ring Theory 7. Introduction to Rings 7.1. Basic Definitions and Examples 7.2. Examples: Polynomials Rings, Matrix Rings, and Group Rings 7.3. Ring Homomorphisms and Quotient Rings 7.4. Properties of Ideals 7.5. Rings of Fractions 8. Euclidean, Principal Ideal, and Unique Factorization Domains 8.1. Euclidean Domains 9. Polynomial Rings 9.1. Definitions and Basic Properties 9.2. Polynomial Rings Over Fields I 9.3. Polynomial Rings That are Unique Factorization Domains 9.4. Irreducibility Criteria 9.5. Polynomial Rings Over Fields II 9.6. Polynomials in Several Variables Over a Field and Gr¨bner Bases o Part III – Modules and Vector Spaces 10. Introduction to Module Theory 10.1. Basic Definitions and Examples 13. Field Theory 13.1. Basic Theory of Field Extensions 13.2. Basic Theory of Field Extensions 13.3. Classical Straightedge and Compass Constructions 13.4. Splitting Fields and Algebraic Closures 13.5. Separable and Inseparable Extensions 13.6. Cyclotomic Polynomials and Extensions 14. Galois Theory 14.1. Basic Definitions 14.2. The Fundamental Theorem of Galois Theory 15. Commutative Rings and Algebraic Geometry 15.1. Noetherian Rings and Affine Algebraic Sets 15.2. Radicals and Affine Varieties 15.3. Integral Extensions and Hilbert’s Nullstellensatz 15.4. Localization 26 26 27 28 28 29 30 30 30 31 31 33 33 33 34 36 39 42 45 45 46 46 47 48 48 50 50 53 53 53 54 54 54 56 56 56 57 57 57 57 58 58 58 58 58 3 0. Preliminaries 0.1. Basics. Proposition 0.1. Let f : A → B. (1) The map f is injective if and only if f has a left inverse. (2) The map f is surjective if and only if f has a right inverse. (3) The map f is a bijection if and only if there exists g : B → A such that f ◦ g is the identity map on B and g ◦ f is the identity map on A. (4) If A and B are finite sets with the same number of elements (|A| = |B|), then f : A → B is bijective if and only if f is injective if and only if f is surjective. Proof. (a) Suppose f is injective so that f −1 (b) contains a single element for b ∈ f (A). Thus g(b) = f −1 (b) is well-defined for b ∈ f (A). Hence g(f (a)) = a for all a ∈ A. Now suppose that f has a left inverse g so that g(f (a)) = a for all a ∈ A. If f (a1 ) = f (a2 ) = b, then g(b) = a1 and g(b) = a2 . But g is a well-defined map so a1 = a2 and thus f is injective. (b) Suppose f is surjective. Take some b ∈ B so we can choose a ∈ A such that f (a) = b. Thus we can define g(b) = a such that f (a) = b. Hence f (g(b)) = f (a) = b. Now suppose that f has a right inverse g so that f (g(b)) = b for all b ∈ B. So f (g(B)) = B and thus f is surjective. (c) Suppose that f is bijective so there is a left inverse, gl , and a right inverse, gr , of f . Let f (a) = b so that a = gl (f (a)) = gl (b) and b = f (gr (b)). Since f is injective, gr (b) = a and thus gl = gr . Thus the right and left inverses of f are the same map. Now suppose that there exists g : B → A such that f ◦ g is the identity map on B and g ◦ f is the identity map on A. Then f is surjective because f (g(B)) = B. Now let f (a1 ) = f (a2 ) so that a1 = g(f (a1 )) = g(f (a2 )) = a2 . Thus f is injective. (d) Suppose that f is injective. Then f takes elements of A to unique elements of B. Since |A| = |B|, f is surjective. Now suppose that f is surjective. Then since |A| = |B|, f takes elements of A to unique elements of B so f is injective. Therefore f is either bijective or neither injective or surjective. Proposition 0.2. Let A be a nonepty set. (1) If ∼ defines an equivalence relation on A then the set of equivalence classes of ∼ form a partition of A. (2) If {Ai | i ∈ I} is a partition of A then there is an equivalence relation on A whose equivalence classes are preciselly the sets Ai , i ∈ I. In Exercises 1 to 4 let A be the set of 2 × 2 matrices with real number entries. Recall that matrix multiplication is defined by a b p q ap + br aq + bs = c d r s cp + dr cq + ds Let M= and let B = {X ∈ A | M X = XM } . 0.1.1. The following elements of A lie in B: 1 0 1 0 , 1 0 0 1 , 0 0 0 . 1 1 0 1 1 0.1.2. Prove that if P, Q ∈ B, then P + Q ∈ B (where + denotes the usual sum of two matrices). Proof. (P + Q)M = P M + QM = M P + M Q = M (P + Q). 0.1.3. Prove that if P, Q ∈ B, then P · Q ∈ B (where · denotes the usual product of two matrices). Proof. (P · Q)M = P · M · Q = M · P · Q = M (P · Q). 4 DAVID S. DUMMIT AND RICHARD M. FOOTE 0.1.4. Find conditions on p, q, r, s which determine precisely when r = 0, p = s. 0.1.5. Determine whether the following functions f are well defined: (1) f : Q → Z defined by f (a/b) = a. (2) f : Q → Q defined by f (a/b) = a2 /b2 . p r q s ∈ B. Proof. (a) f is not a well defined map because 1/2 = 2/4 and 1 = 2. (b) f is a well defined map because if a1 /b1 = a2 /b2 , then a2 /b2 = a2 /b2 . 1 1 2 2 0.1.6. Determine whether the function f : R+ → Z defined by mapping a real number r to the first digit to the right of the decimal point in a decimal expansion of r is well defined. f is well defined because the decimal expansion of a real number is unique. Thus there is one and only one possible first number to the right of the decimal for each real number. 0.1.7. Let f : A → B be a surjective map of sets. Prove that the relation a ∼ b if and only if f (a) = f (b) is an equivalence relation whose equivalence classes are the fibers of f . Proof. f (a) = f (a) so ∼ is reflexive. If f (a) = f (b), then f (b) = f (a) so ∼ is symmetric. If f (a) = f (b) and f (b) = f (c), then f (a) = f (c) so ∼ is transitive. If a1 , a2 ∈ f −1 (b), then f (a1 ) = f (a2 ) so a1 ∼ a2 . If a1 ∼ a2 then f (a1 ) = f (a2 ) so a1 and a2 are in the same fiber of f . 0.2. Properties of the Integers. (1) (Well Ordering of Z) If A is any nonempty subset of Z+ , there is some element m ∈ A such that m ≤ a, for all a ∈ A. (2) If a, b ∈ Z with a = 0, we say a divides b if there is an element c ∈ Z such that b = ac. In this case we write a | b; if a does not divide b we write a b. (3) If a, b ∈ Z\ {0}, there is a unique positive integer d, called the greatest common divisor of a and b, satisfying: (a) d | a and d | b, (b) if e | a and e | b, then e | d. (4) If a, b ∈ Z\ {0}, there is a unique positive integer l, called the least common multiple of a and B, satisfying: (a) a | l and b | l, (b) if a | m and b | m, then l | m. The connection between the greatest common divisor d and the least common multiple l of the two integers a and b is given by dl = ab. (5) The Division Algorithm: if a, b ∈ Z and b = 0, then there exist unique q, r ∈ Z such taht a = qb + r and 0 ≤ r < |b| , where q is the quotient and r the remainder. (6) The Euclidean Algorithm is an important procedure which produces a greatest common divisor of two integers a and b by iterating the Division Algorithm: if a, b ∈ Z\ {0}, then we obtain a sequence of quotients and remainders a = q0 b + r0 b = q1 r0 + r1 r0 = q2 r1 + r2 r1 = q3 r2 + r3 . . . rn−2 = qn rn−1 + rn rn−1 = qn+1 rn where rn is the last nonzero remainder. Such an rn exists since |b| > |r0 | > |r1 | > · · · > |rn | is a decreasing sequence of strictly positive integers if the remainders are nonzero and such a sequence cannot continue indefinitely. Then rn = (a, b). 5 (7) One consequence of the Euclidean Algorithm which we chall use regularly is the following: if a, b ∈ Z\ {0}, then there exist x, y ∈ Z such that (a, b) = ax + by that is, the gcd of a and b is a Z-linear combination of a and b. This follows by recursively writing the element rn in the Euclidean Algorithm in terms of the previous remainders. (8) An element p of Z+ is calle a prime if p > 1 and the only posotive divisors of p are 1 and p. An integer n > 1 which is not prime is called composite. An important property of primes is if p is a prime and p | ab, for some a, b ∈ Z then either p | a or p | b. (9) The fundamental theorem of arithmetic says: if n ∈ Z, n > 1, then n can be factored uniquely into the product of primes, i.e., there are distinct primes p1 , p2 , . . . , ps and positive integers α1 , α2 , . . . , αs such that n = pα1 pα2 · · · pαs . s 1 2 This factorization is unique in the sense that if q1 , q2 , . . . , qt are any distinct primes and β1 , β2 , . . . , βt , are positive integers such that then s = t and if we arrange the two sets of primes in increasing order, then qi = pi and αi = βi , 1 ≤ i ≤ s. Suppose the positive integers a and b are expressed as products of prime powers: a = pα1 pα2 · · · pαs , s 1 2 (a, b) = p1 b = pβ1 pβ2 · · · pβs s 1 2 · · · pmin(αs ,βs ) s β β β n = q1 1 q2 2 · · · qt t , where p1 , p2 , . . . , ps are distinct and the exponents are ≥ 0. Then the gcd of a and b is min(α1 ,β1 ) min(α2 ,β2 ) p2 and the lcm is obtained by taking the maximum of the αi and βi instead of the minimum. (10) The Euler ϕ-function is defined as follows: for n ∈ Z+ let ϕ(n) be the number of positive integers a ≤ n with a relatively prime to n, i.e., (a, n) = 1. For primes p, ϕ(p) = p − 1, and, more generally, for all a ≥ 1 we have the formula ϕ(pa ) = pa − pa−1 = pa−1 (p − 1). The function ϕ is multiplicative in the sense that ϕ(ab) = ϕ(a)ϕ(b) if (a, b) = 1. Together with the formula above this gives a general formula for the values of ϕ: if n = pα1 pα2 · · · pαs , then s 1 2 ϕ(n) = ϕ(pα1 )ϕ(pα2 ) · · · ϕ(pαs ) s 1 2 α α = pα1 −1 (p1 − 1)p2 2 −1 (p2 − 1) · · · ps s −1 (ps − 1). 1 0.2.1. For each of the following paris of integers a and b, determine their gcd, their lcm , and write their gcd in the form ax + by for some integers x and y. (a) (b) (c) (d) (e) (f) a = 20, b = 13. a = 69, b = 372. a = 792, b = 275. a = 11391, b = 5673. a = 1761, b = 1567. a = 507885, b = 60808. Solution. 0.2.2. Prove that if the integer k divides the integers a and b then k divides as + bt for every pair of integers s and t. Proof. 0.2.3. Prove that if n is composite then there are integers a and b such that n divides ab but n does not divide either a or b. Proof. 1.4.2. a2 ∈ Z and b1 . 0. 0. If p is a prime prove that there do not exist nonzero integers a and b such that a2 = pb2 . 0. Determine the value ϕ(n) for each integer n ≤ 30 where ϕ denotes the Euler ϕ-function.2. Let a. . 2. Z/nZ) = {a ∈ Z/nZ | (a. b) in the form ax + by for some integers x and y. Prove that if a = an 10n + an−1 + · · · + a1 10 + a0 is any positive integer then a ≡ an + an−1 + · · · + a1 + a0 n−1 mod 9. Find a formula for the largest power of p which divides n! = n(n−1)(n−2) · · · 2·1. 0.5.9. Proof. Prove for any given positive integer N there exist only finitely many integers n with ϕ(n) = N where ϕ denotes Euler’s ϕ-function. Prove that the distinct equivalence classes in Z/nZ are precisely 0. 0. Write down explicitly all the elements in the residue classes of Z/18Z. then a1 + a2 = b1 + b2 and a1 a2 = b1 b2 .3.3.3. Solution. Proof. 0. Compute the last two digits of 91500 . i.2.3.2.6. 0.3. that is.2.3. n ∈ Z+ . n) = 1}. . Prove that if d divides n then ϕ(d) divides ϕ(n) where ϕ denotes Euler’s ϕ-function.2. Solution. Proof. b) of two integers a and b and to express (a.e. Proof. 1. if a1 . 0. b and N be fixed integers with a and b nonzero and let d = (a. Proof. b2 ∈ Z with a1 = b1 and a2 = b2 . they do not depend on the choices of representatives for the classes involved. Let p be a prime. Proof. b) be the gcd of a and b. . Solution. FOOTE 0.10.2. .2. Conclude in particular that ϕ(n) tends to infinity as n tends to infinity.4. Prove for any integer t that the integers b a x = x0 + t and y = y0 − t d d are also solutions to ax + by = N (this is in fact the general solution). a1 + a2 ≡ b1 + b2 mod n and a1 a2 ≡ b1 b2 Proposition 0. 0.3. 0.2. 0.7. 0. Compute the remainder when 37100 is divided by 29.6 DAVID S. DUMMIT AND RICHARD M.6. Theorem 0. Solution. 0. Prove that the squares of the elements in Z/4Z are just 0 and 1. n − 1.3. The operations of addition and multiplication on Z/nZ are well defined.. Proof. Solution.8. More precisely. if then × a1 ≡ b1 mod n and a2 ≡ b2 mod n mod n. Prove the Well Ordering Property of Z by induction and prove the minimal element is unique.4. Write a computer program to determine the greatest common divisor (a.3. Z/nZ: The Integers Modulo n. .5. Suppose x0 and y0 are particular solutions to ax+ by = N .11. 0. and let a ∈ Z with 1 ≤ a ≤ n. Proof. Solution. 0. (a) a = 13. Prove for any integers a and b that a2 + b2 never leaves a remainder of 3 when divided by 4.7 Proof.11.3.14. Let n ∈ Z. . n = 20. n) = 1.3. b ∈ (Z/nZ)× . for any n given as input. b.16. The output of these operations should be the least residues of the usms and products of two integers. Prove that if a.3. n = 3797. Proof. (b) a = 69. show that a is relatively prime to n and determine the multiplicative inverse of a in Z/nZ. and integer c between 1 and n − 1 such taht a · c = 1 may be printed on request. Proof.3. 0. Prove if a and n are not relatively prime then there exists an itneger b with 1 ≤ b < n such that ab ≡ 0 mod n and deduce that there cannot be an itneger c such that ac ≡ 1 mod n. (c) a = 1891.8.13. For each of the following pairs of integers a and n. and let z ∈ Z with 1 ≤ a ≤ n. Proof. 0. Let n ∈ Z. Prove that the square of any odd integer always leaves a remainder of 1 when divided by 8. Proof.3. 0.12. 0. Prove that the number of elements of (Z/nZ)× is ϕ(n) where ϕ denotes the Euler ϕ-function. Prove that if a and n are relatively prime then there is an integer c such that ac ≡ 1 mod n. n = 77695236973. n) = 1 and hence prove proposition 0.3. 0.3. (d) a = 6003722857.10. Verify this directly in the case n = 12. n > 1.7. Write a computer program to add and multiply mod n.9. 0. Proof. Prove that for any integers a and b that a2 + b2 = 3c2 has no solutions in nonzero integers a. n = 89. n > 1. Proof. Proof.3. then a · b ∈ (Z/nZ)× .15. Conclude from the previous two exercises that (Z/nZ)× is the set of elements a of Z/nZ with (a. Also include the feature that if (a. 0.3. 0. 0. and c.4.3. bd) = (cb + da. Let G be a group and let a. Determine which of the following binary operations are associative: (a) the operation on Z defined by a b = a − b (b) the operation on R defined by a b = a + b + ab (d) the operation on Z × Z defined by (a.4.8 DAVID S.2. b) (c. db) = (c. f ) = (ad + bc)f + bde.1.1. then (1) the identity of G is unique (2) for each a ∈ G. b) (c. If G is a group under the operation . d) (e. the left and right cancellation laws hold in G. d) = (ad + bc. b) (c. Proof. . a2 . . y ∈ G. 1. b ∈ G. Note that 1 − 2 = −1 and 2 − 1 = 1 so (a) is not commutative. bd). We calculate that (b) and (d) are commutative below. Proposition 1. bdf = (a. d) = (ad + bc. so (a) is not associative. i. DUMMIT AND RICHARD M. Introduction to Groups 1. b) (c.1. (a b) c = (a + b + ab) + c + (a + b + ab)c = a + b + c + ab + ac + bc + abc = a + b + c + bc + a(b + c + bc) = a (b c) (a. . FOOTE Part I – Group Theory 1.1. Let G be a group. then u = v. b) 1. bdf = adf + b(cf + de). d) (a. . Basic Axioms and Examples. We calculate that (b) and (d) are associative below. a−1 is uniquely determined (3) (a−1 )−1 = a for all a ∈ G (4) (a b)−1 = (b−1 ) (a−1 ) (5) for any a1 . b) (c. (1) if au = av.2.1. Proposition 1. then u = v. Prove that multiplication of residue classes in Z/nZ is associative. f ) 1.. an ∈ G the value of a1 a2 · · · an is independent of how the expression is bracketed. The equations ax = b and ya = b have unique solutions for x. d) (e. bd). (2) if ub = vb. . Decide which of the following binary operations are commutative: (a) the operation on Z defined by a b = a − b (b) the operation on R defined by a b = a + b + ab (d) the operation on Z × Z defined by (a. Proof. Prove that addition of residue classes in Z/nZ is associative.1.e. a b = a + b + ab = b + a + ba =b a (a.3. d) = (ad + bc. In particular.1. 1. Note that (1 − 2) − 3 = −4 and 1 − (2 − 3) = 2. e. Let G = {x ∈ R | 0 ≤ x < 1} and for x.1. Note that 1/6 + 1/6 = 1/3 so addition is not a binary relation on (b) and thus (b) is not a group. Proof. 13.10.8.1.1. (a) Prove that G is a gorup under multiplication. The inverse of x ∈ G would be given by 1 − x for x = 0 because Note that 1 − x ∈ G for x ∈ G and x = 0.14. Prove that a finite gorup is abelian if and only if its group table is a symmetric matrix. We show that G is abelian by letting x. 5 = 6. 1. such that 0 ≤ x < 1.9. Prove that if x2 = 1 for all x ∈ G then G is abelian. −13 = 6. √ 1. −13. the identity is 0. Find the orders of the following elements of the multiplicative group (Z/36Z)× : 1.1. x.1.9 1.1. x y = x + y − [x + y] = y + x − [y + x] =y x x (1 − x) = x + 1 − x − [x + 1 − x] = 1 − 1 = 0. b ∈ Q . Prove that is a well defined binary operation on G and that G is an abelian group under (called the real numbers mod 1). Proof (b). 5. 1. Prove for all n > 1 that Z/nZ is not a group under multiplication of residue classes. We notice that addition is a well defined binary relation on (e).5. (a) Prove that G is a gorup under addition. 1. Let G = {z ∈ C | z n = 1 for some n ∈ Z+ }. 1 = 1.6.7. Proof.25.1. The inverse of 0 is clearly 0. Therefore (e) is a group. Thus well defined operation that takes x y into G. Proof (b). Let G = a + b 2 ∈ R | a. We show that is associative on G. y ∈ G and computing that is commutative on G. 1. x y = x+y −[x+y] where [a] is the greatest integer less than or equal to a). −1 = 2. y ∈ G let x y be the fractional part of x+y (i. (x y) z = (x + y − [x + y]) + z − [(x + y − [x + y]) + z] = x + y + z − [x + y] + [x + y] − [x + y + z] = x + y + z − [y + z] + [y + z] − [x + y + z] = x (y z) is a = x + (y + z − [y + z]) − [x + (y + z − [y + z])] The identity element of G is clearly 0. 13 = 3. 1. . addition is associative on the rational numbers so this subset must also have the associative property. 17. and the inverse of a is −a. −1. 1. (b) Prove that G is not a group under addition. Proof (a). Proof (a). Also note that −3/2 + 1 = −1/2 so addition is not a binary relation on (d) and thus (d) is note a group. Proof. Determine which of the following sets are groups under addition: (b) the set of rational numbers in lowest terms whose denominators are even together with 0 (d) the set of rational numbers of absolute value ≥ 1 together with 0 (e) the set of rational numbers with denominators equal to 1 or 2.1. (b) Prove that the nonzero elements of G are a group under multiplication. Note that the sum of two real numbers has a unique fractional part. 17 = 2.. Let G be the group of rigid motions in R3 of a tetrahedron. In these exercises. (b) Show that if (3.3. r2 = 3. you can find the number of places to which a given face may be sent and. Proof. once a face is fixed.2.7. [Use the facts that xn = 1 and x3 = 1. and (ab)n = (ssr)n = rn = 1. Dihedral Groups. and it has the same generators and relations as D6 when x is replaced by r and y by s. |r| = 3. b | a2 = b2 = (ab)n = 1 gives a presentation for D2n in terms of the two generators a = s and b = sr of order 2 computed in Exercise 1.9 you can find the order of the group of rigid motions in R3 (also called the group of rotations) of the given Platonic solid by following the proof for the order of D2n : find the number of positions to which an adjacent pair of vertices can be sent. the number of positions to which a vertex on that face may be sent. each face having 3 distinct rotations.1.10 DAVID S. We use the second method described above to show that |G| = 12. there are 4 · 3 = 12 positions to which a vertex on each fixed face may be sent. Notice that sr0 sr0 = 1 and suppose that sri sri = 1 for 0 ≤ i < k.2.2. In this case deduce that X2n has order 2. 1.3 shows b2 = 1. Since x2 = 1 we know that x = x−1 for all x ∈ G. both of which have order 2. [Show that the relations for r and s follow from the relations for a and b and.3 above. First we start with the relations rn = s2 = 1 and rs = sr−1 . conversely. In Exercise 1.2. Let X2n = x. rs = sr−1 . and rs = ssrsrr−1 = sr−1 . FOOTE Proof. Proof. xy = yx2 . If x ∈ D2n is not a power of r.2. Show that a. 1. then rx = xr−1 .2. the relations for a and b follow from those for r and s. Alternatively. then x = srk for some 0 ≤ k < n.2. Exercise 1. Since there are 4 faces of a tetrahedron. rn = (ssr)n = (ab)n = 1. 1. Use the generators and relations above to show that if x is any element of D2n which is not a power of r. 1. Show that |G| = 12. sr2 = 2. 1.2. y | xn = y 2 = 1. then x satisfies the additional relation: x = 1. |sr| = 2. 1.2. Then srk srk = srk−1 sr−1 rk = srk−1 srk−1 = 1. Deduce that D2n is generated by the two elements s and sr. s | rn = s2 = 1. If x is not a power of r. (a) Show that if n = 3k. by induction. 1. Thus xy = (xy)−1 = y −1 x−1 = yx. Compute the order of each of the elements in the following groups: (a) D6 The order of elements from (a) are: |1| = 1. D2n has the usual presentation D2n = r.] . n) = 1. |s| = 2.] Proof. Clearly a2 = 1. Proof. Use the generators and relations above to show that every element of D2n which is not a power of r has order 2.2.17. DUMMIT AND RICHARD M.9. Then clearly s2 = 1. Now suppose that a2 = b2 = (ab)n = 1. then x = srk for some 0 ≤ k < n. Hence the original relations imply the new relations.2. So rsrk = sr−1 rk = srk−1 = srk r−1 . then X2k has order 6. |(34)| = 2. x2 x3 = x−1 x2 2 3 3 1. So x = x3s+nt = (x3 )s (xn )t = 1. Show that σ i is also an m-cycle if and only if i is relatively prime to m.3. But then σ = σ si+tm = σ si σ tm = (σ i )s . (13). |(13)| = 2. Symmetric Groups. c d 1. S3 = x2 . x3 | x2 = x3 = 1. 1.8. The relations xy = yx2 and x3 = 1 show us xy = yx2 = yx−1 . We count the number of non-invertible matrices a b for a matrix in Figure 1.11. |(13)(24)| = 2. αt = m. |(243)| = 3. Then clearly x3 = y 2 = 1. |(132)| = 3. |(123)| = 3. Then clearly x3k = (x3 )k = y 2 = 1. αm−α+1 ) · · · (aα a2α .4.4.3. Thus X2n has order 2. m).3. namely X2n = {1. For elements in S4 we calculate that |1| = 1. . Proof. You may use the fact that a 2 × 2 matrix is not invertible if and only if one row is a multiple of the other. Let p be a prime. Let σ be the m-cycle (1 2 . 1. there exist s. |(1324)| = 4. then σ i must be an m-cycle as well. (23). Proof (b). |(23)| = 2. |(143)| = 3. Now suppose we have the relations given by X2n . |(13)| = 2. Now suppose that i is relatively prime to m so that there are s.3.1 to be p3 + p2 − p. 1. |(124)| = 3. (14). |(12)| = 2. 1. |(234)| = 3. |(1243)| = 4. . 1. (13)(24). . am )]s which can only be an m-cycle if α = 1. |(1234)| = 4.] Proof.20. |(14)| = 2. . Notice that the total number of 2 × 2 matrices over Fp is p4 . Prove that the order of GL2 (Fp ) is p4 − p3 − p2 + p (do not just quote the order formula in this section). Thus the order of GL2 (Fp ) is p4 − p3 − p2 + p. |(1342)| = 4. y}. |(24)| = 2. n) = 1.7. Find a set of generators and relations for S3 .7. |(134)| = 3. |(142)| = 3. . [Subtract the number of 2 × 2 matrices which are not invertible from the total number of 2 × 2 matrices over Fp . (34).11 Proof (a). So σ i = σ αs = [(a1 aα+1 . |(23)| = 2. Since a power of σ i is an m-cycle. t ∈ Z such that si + tm = 1. Since (3. Show that GLn (F ) is non-abelian for any n ≥ 2 and any F . Then assume that α|i and α|m so there are s. 1. |(132)| = 3. t ∈ Z such that 3s + nt = 1. |(1432)| = 4. First suppose that σ i is an m-cycle so that σ i = (a1 a2 . Therefore D6 = X6 and |X6 | = 6. |(12)(34)| = 2.3. |(12)| = 2. (14)(23). Suppose we have the relations given by D6 . Compute the order of each of the elments in the following groups: (a) S3 (b) S4 For elements in S3 we calculate that |1| = 1. .4. (12). |(123)| = 3. |(1423)| = 4. Matrix Groups. . t ∈ Z such that αs = i. Write out the cycle decomposition of each element of order 2 in S4 . . Thus i and m are relatively prime. |(14)(23)| = 2. .4. (24). We also have xy = yx−1 = yx3−1 = yx2 . am ). (12)(34). 0 c1 0 c2 a b (b) Find the matrix inverse of and deduce that G is closed under inverses. Number of non-invertible matrices.12 DAVID S. 0 c (c) Deduce that G is a subgroup of GL2 (R) (cf. Section 1). a = 0. c = 0 . c ∈ R. (d) Prove that the set of elements of G whose two diagonal entries are equal (i. Now notice that extending the matrices with zeros shows the result for n > 2. FOOTE a=0 b=0 b=0 c=0 c=0 d=0 p−1 c=0 d=0 c=0 d=0 d=0 d=0 d=0 d=0 d=0 1 p − 1 (p − 1)2 p − 1 (p − 1)2 0 0 a=0 b=0 c=0 b=0 c=0 d=0 c=0 d=0 c=0 d=0 p−1 d=0 (p − 1)2 d=0 d=0 d=0 d=0 0 0 (p − 1)2 (p − 1)3 Figure 1. 1. Exercise 26. b.10. Let G = a b 0 c | a. DUMMIT AND RICHARD M. First notice 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 = = 1 1 0 1 1 0 1 1 so that GL2 (F ) is non-abelian for any F .. a = c) is also a subgroup of GL2 (R). a1 b1 a2 b2 and to show that G is closed under matrix multiplication. (a) Compute the product of .1.4. Proof.e. then G is closed under inverses. D8 . 1 0 0 1 (c). Computing the product of two matrices in G. Notice that the identity element is in G . this matrix is also in G. Then |1| = 1.5. |−k| = 4.13 Proof (a). c that elements of G have nonzero determinant. k | (−1)2 = 1.5. 1. |k| = 4. Compute the order of each of the elements in Q8 . and G is closed under inverse. Q8 = −1. j. Note matrix multiplication is a binary operation on G. Notice that a 0 b c 1/a −b/ac 0 1/c = = 1/a −b/ac 0 1/c a 0 b c = = 1 0 1 0 1 0 1 0 −b/c + b/c 1 0 1 b/a − b/a 1 0 . c1 c2 Since a1 . Thus G ⊆ GL2 (R).1. Let Q8 = {1. matrix multipliaction is indeed a binary operation G . So G is a subgroup of GL2 (R).5. −k} as given in the textbook. Proof (d). Write out the group tables for S3 . |j| = 4. −i. k. −j. i. i. we must first show a 0 b . c2 = 0.2. First notice that G = ∅ because ∈ G. Find a set of generators and relations for Q8 . To show that G is a subgroup of GL2 (R). 1. (b). 1. We first show that matrix multiplication is a binary operation on G = {G | a = c}. the identity is in G. a1 a2 Since a1 a2 = a1 a2 . we must have G as a subgroup of GL2 (R). 1 Since 1/a −b/ac 0 1/c is in G.5. |−1| = 2. Quaternion Groups. |−j| = 4.3. j. 1. So a1 0 b1 a1 a2 0 b2 a2 = a1 a2 0 a1 b2 + b1 a2 . we find a1 0 b1 c1 a2 0 b2 c2 = a1 a2 0 a1 b2 + b1 c2 . |−i| = 4. and Q8 . The inverse is given by a 0 b a −1 = 1/a −b/a2 0 1/a due to the calculation of inverse in G. i2 = j 2 = k 2 = ijk = −1 . |i| = 4. So let M = det M = ac = 0. a2 = 0 and c1 . −1. Thus G is closed under inverse and since G ⊆ G ⊆ GL2 (R). is the fiber over the identity of H). φ(G). Notice that 1 · a = a for all a ∈ A so the kernel of the action is nonempty. −1 Now suppose that ker φ = {1G } and let φ(g1 ) = φ(g2 ). Then φ(g1 g2 ) = φ(g1 )φ(g2 ) = 1 so that g1 g2 ∈ ker φ. 1. Then g −1 a = g −1 (ga) = (g −1 g)a = a so the kernel of the action is closed under inverse. Prove that the kernel of φ is a subgroup (cf. Let g1 . Suppose that g ∈ ker φ so that φ(g) = 1. 1. 2 2 Now suppose that G is abelian. Prove that the image of φ. 1. Prove that the map from G to itself defined by g → g 2 is a homomorphism if and only if G is abelian. Show that the following sets are subgroups of G (cf. Let |φ(x)| = n so that φ(x)n = 1 and φ(x)l = 1 for 0 < l < n. Let g1 . Is the result true if φ is only assumed to be a homomorphism? Proof.6.7.14 DAVID S. Prove that φ is injective if and only if the kernel of φ is the identity subgroup of G.5.4. Then (g1 g2 )(a) = g1 (g2 a) = g1 a = a so the group operation on G is a binary operation on the kernel of the action. g2 ∈ {g ∈ G | ga = a} so that (g1 g2 )a = g1 (g2 a) = g1 a = a and hence the group operation on G is a binary operation on {g ∈ G | ga = a}.18. Exercise 26 of Section 1). 2 2 Proof. Let G be a group acting on a set A and fix some a ∈ A.13. g2 be two elements in the kernel of the action. Let G be any group. g2 ∈ ker φ so that φ(g1 ) = φ(g2 ) = 1. Let G and H be groups and let φ : G → H be a homomorphism.. Proof (a). Now suppose that xl = 1. Let g1 .7. Suppose that φ is injective. is a subgroup of H (cf. Hence any two isomorphic groups have the same number of elements of order n for each n ∈ Z+ because of the bijectivity of φ. (b) {g ∈ G | ga = a} –this subgroup is called the stabilizer of a in G. Exercise 26 of Section 1) of G. ker φ = {1G }. Exercise 14 in Section 6). Then 1 = φ(xl ) which is a contradiction. Then φ(g −1 ) = φ(g)−1 = h−1 .5. Let G and H be groups and let φ : G → H be a homomorphism. Now let g ∈ {g ∈ G | ga = a} so that g −1 a = g −1 (ga) = (g −1 g)a = a and hence {g ∈ G | ga = a} is closed under inverse. Therefore the kernel of the action is a subgroup of G. Thus Exercise 1. Prove that if φ is injective then G ∼ φ(G). 1. Note that |s| = r2 = 2. we must have g1 g2 = 1G and thus g1 = g2 . Now let g be in the kernel of the action. 1. Exercise 26 of Section 1): (a) the kernel of the action. Proof. . 1. 1. Group Actions.14.6. Deduce that any two isomorphic groups have the same number of elements of order n for each n ∈ Z+ . Let h1 . Prove that the kernel of an action of the group G on the set A is the same as the kernel of the corresponding permutation representation G → SA (cf. Proof. Notice that φ(1) = 1 so H is nonempty. Since φ(1G ) = 1H .6. Thus ker φ ⊆ H is a subgroup. Therefore {g ∈ G | ga = a} is a subgroup of G. Define the kernel of φ to be {g ∈ G | φ(g) = 1H } (so the kernel is the set of elements in G which map to the identity of H. g2 ∈ G so that g1 g2 g1 g2 = (g1 g2 )2 = g1 g2 .e.1 shows there are not two elements with order two in Q8 . h2 ∈ H so that there are g1 . Prove that D8 and Q8 are not isomorphic. Let h ∈ H and let φ(g) = h. Notice that 1 · a = a for all a ∈ A so the stabilizer of a in G is nonempty. Thus |x| = n. so H is closed under inverse. prove that |φ(x)| = |x| for all x ∈ G. Now let g1 . DUMMIT AND RICHARD M. First note that 1 ∈ ker φ so that ker φ = ∅. i. Then φ(g1 )φ(g2 )−1 = 1. g2 ∈ G such that φ(g1 ) = h1 and φ(g2 ) = h2 . So φ(g1 g2 ) = 1 and since −1 ker φ = {1G }.5.7. If φ : G → H is an isomorphism. Therefore H ⊆ G is a subgroup. Thus there is only one element that maps to 1H .6. The result that |φ(x)| = |x| is not true for φ any homomorphism. For example the trivial map φ(x) = 1 for all x ∈ G for G non-trivial. Proof (b). Therefore φ is injective. But 1 = φ(x)n = φ(xn ) and thus xn = 1. g2 ∈ G so that (g1 g2 )2 = g1 g2 g1 g2 = g1 g2 and thus g → g 2 is a homomorphism. Let g1 . = Proof.6.7. First suppose that g → g 2 is a homomorphism. Thus φ(g1 g2 ) = φ(g1 )φ(g2 ) = h1 h2 so h1 h2 ∈ H.6.2. Homomorphisms and Isomorphisms. Then φ(g −1 ) = φ(g)−1 = 1 so g −1 ∈ ker φ and ker φ is closed under inverse. Hence ∗H×H : H × H → H. FOOTE 1. 1.2 shows that D8 and Q8 are not isomorphic. Thus g2 g1 = g1 g2 showing that G is abelian. But Exercise 1. . . Does Lagrange’s Theorem (Exercise 19 in Section 1. a ∈ G do not satisfy the axioms of a (left) group action of G on itself. . CQ8 (k) = {1. Now let σ ∈ Sn be a non-identity element for n > 1.2. −j}.15 Proof. .7. .7) simplify your work? S3 . s. CS3 (1) = S3 . Note that Lagrange’s theorem lets us check our answers because the centralizer is a subgroup and thus must divide the order of the group. CQ8 (j) = {1. ak }. If k < n. First suppose that ga = a for all a ∈ A. CS3 ((2 3)) = {1. sr. ak } . s. 2. r2 . If g is in the kernel of the left regular action. . r2 . (2 3)}.2. . .14. then the group action is trivially faithful. If k = n. . sr.10. . . then σ is just a bijection of n-element sets so the homomorphism representing the group action is trivial. For each of S3 . CS3 ((1 2 3)) = {1. Therefore the values of k giving rise to a faithful group action are all values of k < n. Then σg (a) = a for all a ∈ A hence σg = 1SA . r2 .2. (1 2)}. then we can choose {a1 . sr3 . Subgroups 2. sr2 . . ak } such that ai ∈ {a1 . . i}. ∀a ∈ Z(G) by definition. 1. 2.5. . (c) G = D10 and A = 1. CQ8 (1) = Q8 . Since G is non-abelian we can find g1 . Then ga = σg (a) = a. But a ∈ Z(G) if and only if ag = ga for all g ∈ G. CQ8 (−j) = {1. Hence the homomorphism representing the group action is injective for k < n. if H is finite. (b) G = D8 and A = 1. CQ8 (−k) = {1. Let Sn act on subsets of order k for some 1 ≤ k ≤ n. (1 2 3). .7.1. k}. ak } = {a1 . CS3 ((1 3)) = {1. sr2 . Find the kernel of the left regular action. y ∈ H. Proof. Thus g = 1 so the kernel of the left regular action is {1}.7. then ga = a for all a ∈ G. r. Proposition 2.13. We assume the axioms of a group action hold. then it suffices to check that H is nonempty and closed under multiplication. (1 3 2)}. if and only if a = gag −1 for all g ∈ G. . sr3 . s. CQ8 (−i) = {1. Therefore these kernels are equivalent. Since CG (Z(G)) ≤ NG (Z(G)) and NG (Z(G)) ⊆ G. CD8 (1) = D8 . (1 2 3). a contradiction. Prove that CG (Z(G)) = G and deduce that NG (Z(G)) = G. 2. CQ8 (i) = {1. CD8 (sr2 ) = 1. −k}.4. r4 . Now suppose that σg (a) = a for all a ∈ A. g2 ∈ G such that g1 g2 = g2 g1 . Note if n = 1. Since σ is not the identity. Centralizers and Normalizers. 1. sr2 . D8 . Stabilizers and Kernels. A subset H of a group G is a subgroup if and only if (1) H = ∅ (2) for all x. Thus CG (Z(G)) = G. But g1 g2 = a−1 ag1 g2 = a−1 ((g1 g2 ) · a) = a−1 (g1 · (g2 · a)) = a−1 (g1 · (ag2 )) = a−1 (ag2 g1 ) = g2 g1 . CD8 (s) = 1. j}. r2 . xy −1 ∈ H Furthermore. Proof. CQ8 (−1) = {1. Thus / σ · {a1 . CD8 (sr) = 1. CD8 (sr3 ) = 1. CD8 (r3 ) = r . and Q8 compute the centralizers of each element and find the center of each group. CD8 (r) = r . r3 .2. . CD8 (r2 ) = D8 . Definition and Examples. ak } and aj ∈ {a1 . (1 3)}. Notice CG (Z(G)) = g ∈ G | gag −1 = a. With reference to the preceding two exercises determine: (a) for which values of k the action of Sn on k-element subsets is faithful. . (a) G = S3 and A = {1. CG (A) = A and NG (A) = G. Q8 . CS3 ((1 2)) = {1. . let σ(ai ) = aj for ai = aj . r2 . 2. Therefore this is not a group action. (1 3 2)}. In each of parts (a) to (c) show that for the specified group G and subgroup A of G. . Let G be a group and let A = G. −1}. r2 . D8 .1 (The Subgroup Criterion). (1 3 2)}. 2. (1 2 3).2. −i}. Show that if G is non-abelian then the maps defined by g · a = ag for all g. we have NG (Z(G)) = G as well. 1. . CS3 ((1 3 2)) = {1. 4. then |xa | = (n. Let H = x . if K ≤ H. (2) Assume |x| = n < ∞. Let H = x be a cyclic group. Since CD10 (r) = r .2. x ∈ NG (H) so that x ∈ G and xHx−1 = H.16 DAVID S. Similarly to part (a) by Exercise 2. FOOTE Proof (a). Cyclic Groups and Cyclic Subgroups.4. Now let x ∈ NG (A) ∩ H so that x ∈ HandxAx−1 = A. h} x−1 = {1. Proposition 2. the map ϕ: Z → x is well defined and is an isomorphism. sr2 ∩ D8 ∩ 1. Proposition 2. Deduce that if NG (H) = G then H ≤ Z(G). then the map ϕ: x → y xk → y k is well defined and is an isomorphism (2) if x is an infinite cyclic group. Since H and Z(G) are subgroups we get H ≤ Z(G). n).3. Let G be an arbitrary group. More specifically (1) if |H| = n < ∞. Let G be a group. (3) In particular.a) . Proof (c).6. then either K = {1} or K = xd . Let H be a subgroup of order 2 in G. Thus x ∈ CG (H). x2 .3. Proof (b). If xn = 1 and xm = 1. Let x ∈ NH (A) so x ∈ H ≤ G and xAx−1 = A.9. Let H = {1. Note that NG (A) and H are both subgroups of G and the intersection of two subgroups is again a subgroup. Thus any element of G commutes with any element in H so H ⊆ Z(G). . x.2. Proposition 2. Proposition 2. where d is the smallest positive integer such that xd ∈ K. so is the other). h}. Theorem 2. sr2 = A. (1 2 3). if |x| = n < ∞ and a is a positive integer dividing n. Thus x ∈ GH (A) and hence NH (A) = NG (A) ∩ H. Note that any power of r commutes with any power of r so A ⊆ CD10 (A). then x1x−1 = 1 and xhx−1 = h so xHx−1 = H. then |xa | = k → xk n a. Thus x ∈ NG (A) ∩ H.10. n (2) If |x| = n < ∞. n ∈ Z. DUMMIT AND RICHARD M. s. (1) if n ∈ Z+ and x and y are both cyclic groups of order n. Theorem 2. Proof. In particular. Then H = xa if and only if a = ±1. (1 3 2)} = A. xn−1 are all the distinct elements of H (2) if |H| = ∞. . then xd = 1. then |xa | = ∞. Thus x ∈ NG (H) and hence NG (H) = CG (H). then |x| divides m. (1 2 3). then G = CG (H). Note if NG (H) = G. 2. CS3 (A) = S3 ∩{1. then xn = 1 and 1.4.2. Any two cyclic groups of the same order are isomorphic. For any subgroup H of G and any nonempty subset A of G define NH (A) to be the set h ∈ H | hAh−1 = A . then xn = 1 for all n = 0 and xa = xb for all a = b in Z. where d = (m. More specifically. r2 . Then H = xa if and only if (a. Show that NH (A) = NG (A) ∩ H and deduce that NH (A) is a subgroup of H (note that A need not be a subset of H). Since x1x−1 = 1. 2. we must have xhx−1 = h so that x {1. x ∈ G and let m. r2 .5. . n) = 1. Proof. (1) If |x| = ∞. .2. if xm = 1 for some m ∈ Z.2. (1) Every subgroup of H is cyclic. Now if x ∈ CG (H). we also must have CD10 (A) ⊆ A. .7. h}. In particular. then number of generators of H is ϕ(n). let x ∈ G and let z ∈ Z\ {0}. CD8 (A) = D8 ∩ 1. then |H| = |x| (where if one side of this equality is infinite. 2. Show that NG (H) = CG (H). More precisely. Note that CG (A) = ∩a∈A CG (a) so by Exercise 2. (1 3 2)}∩{1. Therefore NH (A) ≤ G. s. (1) Assume |x| = ∞. If H = x . y) = (x. This is given by Euler’s ϕ function. ky). This subgroup is the cyclic group xd . we count the number of relatively prime integers to 49000. But then y = ky shows that y = 0. Show that if H is any group and h is an element of H with hn = 1. 2. Now suppose that ϕ and ψ are two homomorphisms such that ϕ(x) = h = ψ(x). x3 . . x9 . This is a contradiction so this group is not cyclic. xm = x|m| .8. Suppose that ϕ : (Q × Z2 ) → Q is an isomorphism. Proposition 2. 2. Proposition 2. (3) If |H| = n < ∞.12. 31. Describe the containments between these subgroups. y) be a generator. 5. Then there exists k ∈ Z such that (1.3. If xk = xl so that k ≡ l (mod n). Prove that the following pairs of groups are not isomorphic: (b) Q × Z2 and Q. 48) = 1. then for any distinct nonnegative integers a and b. xa = xb . The subgroups of Z45 are given by x0 . then the intersection of all members of A is also a subgroup of G. 2. . 47}.3. (1 2 4 3) .. But then hk = hαn+l = hl so ϕ is well-defined. 29. thus no such isomorphism exists. then there is a unique homomorphism from Zn = x to H such that x → h. 2. Since hn = 1. 2. 13. Suppose Z2 × Z is cyclic and let (x. for every integer m. So ϕ((0.1. x15 . Proof.3.8. Let Z48 = x . A = A . Since Z/49000 Z is cyclic of order 49000. y)k = (xk .4. Note that x. 7.13. 35. . where |m| denotes the absolute value of m. there is an m ∈ N such that | h | = m so m|n. xl ∈ x so that ϕ(xk )ϕ(xl ) = hk hl = hk+l = ϕ(xk+l ) shows that ϕ is such a homomorphism. for every integer m. 2. x5 . The containments between these subgroups are given by x ⊇ x3 ⊇ x5 ⊇ x9 ⊇ x15 ⊇ x0 . then for each positive integer a dividing n there is a unique subroups of H of order a.3. x)) = y = 0 but 0 = ϕ((0. Find all subgroups of Z45 = x . xm = x(n. 37. x .4. Therefore ϕ = ψ shows there is a unique map with these properties.9. Furthermore. 17. 3. Subgroups Generated by Subsets of a Group.8. For which integers a does the map ϕa defined by ϕa : ¯ → xa extend to an isomorphism from 1 Z/48Z onto Z48 . so ϕ(49000) = ϕ(23 53 72 ) = ϕ(23 )ϕ(53 )ϕ(72 ) = 22 (1) · 52 (4) · 7(6) = 16800.5. 1)) = ϕ((0. Proof. Since y = 0 this is a contradiction. then k − l = αn for some α ∈ Z. 2. a so that the subgroups of H correspond bijectively with the positive divisors of n. If A is any nonempty collection of subgroups of G. Prove that S4 = (1 2 3 4). So a ∈ {1. Take some xk ∈ x so that ϕ(xk ) = ϕ(x)k = hk = ψ(x)k = ψ(xk ).3. so that the nontrivial subgroups of H correspond bijectively with the integers 1.3. Define a map ϕ(xk ) = hk . y are not identity elements in their respective groups. 41. 2. giving a generator for each. Furthermore. 25.17 (2) If |H| = ∞. Note k = 1 because x = 1. Now take xk . x) ∈ Q × Z2 be a non-identity element. . 2. where d = n . All integers a such that (a.18. Prove that the following groups are not cyclic: (b) Z2 × Z Proof. Let (0.m) . 43. x)2 ) = 2y. 23. 11. 19. Find the number of generators for Z/49000 Z. = .] Proof. (1 2 3 4)(1 3 4 2) = (1 4 3). So we have . (1 3)(2 4)(2 4) = (1 3). Since the diagonal must be have nonzero entries. We will show that the subgroup of upper triangular matrices in GL3 (F2 ) is equal to     1 1 1 1 1 0 A = 0 1 0 . = 0 1 0 1 0 1 1 0 1 1 1 0 2 1 1 2 1 0 0 2 1 . So d(a + d) = 2 shows that cd(a + d) = 2c = 0 hence c = 0. Prove that SL2 (F3 ) and S4 are two nonisomorphic groups of order 24. namely (1 2) and (1 3). 0 Thus the relations of the subgroup of SL2 (F3 ) generated by the matrices are contained in the presentation of Q8 . Since we have found 5 matrices in this subgroup and |SL2 (F3 )| = 24. (1 2 4 3)3 = (1 3 4 2).8 to show this subgroup 3 2 1 1 1 0 1 0 1 0 . (1 2 4 3)2 = (1 4)(2 3). 1 2 1 2 1 1 1 0 1 0 2 1 2 1 1 1 1 2 = . 0 −1 2. Prove that the subgroup of upper triangular matrices in GL3 (F2 ) is isomorphic to the dihedral group of order 8. First note that |S4 | = 4! = 24. (1 4 3 2)(1 2 4 3) = (2 3 4). (1 2 3 4)3 = (1 4 3 2). [Use a presentation for Q8 . (1 2 3 4)(1 4)(2 3) = (2 4). = . Thus the order of this subgroup is 23 = 8.4. 0 2 2. Notice that 0 1 −1 0 4 1 0 1 1 and 1 1 0 .4. j | i4 = 1 = j 4 . So c d a c b d 2 = a2 + bc ac + cd ab + bd bc + d2 = 1 0 0 1 and ad − bc = 1. 2. Proof. the diagonal is a string of 1’s. = . DUMMIT AND RICHARD M. = . 2.4. (1 2 3 4)2 = (1 3)(2 4). Therefore these groups are isomorphic. So the order of the matrix subgroup is less than or equal to 8. (1 2 4 3) ≤ S4 . First notice that there is more than one element of order two in S4 . 2 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 0 1 2 1 2 1 0 1 1 1 1 1 . 1 is = = = to exercise 2. (1 2 3 4)(1 2 4 3) = (1 3 2). So we have (1 2 3 4). Now let a b ∈ SL2 (F3 ) be an element of order 2. (1 3)(2 4)(1 2 4 3) = (1 4). [First find the order of this subgroup.11. 1 1 1 2 0 1 2 1 0 0 1 −1 0 and 1 1 1 −1 is isomorphic to the quaternion =I= 0 1 −1 0 1 1 1 1 1 −1 1 −1 4 .] Proof. 0 1 0 1 1 1 1 1 1 1 0 1 1 1 0 1 1 = . First note that |SL2 (F3 )| = 24 so we use a method similar 2 1 1 1 1 1 2 generated by the two matrices.10. and = −1 1 1 1 =− 1 −1 −1 −1 =− 1 1 1 −1 0 1 −1 . FOOTE Proof. Let Q8 be given by the presentation Q8 = i. .9. (1 4)(2 4) = (1 4 2). (1 2 4 3). 1 2 1 1 0 1 1 0 2 1 0 1 2 0 2 2 2 . So ad = a2 = d2 = 1 together with b(a + d) = 0 shows that a2 b + abd = 0 hence b = 0. (1 2 4 3) = S4 . = .4. B = 0 1 1 0 0 1 0 0 1 . Prove that the subgroup of SL2 (F3 ) generated by group of order 8. Prove that SL2 (F3 ) is the subgroup of GL2 (F3 ) generated by Proof. Since (1 2 3 4).18 DAVID S. Since bc = ad − 1 we see ad − 1 + d2 = 1. Lagrange’s theorem shows that the order of this subgroup is indeed 8. Therefore the only −1 0 element of order 2 is and hence these groups are not isomorphic. Lagrange’s theorem states we need to find 13 elements generated by this set so that (1 2 3 4).12. ij = −ji .4. Proof. Let g1 . Deduce that ker ϕ G.19 where AB = B −1 A and B 4 = I = A2 . Note that r2 = 2 in D8 but r2 r2 = 1 in 8 D / r2 . αβ < 0. Calculating we find  2  1 1 1 1 0 0 1 0 = 0 1 0 0 1 0 0  4  1 1 0 1 1 0 1 1 = 0 1 0 0 1 0 0  1 1 = 0 1 0 0  1 0 = 0 1 0 0  0 0 1 2   0 1 0 1 1 0 1 0 1 0 0 1   0 1 1 1 1 0 1 1 1 0 0 1  0 0 . Define ϕ : C× → R× by ϕ(a + bi) = a2 + b2 . 2.6. h2 ∈ E. Definitions and Examples. ϕ((a + bi)(c + di)) = ϕ(ac + adi + bci − bc) = (ac − bd)2 + (ad + bd)2 ϕ(a + bi)ϕ(c + di) = (a2 + b2 )(c2 + d2 ) = a2 c2 − 2abcd + b2 d2 + a2 d2 + 2abcd + b2 d2 = a2 c2 + a2 d2 + b2 c2 + b2 d2 . αβ > 0. αβ < 0. 1 Thus we found five matrices generated by A and B to show that it generates the whole group by Lagrange’s theorem.5. Prove that ϕ is a homomorphism and find the image of ϕ. Thus g ϕ−1 (E) g −1 ⊆ −1 ϕ (E) shows ϕ−1 (E) G. Let G and H be groups. 2.9. The fibers of ϕ are given by ϕ−1 (1) = {α ∈ R× | α > 0} and ϕ−1 (−1) = {α ∈ R× | α < 0}. If E H then hEh−1 ⊆ E for all h ∈ H. Proof. −1 Proof. Since 1H is normal in H then ker ϕ G. But ϕ(1G ) = ϕ(1H ) and ϕ(g1 g2 ) = −1 h1 h2 ∈ E so ϕ−1 (E) ≤ G. Give an example to show that the order of gN in G/N may be strictly smaller than the order of g in G. So take ϕ g ϕ−1 (E) g −1 ⊆ ϕ(g) E ϕ(g)−1 ⊆ E. the preimage or pullback of a subgroup under a homormorphism is a subgroup). Proof. First note if 1 < m < n then (gN )m = g m N N for g ∈ N so the order of gN ≥ n.5.5. But (gN )n = g n N = N / so the order of gN is n. To show ϕ is a homomorphism we calculate ϕ(αβ) = ϕ(α)ϕ(β) = 1 −1 αβ > 0. where n is the smallest positive integer such that g n ∈ N (and gN has infinite order if no such positive integer exists). We calculate. Describe the kernel and the fibers of ϕ geometrically (as subsets of the plane).5. 2. Define ϕ : R× → {±1} by letting ϕ(x) be x divided by the absolute value of x.5..1. 1 −1 2. Use the preceding exercise to prove that the order of the element gN in G/N is n. Let ϕ : G → H be a homomorphism and let E be a subgroup of H. Therefore these two groups are isomorphic. Describe the fibers of ϕ and prove that ϕ is a homomorphism. An example with the desired property would be D8 / r2 . If E H prove that ϕ−1 (E) G.5. g2 ∈ ϕ−1 (E) so ϕ(g1 ) = h1 and ϕ(g2 ) = h2 where h1 . Prove that ϕ−1 (E) ≤ G (ie. 2. Theorem 3. Describe 0 c the fibers and kernel of ψ. (4) ker ϕ is a subgroup of G. c = 0 . X = ϕ−1 (a).1.e. 2. 1 b (c) Let H = | b ∈ F .e.6. Describe the fibers and kernel of ϕ. ac = 0 ≤ GL2 (F ). The same statement is true with “right coset” in place of “left coset.20 DAVID S. a b (b) Prove that the map ψ : → (a. c) is a surjective homomorphism from G onto F × × F × . Describe similarly the elements in the fibers of ϕ above the points −1. 2.. X = {uk | k ∈ K} (2) For any u ∈ X.” b1 1 1 0 b2 1 = 1 0 b1 + b2 1 . this operation is well defined in the sense that if u1 is any element in uK and v1 is any element in vK. Proof (b).11.12.. let H be the multiplicative group of complex numbers of absolute value 1 (the unit circle S 1 in the complex plane) and let ϕ : G → H be the homomorphism ϕ : r → e2πir . i. In particular.1. ψ is clearly surjective and the calculation from proof (a) shows it is a homomorphism. u1 v1 K = uvK so that the multiplication does not depend on the choice of representatives for the cosets.2. G/K. Let X ∈ G/K be the fiber above a. Prove that H is isomorphic to the additive group F . Definitions and Examples. Draw the points on a real line which lie in the kernel of ϕ. FOOTE so ϕ is a homomorphism. Quotient Groups and Homomorphisms 3. ϕ(e4πi/3 ) = (2/3)(R+ /Z). The points of the real line in the kernel of ϕ are the integers. Let G be a group and let K be the kernel of some homomorphism from G to another group. The kernel of ϕ is elements of G with a = 1. c ∈ F. then u1 v1 ∈ uvK. (1) ϕ(1G ) = 1H (2) ϕ(g −1 ) = ϕ(g)−1 for all g ∈ G. ϕ is clearly surjective so we show it is a homomorphism by a1 0 The fibers of ϕ are given by ϕ−1 (a) = b1 c1 a 0 a2 0 b c b2 c2 = a1 a2 0 a1 b2 + b1 c2 . a b → a is a surjective homomorphism from G onto F × (recall that F × is the 0 c multiplicative group of nonzero elements in F ). Let G and H be groups and let ϕ : G → H be a homomorphism. DUMMIT AND RICHARD M. The fibers are given by ϕ−1 (−1) = (1/2)(R+ /Z). (3) ϕ(g n ) = ϕ(g)n for all n ∈ Z. The calculation 1 0 makes it clear that H is isomorphic to F + .5. The kernel of ψ is elements of G with a = c = 1. The fibers of ψ a b are given by ψ −1 ((a. −1 ϕ (i) = (1/4)(R+ /Z). 2. i. Proposition 3. 0 c Proof (c). e.5. Then (1) For any u ∈ X.3. X = {ku | k ∈ K}. 3. Let G be the additive group of real numbers. c1 c2 | b ∈ F. Let F be a field and let G = (a) Prove that the map ϕ : a 0 b c | a. 0 1 Proof (a). (5) im (ϕ) is a subgroup of H. Then the set whose eelements are the left cosets of K in G with operation defined by uK ◦ vK = (uv)K forms a group. Let ϕ : G → H be a homomorphism of groups with kernel K. Proposition 3. and e4πi/3 of H. b. The Lattice of Subgroups of a Group. c)) = | b ∈ F . .15.10. then if gZ(G) ∈ G/Z(G) we have gZ(G) = (xZ(G))n = xn Z(G) for some n ∈ N. In particular the identity of this group is the coset 1N and the inverse of gN is the coset g −1 N i. uN = vN if and only if v −1 u ∈ N and in particular. Proof. If G/Z(G) is cyclic with generator xZ(G).21 Proposition 3. for all u. Proposition 3. .1.8 (Lagrange’s Theorem). uN = vN if and only if u and v are representatives of the same coset. if K G then HK ≤ G for any H ≤ G. v ∈ G.. Furthermore. (1) The operation on the set of left cosets of N in G described by is well defined if and only if gng ∈ N for all g ∈ G and all n ∈ N . Thus there is some z ∈ Z(G) such that g = xn z. HK is a subgroup if and only if HK = KH.9. Now take x ∈ H ∩ K so |x| |m and |x| |n so x = 1. (gN )−1 = g −1 N .12 (Sylow). then it makes the set of left cosets of N in G into a group. Then g1 g2 = xm z1 xn z2 = xm xn z2 z1 = xn xm z2 z1 = xn z2 xm z1 = g2 g1 . Theorem 3. If H and K are finite subgroups of a group then |HK| = |H| |K| . Proof. then HK is a subgroup of G. A subgroup N of the group G is normal if and only if it is the kernel of some homomorphism. Proposition 3.6. Let |H| = m and |K| = n with (m.5 makes the set of left cosets into a group (5) gN g −1 ⊆ N for all g ∈ G. Prove that if G/Z(G) is cyclic then G is abelian.14.36. |H| | |G|) and the number of left cosets of H in G equals |H| . In particular. Corollary 3. then the roder of x divides the order of G. If G is a finite group and p is a prime dividing |G|.4. Theorem 3. then G has a subgroup of order pα . Proposition 3. Let N be any subgroup of the group G.11 (Cauchy’s Theorem). g2 ∈ G such that g1 = xm z1 and g2 = xn z2 . then G has an element of order p.13. Corollary 3. (2) If the above operation is well defined.2. Corollary 3. 3. In particular x|G| = 1 for all x in G.e. If H and K are subgroups of a group. hence G ∼ Zp . Theorem 3. If H and K are subgroups of G and H and H ≤ NG (K). More on Cosets and Lagrange’s Theorem. Let N be a subgroup of the group G. If G is a finite group of order pα m.7. n) = 1.8. If G is a group of prime order p. Now take g1 . Prove that if H and K are finite subgroups of G whose orders are relatively prime then H ∩ K = 1. then the order of H divides |G| the order of G (i. |H ∩ K| Proposition 3. then G is cyclic. where p is a prime and p does not divide m.5. −1 uN · vN = (uv)N 3.e. = Theorem 3.2. Let G be a group and let N be a subgroup of G. If G is a finite group and x ∈ G. 3. The following are equivalent: (1) N G (2) NG (N ) = G (3) gN = N g for all g ∈ G (4) the operation on left cosets of N in G described in proposition 3. If G is a finite gorup and H is a subgroup of G. The set of left cosets of N in G form a partition of G. Show that a cyclic permutation of an element of S is again an element of S. . The first p − 1 coordinates have |G| choices and the pth coordinate must be the inverse of the product p−1 of the first p − 1 coordinates. [G : H ∩ K] = |G/(H ∩ K)| ⇐ |G/H × G/K| = |G/H||G/K| = n · m. j = 0 because |σ| = p (for p prime. Reflexive is obvious. Thus |S| = |G| . 3. (a) (b) (c) (d) p−1 Show that S has |G| elements. Let S denote the set of p-tuples of elements of G the product of whose coordinates is 1: S = {(x1 . Suppose H and K are subgroups of finite index in the (possibly infinity) group G with |G : H| = m and |G : K| = n. Let (x1 . then γ is a cyclic permutation of α so ∼ is transitive. you have [G : H ∩ K] = [G : K][K : H ∩ K].e. xp ) ∈ S and notice xp x1 x2 · · · xp−1 = 1 just by multiplication on the right by (xp )−1 . × Proof. Since equivalence classes are p−1 disjoint. It is clear that every equivalence class has order less than or equal to p. xp−1 ) ∈ S and any cyclic permutation can be realized by repeating this process. Proof (f ). Since a is relatively prime to n. Its kernel is H ∩ K. Proof (b). then the left by xp . . xp ) | xi ∈ G and x1 x2 · · · xp = 1} . i. Deduce that p−1 |G| = k + pd. . If all the x are the same. . . Then xi = xσ(i) = xσ2 (i) = · · · = xσp−1 (i) for all i. then any cyclic permutation is indeed the same element of S.9. . Apply this to the map G → G × G → G/H × G/K sending g → (g. This exercise outlines a proof of Cauchy’s Theorem due to James McKay. . 1. Now note that the first isomorphism theorem works also for non-normal subgroups. Thus. . then β is a cyclic permutation of α so ∼ is symmetric. . . Thus k > 1 and hence there is an 3. . . By Problem 11. .22 DAVID S. . . |G| = k + pd. g) → (gH. If β is a cyclic permutation of α and γ is a cyclic permutation of β. . G contains an element of order p. but you get a statement on maps of sets (not groups). Suppose the equivalence class is not of the form from part (d) and let σ ∈ Sp be given by σ(i) ≡ i + j mod p for 0 ≤ j < p and suppose that (x1 . x2 . a ∈ (Z\nZ)× . Thus the number of equivalence classes is the sum of size 1 and size p classes. . Prove that ∼ is an equivalence relation on S. x. Thus ∼ is an equivalence relation. so the (set version of the) first isomorphism theorem gives an injective map G/(H ∩ K) → G/H × G/K. x1 . Deduce that if m and n are relatively prime then |G : H ∩ K| = |G : H| · |G : K|. then p divides |G| element x in G with order p.2.2. gK). Prove that lcm (m. Proof (c). (f) Since {(1. (e) Prove that every equivalence class has order 1 or p (this uses the fact that p is a prime). xσ(p) ). n) ⇐ [G : H ∩ K]. xp ) ∼ (xσ(1) . and also m divides by the same argument. . If some x is changed to a y = x. then a cyclic permutation would yield a different element of the equivalence class. FOOTE 3. So n divides [G : H ∩ K]. Corollary 9 states that aϕ(n) = a|(Z\nZ) | = 1|(Z\nZ)× | ≡ 1 mod n. If α is a cyclic permutation of β. Since p divides |G|. Proof (e). n) ≤ |G : H ∩ K| ≤ mn.2. Prove that an equivalence class contains a single element if and only if it is of the form (x. . where ϕ denotes Euler’s ϕ-function. Define the relation ∼ on S by letting α ∼ β if β is a cyclic permutation of α. hence has order divisible by p. Use Lagrange’s Theorem in the multiplicative group (Z/nZ)× to prove Euler’s Theorem: aϕ(n) ≡ 1 mod n for every integer a relatively prime to n. . where k is the number of classes of size 1 and d is the number of classes of size p.10. . conclude from (e) that there must be a nonidentity element x in G with xp = 1. Thus (xp . σ = 1). Therefore lcm(m. 1)} is an equivalence class of size 1. . .] Proof (a). . x) with xp = 1. . Let G be a finite group and let p be a prime dividing |G|.. . Proof (d). DUMMIT AND RICHARD M. Schulze.22. . x2 . [Show p|k and so k > 1. Proof by Dr. Since not every coordinate is allowed to be equal. p−1 and specifically p divides k. . . 4. B AB. But this divides pa m so |P ∩ N | = pb α. Prove that (C × D) (A × B) and (A × B)/(C × D) ∼ (A/C) × (B/D). The fact that |K : H ∩ H| = p is immediate by the diamond isomorphism theorem. (The subgroup P of G is called a Sylow p-subgroup of G. Let ϕ : G → H be a homomorphism of groups. Thus ker Φ = C × D and thus (C × D) (A × B). |HK| > H so HK = G because |G : H| = p and Lagrange’s theorem. Let G be a group. Let G be a group and let H and K be normal subgroups of G with H ≤ K.9. The kernel of ϕ is clearly M ∩ N because g ∈ M and g ∈ N if and only if (g. But now pb α divides pa so α = 1 and hence |P ∩ N | = pb . = 3. = Corollary 3. n) so ϕ is surjective. Suppose that K K so since K ≤ NG (H).19 (The Third Isomorphism Theorem). the first isomorphism theorem shows that (A × B)/(C × D) ∼ (A/C) × (B/D). Proof. this can be written G/K ∼ G/K. Then K/H G/H and (G/H)/(K/H) ∼ G/K. where p does not divide m. Let π1 : A → A/C and π2 : B → B/D which exist by the normality of C and D. (2) if A ≤ B. Let M and N be normal subgroups of G such that G = M N . Prove that |P ∩ N | = pb and |P N/N | = pa−b .23 3. (1) ϕ is injective if and only if ker ϕ = 1. B ≤ G with N ≤ A and N ≤ B. 3. Then nm n−1 = m so take nm = g = mn so ϕ(g) = (g. Define a homomorphism ϕ : G → (G/M ) × (G/N ) by ϕ(g) = (g.18 (The Second Isomorphism Theorem). If ϕ : G → H is a homomorphism of groups.3. Since |K| = 1. = [Draw the lattice. = Theorem 3. HK = KH ≤ G. where p does not divide n. every subgroup of G is of the form A/N for some subgroup A of G containing N (namely. g) = (m.3. let A and B be subgroups of G and assume A ≤ NG (B).] Proof. Then there is a bijection from the set of subgroups A of G which contain N onto the set of subgroups A = A/N of G/N . n) ∈ (G/M ) × (G/N ).3.3. then |B : A| = B : A . (3) A. = Proof.) Proof. G . Then AB is a subgroup of G. This bijection has the following properties: for all A. Therefore the first isomorphism theorem states that G/(M ∩ N ) ∼ (G/M ) × (G/N ). its preimage in G under the natural projection homomorphism from G to G/N ). The diamond isomorphism theorem shows that |P N/N | = pa−b . Take some (m.17. The Isomorphism Theorems. Prove that G/(M ∩ N ) ∼ (G/M ) × (G/N ).7. Theorem 3. = 3. g). B . (2) |G : ker ϕ| = |ϕ(G)|. Let G be a group and let N be a normal subgroup of G. = If we denote the quotient by H with a bar.3. (1) A ≤ B if and only if A ≤ B. Let p be a prime and let G be a group of order pa m. This exercise shows that the intersection of any Sylow p-subgroup of G with a normal subgroup N is a Sylow p-subgroup of N . (4) A ∩ B = A ∩ B. then ker ϕ and G/ ker ϕ ∼ ϕ(G). Now let Φ : (A × B) → (A/C × B/D) be defined by π1 and π2 .3. Since Φ is surjective. = Theorem 3. But then pb α divides pb n so α divides n shows α does not divide p. A ∩ B A and AB/B ∼ A/A ∩ B. Assume P is a subgroup of G of order pa and N is a normal subgroup of G of order pb n.16 (The First Isomorphism Theorem).20 (The Fourth Isomorphism Theorem). g) = 1G/M ×G/N . Theorem 3. Let C be a normal subgroup of the group A and let D be a normal subgroup of the group B. Prove that if H is a normal subgroup of G of prime index p then for all K ≤ G either (i) K ≤ H or (ii) G = HK and |K : K ∩ H| = p. First note that |P N | = (|P | |N |)/ |P ∩ N | = (pa+b n)/ |P ∩ N |. 3. and (5) A G if and only if A G. In particular. B = A. Since the order of each composition factor is prime. Simple calculations show that 1 = N0 N1 N2 N3 = A4 . (1 2 4). Therefore this set generates Sn . 3.17. suppose σ = (1 2) and τ = (1 · · · p). Proof. (1 3 2). The permutation σ is odd if and only if the number of cycles of even length in its cycle decomposition is odd.21. 3. Proposition 3. or Z3 × Z3 . Sp = σ. (1 4)(2 3)}. (2 4 3).5. A5 . (1 3)(2 4). (1 2 3).6. If 1 ≤ i ≤ n − 1. then (i j) = (j − 1 j) · · · (i i + 1) · · · (j − 1 j). Show that (1 3). this is A4 . (1 3)(2 4)} and N2 = {1. of {1. Then (1) G has a composition series and (2) The composition factors in a composition sereis are unique. then G contains an element of order p. prove that x. namely. Proof. Notice that (1 3)(1 2 3 4)2 (1 3) = (1 2 3 4)2 shows that (1 2 3 4)2 ∈ Z( (1 3).5. 3. (1 2)(3 4). But if (i j) is any transposition. Proposition 3. Deduce that A4 is solvable.3 shows this generates Sp . (1 4 2). 3. Now we represent the group of rigid motions of a tetrahedron as a subgroup of Sn . (1 3)(2 4). = 1 ≤ i ≤ r. Exercise 3. .5. Let σ ∈ Sn and decompose σ into a product of transpositions. (1 2 3 4) < S4 . 2.5. Transpositions are all odd permutations and is a surjective homomorphism. If x and y are 3-cycles in Sn . π. Let N1 = {1.7. Therefore (1 3).23. (1 2)(3 4).5. Proposition 3. Let G be a finite group with G = 1.4. 3. (1 4 3). (1 2 3 4) is a proper subgroup of S4 .22 (Jordan-Holder). . Without loss of generality. Proposition 3.24. The map : Sn → {±1} is a homomorphism (where {±1} is a multiplicative version of the cyclic group of order 2). What is the isomorphism type of this subgroup? Proof. (1 3 4). τ where σ is any transposition and τ is any p-cycle. Composition Series and the Holder Program. Prove that Sn = {(i i + 1) | 1 ≤ i ≤ n − 1} . Proof. Prove that the group of rigid motions of a tetrahedron is isomorphic to A4 . Solution. The isomorphism type of this subgroup is D8 because we can construct 8 elements and we have (1 3)2 = (1 2 3 4)4 = (1 3)(1 2 3 4)(1 3)(1 2 3 4) = 1. A4 .5. Theorem 3. 3. while |Ni+1 /Ni | = p shows that each composition factor is simple. Find a composition series for A4 .25. 3. DUMMIT AND RICHARD M. But (1 2 3 4)2 (1 4) = (1 2 4 3) = (1 3 4 2) = (1 4)(1 2 3 4)2 . then (i i + 1) = τ i−1 στ −i+1 . r} such taht Mπ(i) /Mπ(i)−1 ∼ Ni /Ni−1 . (2 3 4).5. Transpositions and the Alternating Group.5. (1 2 3 4) ). If G is a finite abelian group and P is a prime dividing |G|. . The elements in this subgroup are 1. FOOTE 3.3.5. then r = s and there is some permutation. . y is isomorphic to Z3 . they are also abelian and hence A4 is solvable. .10. if 1 = N0 ≤ N1 ≤ · · · ≤ Nr = G and 1 = M0 ≤ M1 ≤ · · · ≤ Ms = G are two composition series for G. Show that if p is prime. First notice that both groups are order 12. (1 4)(2 3).24 DAVID S. Since all 12 of these elements are even. 6. Let p be a prime and let G be a group of order pα . (i k l) and (i j)(k l) are all generated by x and y. (i j k)(i l) = (j k l) = [(i j k)(k l m) ](i j k) . 4. g2 . 4. . (1 2). for every β with 0 ≤ β ≤ α. then we are done because any permutation can be represented by a transposition.1. Use this to classify groups of order 6. (1 3).3. and (i j k)(k m) = (i j k)(k m l) . then x. Groups Acting on Themselves by Left Multiplication – Cayley’s Theorem.13.3. Proof. [Use theorem 8 and induction on α. Solution. (j k l). gr be representatives of the conjugacy classes of the finite group G and assume these elements pairwise commute.2. Prove that every non-abelian group of order 6 has a nonnormal subgroup of order 2. Solution.2.3.8.2. (i j k)(k l) = (i j k)(k l m) .2.3. (i j k).1.] Proof. Groups Acting on Themselves by Conjugation – The Class Equation. If we can show conjugating (i j k) by any transposition remains in x.10. (2 3). Use the left regular representation of Q8 to produce two elements of S8 which generate a subgroup of S8 isomorphic to the quaternion group Q8 .4. . Find all conjugacy classes and their sizes in the following groups: (c) A4 . Solution (c). If x and y are disjoint 3-cycles. [Use exercise 36. Since 1. If x = y. then it is clear that x. (i j k)(i m) = [(i j k)(k m l) ](i j k) . 4. Let g1 . 4.6 respectively. . 4. 4. 4. Use the fact that g ≤ CG (g) for all g ∈ G to show that there is at most one possible class equation for G. 4. Prove that G has a subgroup of order pβ . (1 3 2) and label these with the integers 1. y ∼ Z3 × Z3 . 4. Now let x = (i j k).] Proof. List the elements of S3 as 1. y = x ∼ Z3 by order argument. .3. Now let x = (i j k) and y = (j k l) and notice these = are both even permutations hence generate even permutations. y = (k l m) = and we will show that x.] Proof.2.2. (i j k)(j l) = [(j i k)(k l m) ](j i k) . 4. section 3. Proof.29. Group Actions 4. Exhibit the image of each element of S3 under the left regular representation of S3 into S6 . Solution. y . y A5 . y ∼ A4 by Lagrange’s theorem. (i j k)(l m) = (i j k).3. Group Actions and Permutation Representations.4. then x. = 4. (i k j). (i j k)(j m) = [(j i k)(k m l) ](j i k) . .27.3.25 Proof. But (i j k)(i j) = (i j k)(j k) = (i j k)(i k) = (i j k)−1 .2. Prove that if H has finite index n then there is a normal subgroup K of G with K ≤ H and |G : K| ≤ n!. 4. so they are equal because A5 is simple.5. (j l k). Find all finite groups which have exactly two conjugacy classes. Assume G is a non-abelian group of order 15.3. Prove that Z(G) = 1. and the argument for (k l m) will be similar. (1 2 3). [Produce an injective homomorphism into S3 . Prove that G is abelian. .3. = Proof. σ : (1 3) → (a b3 ). So σ(x) = σ(srk ) = σ(s)σ(r)k and thus the automorphism is uniquely determined by the value on s and r. (b) Let K be the conjugacy class of transpositions in Sn and let K be the conjugacy class of any element of order 2 in Sn that is not a transposition.5. Proof. . If p is a prime and P is a subgroup of Sp of order p. Then xl = p. FOOTE 4. Let 1 = H G and p |G|. Then pk |H| |G|. where p and q are distinct primes. Since there is only one subgroup of order 4. . . .5. Let G be a group.5.4. If x ∈ D8 . Proof. then x = srk for some 0 ≤ k ≤ 3. r is characteristic so if σ ∈ Aut(D8 ). Now take x ∈ P \1 so |x| pk and let |x| = pl .3. Then 1 ∈ Sylp (H) and 1 ∈ Sylp (G). This exercise shows that for n = 6 every automorhpism of Sn is inner. for each σ ∈ Aut(G) and each conjugacy class K of G the set σ(K) is also a conjugacy class of G. Use Sylow’s Theorem to prove Cauchy’s Theorem. 4.1.1. Aut(G). sr2 . . a Sylow p-subgroup of a subgroup of G need not be a Sylow p-subgroup of G. σ : (1 n) → (a bn ) for some distinct integers a. Fix an integer n ≥ 2 with n = 6.e. We know from a previous exercise that NSp (P ) = p(p − 1) and from Corollary 15 that NSp (P )/CSp (P ) ∼ = H ≤ Aut(P ). Use (c) to show that Sn has at most n! automorphisms and conclude that Aut(Sn ) = Inn(Sn ) for n = 6. . Since x = y −1 . sr. Since conjugating an element of Inn(G) with an element of Aut(G) is contained in Inn(G). in general. Exhibit all Sylow 3-subgroups of SL2 (F3 ). . sr3 . 4. Let G be a finite group and let p be a prime. so σϕg σ −1 (x) = σ(gσ −1 (x)g −1 ) = σ(g)xσ(g)−1 = ϕσ(g) (x) shows the desired equation. then k ≥ pq. But if 1 = (xy)k = xk y k for k ≥ 1. b2 . / 4.. . n}. DUMMIT AND RICHARD M. then σ(r) ∈ r. (1 3).4.9. Prove that |K| = |K |. i. 4. Let x ∈ G. r3 . y ∈ G with |x| = p and |y| = q. Let |P | = pk . Since σ( r ) = r . Since |r| = 4. Prove that if P ∈ Sylp (G) and H is a subgroup of G containing P then P ∈ Sylp (H). If σ ∈ Aut(G) and ϕg is conjugation by g prove σϕg σ −1 = ϕσ(g) .5. Proof. (Note that we only used Cauchy’s Theorem for abelian groups – Proposition 3. We use Cauchy’s theorem to pick x. Example. Prove that under any automorphism of D8 . . then G is cyclic. so P ∈ Sylp (H). prove NSp (P )/CSp (P ) ∼ Aut(P ). Give an example to show that.) Proof. . . normality must hold. Since CSp (P ) = P and |Aut(P )| = p − 1.4.4. (c) Prove that for each σ ∈ Aut(Sn ) σ : (1 2) → (a b2 ).. Prove that if G is an abelian group of order pq. Deduce that Inn(G) Proof. bn ∈ {1.21 – in the proof of Sylow’s Theorem so this line of reasoning is not circular. Suppose that p |G| and let P ∈ Sylp (G) with |P | = pk . (d) Show that (1 2). . then σ( r ) = r . 4. r has at most 2 possible images and s has at most 4 possible images (r and s are the usual generators).4. Automorphisms. 2. (1 n) generate Sn and deduce that any automorphism of Sn is uniquely determined by its action on these elements. we have the desired isomorphism. we must have σ(s) ∈ s. 4.26 DAVID S. The Sylow Theorems. b3 .18. 4. Thus xy = G and therefore G is cyclic. Deduce that |Aut(D8 )| ≤ 8..11. So |Aut(D8 )| ≤ 2 · 4 = 8.4. then k | p and k | q.2.. Proof (a). 4. 4. (a) Prove that the automorphism group of a group G permutes the conjugacy classes of G.. Deduce that any automorphism of Sn sends transpositions to transpositions. 5. 1 2 1 0 . 1 Since the upper diagonal matrices have nonzero diagonal elements and any upper right entry. n3 ∈ {1.30. 4. we must have n11 = 1. b.5. prove that P is Syl7 (G))\1 ≥ 48. But since n11 ≡ 1 mod 11 and n11 | 595. H and H K. G. Thus (p − 1)2 p | NG (P1 ). Note 56 = 23 · 7. Therefore np = p + 1. Sylow’s theorem states that np = [G : NG (P1 )]. for γ ∈ Fp \ {0. Suppose n2 . then P is characteristic in H. P2 = 1 γ 0 1 ∈ Sylp (G). then g P = P so g ∈ NG (P ) = H.5.5. of order 56 has a normal Sylow p-subgroup for some prime p dividing its order. Let P be a Sylow p-subgroup of H and let H be a subgroup of K. Note 200 = 23 · 52 . Note |GL2 (Fp )| = (p − 1)2 p(p + 1) and np ≡ 1 mod p. 4. we have a 0 Now if γ ∈ Fp \ {0. 4. Proof. 4. then P ∈ Sylp (H) and P is characteristic in H. 1 1 0 0 . But n5 ∈ {1. 4. The Simplicity of An . Proof. 1} we get a 0 b c −1 b c −1 = a−1 0 −a−1 bc−1 . Prove that the number of Sylow p-subgroups of G = GL2 (Fp ) is p + 1. we must have np ≥ p + 1. Note 168 = 23 · 3 · 7. These are clearly 1 1 . Thus if H g = H. we must have n7 = 8. then NG (H) = H. Simple calculations show that for a. 4}. so np | p + 1. Proof. Proof.5. Prove that if |G| = 6545 then G is not simple. But 1 0 all generate different subgroups of order 3. 4. If P H. then P K. 2 Proof. Therefore NG (H) = H. How many elements of order 7 must there be in a simple group of order 168? Solution. so n2 or n7 is equal to 1. 4.32. Since n3 ≡ 1 mod 3. Since n7 ≡ 1 mod 7 and n7 | 24. Note |SL2 (F3 )| = 24 so if P ∈ Syl3 (SL2 (F3 )) then |P | = 3.27 Solution.13. Now we show that the upper triangular matrices are a subgroup of NG (P1 ) with order (p − 1)2 p. n7 > 1. Therefore there are 6 · 8 = 48 elements of order 7. . Now if P ∈ Sylp (G) and H = NG (P ).40. c ∈ Fp . c−1 1 0 γ 1 a 0 b c = 1 0 a−1 γb − a−1 bc−1 b . Prove that a group.5.6. Since P is characteristic in H and H K. Since P1 = 1 0 γ 1 .18. 2 1 2 . 1}. the order must be (p − 1)2 p. Deduce that if P ∈ Sylp (G) and H = NG (P ). n7 > 1.19. 6} and since 6 8 we must have n5 = 1. Then ( Syl2 (G))\1 ≥ 14 and ( all distinct elements. Note 6545 = 5 · 7 · 11 · 17. If P normal in K. Prove that a group of order 200 has a normal Sylow 5-subgroup. Since the group is simple. In each of (a) to (e) give an example of a group with the specified properties: (a) an infinite group in which every element has order 1 or 2 (b) an infinite group in which every element has finite order but for each positive integer n there is an element of order n (c) a group with an element of infinite order and an element of order 2 (d) a group G such that every finite group is isomorphic to some subgroup of G (e) a nontrivial group G such that G ∼ G × G. so there are pn − 1 many elements of order p. respectively. Z2 and the isomorphism between them. = Solution (a).. FOOTE 5. xy p−1 . each distinct subgroup intersects trivially.e.1. Prove that the distinct subgroups of E of order p are x .1. (xy)p = xp y p = 1. then we are done. Set E = A × B so that E is the elementary abelian groups of order p2 : Ep2 . (Think of A ∗ B as the direct product of A and B “collapsed” by identifying each element xi ∈ Z1 with its corresponding element yi ∈ Z2 . Solution (d). Prove that Z4 ∗ D8 ∼ Z4 ∗ Q8 . Assume Z(A) contains a subgroup Z1 and Z(B) contains a subgroup Z2 with Z1 ∼ Z2 . 5. xy 2 . 5. But xp = 1. . Solution. Z2 = r2 and the isomorphism is x2 → r2 ) and let Z4 ∗ Q8 be the central product of Z4 and Q8 which identifies x2 and −1. Let D8 = r. (note that there are p + 1 of them). Let p be a prime.1. and that these images intersect in a central subgroup isomorphic to Z1 . DUMMIT AND RICHARD M. S2 × S3 × · · ·. 5. (xy k )p = xp (y k )p = 1. Direct and Semidirect Products and Abelian Groups 5. G = Z2 × Z2 × · · ·. Find a formula for the number of subgroups of order p in the elementary abelian group Epn . 5. respectively.) (a) Prove that the images of A and B in the quotient group A ∗ B are isomrophic to A and B. . Solution (e). y p = 1..18. Let A and B be groups. Proof (b).1.12. yi ) | xi ∈ Z1 xy . Note that these p + 1 subgroups are all distinct and nontrivial. So there are (pn − 1)/(p − 1) = pn−1 + pn−2 + · · · + 1 many distinct subgroups of order p. Let A and B be two cyclic groups of order p with generators x and y. Proof. s and Q8 = i.. Z × Z2 . y and is denoted by A ∗ B – it is not unique since it depends on Z1 .10. = Proof (a). so if the pth power of each generator is the identity element.11. The direct sum Z2 × Z3 × · · ·. Find |A ∗ B|. Solution (c). The direct sum Z2 × Z2 × · · ·. A central product of A and B is a = quotient (A × B)/Z where −1 Z = (xi . Let Z4 ∗ D8 be the central product of Z4 and D8 which identifies x2 and r2 (i. Solution (b). Let this isomorphism be given by the map xi → yi for all xi ∈ Z1 . (b) Let Z4 = x . j be given by their usual generators and realtions.1. Example 3 in this section shows that there are p + 1 subgroups of order p.28 DAVID S.. Let p be a prime and let n ∈ Z+ . Direct Products. By Lagrange’s theorem. Every nonidentity element has order p. Z1 = x2 . . Since An is simple. . Since 270 = 2 · 33 · 5. . while A/im ϕ is of type (¯1 . we must have An = Sn . 5.5. i i Proof (c). . (a) Prove that ϕ is a homomorphism.29 5.e.. 5. This map is clearly well-defined. 3. . Solution (b).2. . . have the same rank as A) and prove these groups are both isomorphic to the elementary abelian group. Sn is characteristic in Sn . (b) Describe the image and kernel of ϕ in terms of the given generators. = 5. Proof. Since ϕ : A → A and ker ϕ| Therefore A/im ϕ ∼ Epn . βn ) | βi = (xγ )p . Prove that An is the commutator subgroup of Sn for all n ≥ 5. . with each element to the power of p clearly one.2. an ) so also has a ¯ rank n. Let p be a prime and let A = x1 × x2 × · · · × xn be an abelian p-group. = xi = p. . 3. .11. . γ ∈ Z} .2. 5. proposition 7 part (4) states Sn ≤ An . . (c) Prove both ker ϕ and A/im ϕ have rank n (i.2. 9. 3. . 5) with lists corresponding to respective orderings.3. Prove that if G = HK where H and K are characteristic subgroups of G with H ∩ K = 1 then Aut(G) ∼ = Aut(H) × Aut(K). where |xi | = pαi > 1 for all i.2. . The image of ϕ is given by ϕ(A) = {(β1 . Proof (a). and Sn is non abelian.2. 5. Also by proposition 7 = part (3). 3). kn ) so it has rank n. then |xγ | p so i i xp i αi −1 = p. (90. If (xγ )p = 1. γ ∈ Z} . The elementary divisors of an abelian group of order 270 are either (2.7. . Sn An . the list of invariant factors is given by (30. In each of parts (a) to (e) give the lists of elementary divisors for all abelian groups of the specified order and then match each list with the corresponding list of invariant factors found in the preceding exercise: (a) order 270 Solution (a). 3). Thus xi /im ϕ| xi ∼ Zp . Define the pth-power map ϕ: A → A by ϕ : x → xp . Since An Sn and Sn /An ∼ Z2 is abelian. Deduce that if G is an abelian group of finite order then Aut(G) is isomorphic to the direct product of the automorphism groups of its Sylow subgroups. The Fundamental Theorem of Finitely Generated Abelian Groups. i The kernel of ϕ is given by {(k1 . 5) or (2. then im ϕ| xi p−1 = αi . The kernel of ϕ is of type (k1 . In each of parts (a) to (e) give the lists of invariant factors for all abelian groups of the specified order: (a) order 270 Solution (a). of order pn . . Epn . Thus n ker ϕ ∼ = i=1 xp αi −1 so it is isomorphic to Epn .2. kn )} where ki ∈ {xγ | (xγ )p = 1. so we calculate ϕ(xy) = (xy)p = xp y p = ϕ(x)ϕ(y). . 4. This exercise describes thirteen isomorphism types of groups of order 56. B ∈ G such that    a a12 · · · a1n b   a · · · a2n     A= . then it is the direct product of its (characteristic) Sylow subgroups. where D is the group = of all nonzero multiples of the identity matrix and U is the group of upper triangular matrices with 1’s down the diagonal. (c) Construct the following non-abelian groups of order 56 which have a normal Sylow 7-subgroup and whose Sylow 2-subgroup S is as specified: one group when S ∼ Z2 × Z2 × Z2 = two non-isomorphic groups when S ∼ Z4 × Z2 = one group when S ∼ Z8 = two non-isomorphic groups when S ∼ Q8 = three non-isomorphic groups when S ∼ D8 . (a) Prove that there are three abelian groups of order 56. Now if (φ|H . So f (φ) = (φ|H . φ|K )(ψ|H . φ|K ) = (φH . . and a11 = a22 = · · · = ann }. Therefore f shows Aut(G) ∼ Aut(H) × Aut(K).  . Prove that G ∼ D × U .5. DUMMIT AND RICHARD M. Recognizing Direct Products. Therefore Aut(G) is isomorphic to the direct product of the automorphism groups of the Sylow subgroups.30 DAVID S. 5. then ab = 1. Table of Groups of Small Order. If λI ∈ D for λ ∈ F . = ··· ··· . φK ) shows f is surjective. 5. If U ∈ U . φ|K ) = (ψ|H . where F is a field. (b) Prove that every group of order 56 has either a normal Sylow 2-subgroup or a normal Sylow 7-subgroup. FOOTE Proof. then let φ ∈ Aut(G) be given by φ|H = φH and φ|K = φK . Thus D and U are normal in G and G ∼ D × U . then for g = hk ∈ G. . Now let f : Aut(G) → Aut(H) × Aut(K) be given by f (φ) = (φ|H .. (φ ◦ ψ)|K ) = (φ|H . ψ ∈ Aut(G) and notice that φ|H ∈ Aut(H) and φ|K ∈ Aut(K) since H. b = (φ|H ◦ ψ|H .. It is clear that D ∩ U = 1. UB ∈ U such that A = a UA and B = b UB and A UB = (a UA ) U(b UB ) = ab (UA U UB ) ∈ U. so φ = ψ.  b1n b2n   . φ|K ). Semidirect Products. ψ|K ) = f (φ)f (ψ). then note that there exist UA .  . . Now we calculate f (φ ◦ ψ) = ((φ ◦ ψ)|H . B =  . If φ = ψ.14. 5. be the group of upper triangular matrices all of whose diagonal entries are equal. . Now let A.2. ψ|K ) so f is well-defined. Now if φH ∈ Aut(H) and φK ∈ Aut(K). Let φ. K are characteristic in G. then (φ|H . Let G = {(aij ) ∈ GLn (F ) | aij = 0 if i > j. φ|K ) = (ψ|H . ψ|K ). Proof.3. 5. a and note if AB = I. φ(hk) = φ(h)φ(k) = φ|H (h)φ|K (k) = ψ|H (h)ψ|K (k) = ψ(h)ψ(k) = ψ(hk). then AλIB = λAIB = λAB = λI.5. φ|K ◦ ψ|K ) b12 b . = 5.    . = If G is finite abelian.7. so f is a homomorphism. 6. The invariant factors are given by (23 · 7). Let P1 . P/Z.3. Suppose G is nilpotent so theorem 3 states G ∼ P1 ×· · ·×Ps for Pi ∈ Sylpi . If n7 > 1. Prove that if S is the Sylow 2-subgroup of G then S ∼ Z2 × Z2 × Z2 . 2 If S ∼ Z4 × Z2 . Therefore [G : M ] = p.1. Let |P | = pn with P non-abelian. where Aut(P ) ∼ Z6 . then ker ϕ ∈ Z4 . then the map ϕ : P → Aut(S) is non-trivial so that P is not normal. Thus there are two isomorphism types. exercise 6.4. Now suppose there exists = U P with U ∼ Zp × Zp for all k < n. = (e) Prove that there is a unique group of order 56 with a non-normal Sylow 7-subgroup. If G is finite prove that G is nilpotent if and only if it has a normal subgroup of each order dividing |G|. = ∼ Z8 . Since p | |Z(P )|. we have P ∩ S = 1 and P G so G is a semidirect product. Proof (d). Then each Sylow subgroup is normal in G. Since np ≡ 1 mod p. Thus ker ϕ = 1 and k ϕ(P ) ∼ Z7 . Since G is nilpotent. If P M ≤ P ≤ G so P = G by maximality of M .7 states it has a unique subgroup of each order dividing |G|. then shows it has a normal subgroup of each order dividing |G/M |. then the trivial map gives rise to the non-abelian Z7 × Q8 . = s 1 Theorem 1 states each Pi has a normal subgroup. 2. = = Thus |ker ϕ| = 4 and |ϕ(S)| = 2 for any such ϕ. then we consider the non-trivial homomorphisms ϕ : S → Aut(P ). the comment in part (c) states there = = are three isomorphism types. Proof (a). P2 ∈ Syl7 (G) so that for any = 3 = σ non-trivial ϕ : Pi → Aut(S). Z2 . 6. then the trivial map gives rise to the non-abelian Z7 × D8 .1.1. Since these kernels are not isomorphic. Thus ϕ(Pi ) ∈ Syl7 (Aut(S)). Let ϕ(P ) = x so if s ∈ S\1. if P ∈ Syl7 (G). M G. so group. But if n | |G|. Now we have the non-cyclic U P with |U | = p3 . If n = 2. S ϕ P1 ∼ S ϕ P2 .8.5. Two possible kernels of non-trivial maps = are r ∼ Z4 and r2 . Let M < G be a maximal subgroup of G. then ker ϕ = Z4 giving rise to one isomorphism type. If S = If S ∼ Q8 . let Z ≤ Z(P ) with |Z| = p. 2). Prove that if P is a non-cyclic p-group then P contains a normal subgroup U with U ∼ Zp × Zp . (2 · 7. = . S ∼ Z2 . Therefore P ϕ S is the unique isomorphism type. n7 > 1. Thus the non-cyclic quotient = P/Z with U ∼ Zp × Zp . Suppose that n2 . there is a U P with U = 6. Since G/M is nilpotent.31 (d) Let G be a group of order 56 with a non-normal Sylow 7-subgroup. Therefore every group of order 56 has either a normal Sylow 2-subgroup or a normal Sylow 7-subgroup. Since every = non-identity element in S must have order 2. Thus |P1 × · · · × Ps | = n with the desired i normality property. Now suppose G has a normal subgroup of each order dividing |G|. Counting | Syl2 (G)\1| + | Syl7 (G)\1| = 3 · 7 + 8 · 6 > 56 reaches a contradiction. If S ∼ D8 . Note for all of the Sylow 2-subgroup isomorphism types. Proof. s ∼ Z2 × Z2 .20. If S ∼ Z2 × Z2 × Z2 . Solution (c). Z28 × Z2 . ker ϕ = 1. (22 · 7. = Proof. then sx shows all elements in S\1 have the same order. Let p be an odd prime. If G has a unique subgroup of each order dividing |G|. If S ∼ Z2 . and Solvable Groups. Suppose that G is an abelian group of order 56 = 23 · 7. and is cyclic if and only if it has a unique subgroup of each order dividing |G|. so theorem 3 states G is nilpotent. we have n2 ≥ 3 and n7 ≥ 8. so P1 = P2 for some σ ∈ Aut(S). p-groups. Thus there are three abelian groups of order 56 given by Z56 . Z14 × Z2 × Z2 . then n = pk1 · · · pks . has a subgroup U = ∼ Zp × Zp . Prove that a maximal subgroup of a finite nilpotent group has prime index. then theorem 2. = 3 Proof (e). The non-trivial maps ϕ : S → Aut(P ) give = rise to a second isomorphism type. then P ∼ Zp × Zp since P is not cyclic. 2). Nilpotent Groups. Therefore by exercise 6. then proposition 5 shows G is cyclic. Proof. with order pki . 6. Prove that if p is a prime and P is a non-abelian group of order p3 then |Z(P )| = p and P/Z(P ) ∼ Zp × Zp . Deduce that for odd primes p a p-group that contains a unique subgroup of order p is cyclic. it is given by the elementary abelian group.3 G/M with P = p for p prime.1. Proof (b). Further Topics in Group Theory 6. Pi . = 5. then Aut(S) ∼ GL3 (F2 ) with |Aut(S)| = 7 · 6 · 4 = 168. If G is cyclic.1. DUMMIT AND RICHARD M. Thus |Z(P )| = p. p2 . Now N = {1. Solution. The same argument shows S4 = A4 . Since S4 = A4 and S4 = A4 . then P/Z(P ) is cyclic contradicting P being non-abelian.12.32 DAVID S.1. = 6. 1 Since s−1 a−1 sa is an even permutation. we can choose s ∈ S4 such that s−1 a−1 s = αa 1 2 1 2 so that α ∈ S4 . Since |P | = p3 we know P is nilpotent. . (14)(23)} is the only proper normal subgroup of A4 . Thus |Z(P )| ∈ p. (13)(24). Since P/Z(P ) is not cyclic. we have found the lower central series for S4 . But if α ∈ A4 . FOOTE Proof. (12)(34). Thus Ai = N for all i so 4 this is the lower central series for A4 . If |Z(P )| = p2 . Z(P ) = P . S4 ≤ A4 . we must have P/Z(P ) ∼ Zp × Zp . Since P is a p-group Z(P ) = 1. since P is non-abelian. The upper central series of A4 is Zi (A4 ) = 1 and of S4 is Zi (S4 ) = 1 for all i since the center of S4 is trivial. Find the upper and lower central series for A4 and S4 . 3. Show by example that the converse if false. Exercise 7. show that the element a ∈ Z/nZ is nilpotent if and only if every prime divisor of n is also a divisor of a. n ∈ Z+ . is the set of all elements which commute with every element of R). Solution. (a) Show that if n = ak b for some integers a and b then ab is a nilpotent element of Z/nZ. Proposition 7. (b). Any finite integral domain is a field. Proposition 7. Let x be a nilpotent element of the commutative ring R (cf. (a) Prove that x is either zero or a zero divisor. Decide which (a) the set of all (b) the set of all (c) the set of all (d) the set of all (e) the set of all (f) the set of all of the following are subrings of the ring of all functions from the closed interval [0. (b) If a ∈ Z is an integer. In particular. Prove that if u is a unit in S then u is a unit in R.5.1. (a). (3) (−a)(−b) = ab for all a. Prove that any subring of a field which contains the identity is an integral domain.e. (c). Let F be a field and let R ⊆ F be a subring of F with 1 ∈ R. Proof. every element of R is a unit. b and c are element of any ring with A not a zero divisor.12. determine the nilpotent elements of Z/72Z explicitly. (2) (−a)b = a(−b) = −(ab) for all a. b ∈ R.14.3.1. then either a = 0 or b = c. Then (1) 0a = a0 = 0 for all a ∈ R.1. 1] polynomial functions functions which have only a finite number of zeros.13). Proof. . Assume a. Prove that R contains no nonzero nilpotent elements.1. 7.2. functions f (x) such that f (q) = 0 for all q ∈ Q ∩ [0.1. 7. ∀r ∈ R} (i. if a. Decide which of the following (a)-(f) are subrings of Q: (a) the set of all rational numbers with odd denominators (when written in lowest terms) (b) the set of all rational numbers with even denominators (when written in lowest terms) (c) the set of nonnegative rational numbers (d) the set of squares of rational numbers (e) the set of all rational numbers with odd numerators (when written in lowest terms) (f) the set of all rational numbers with even numerators (when written in lowest terms).13.1. 7.6.1.7. Basic Definitions and Examples. (f).33 Part II – Ring Theory 7. where m. Let R be a ring. But Z ⊆ Q is a subring of Q and 2 is not a unit. Introduction to Rings 7. In particular. Prove that the center of a division ring is a field. Proof. b ∈ R. (4) if R has an identity 1.1. together with the zero function functions which have an infinite number of zeros functions f such that limx→1− f (x) = 0 rational linear combinations of the functions sin nx and cos mx. If u is a unit in S. (a). 7. Then since R ⊆ F . An element x in R is called nilpotent if xm = 0 for some m ∈ Z+ . The center of a ring R is {z ∈ R | zr = rz. If ab = ac. Therefore there are no zero divisors in R.1.1. then either a = 0 or b = c. (e). Let R be a ring with identity and let S be subring of R containing the identity. 7.. But then v ∈ R and hence u is a unit in R. Corollary 7. Now consider 2 ∈ Q and notice that it is a unit because 1/2 ∈ Q. 7. c are any elements in an integral domain and ab = ac. 1] to R. then there is a v ∈ S such that uv = 1 = vu. (c) Let R be the ring of functions from a nonempty set X to a field F . then the identity is unique and −a = (−1)a. Solution. 7. Thus the converse is false. b. ν(y)} for all x. Let R be a commutative ring with 1. (b) Prove that for each nonzero element x ∈ K either x or x−1 is in R. Proof (a). (2) the units of R[x] are just the units of R. Proof (c). But this is a contradiction because f would have to divide b − b . Let p(x) = 2x3 − 3x2 + 4x − 5 and let q(x) = 7x3 + 33x − 4. we must have Of = S. of O contains the identity and has finite index f . Proposition 7. The set R = {x ∈ K × | ν(x) ≥ 0} ∪ {0} is called the valuation ring of ν. (b). Now we have a + bf ω ∈ S for all a. Let a + bω + Of = a + b ω + Of . √ 7. S. Matrix Rings. Then xxn−1 = 0 where xn−1 = 0 so x is a zero divisor. 7.2. Now let b ≡ b + α for 1 ≤ α < f and suppose ∃a. (c) Prove that 1 + x is a unit in R. Then (a + bω) − (a + b ω) ∈ Of and (a − a ) + (b − b )ω ∈ Of . Since S is closed under addition then bf ω ∈ S for all b ∈ Z.4. Let K be a field. But then S = f (ω + S) = f ω + S so f ω ∈ S. and Group Rings. Let u be a unit so vu = 1 and let x be our nilpotent element.34 DAVID S. (d) Deduce that the sum of a nilpotent element and a unit is a unit. let n be the smallest positive integer such that xn = 0. Proof (b). we have (rx)m = rm xm = 0. (c) Prove that an element x is a unit of R if and only if ν(x) = 0. Proof (a). Let R be an integral domain and let p(x). |ω + S| divides f . Since 1 ∈ S then a ∈ S for all a ∈ Z because S is closed under addition. b ∈ Z} is a subring of O containing the identity. Examples: Polynomials Rings. Proof. 7. Observing the calculation shows that 1 + x is a unit. For any positive integer f prove that the set Of = Z[f ω] = {a + bf ω | a. DUMMIT AND RICHARD M. and let O be the ring of integers in the quadratic field Q( D). a ring R is called a discrete valuation ring if there is some field K and some discrete valuation ν on K such that R is the valuation ring of ν. FOOTE (b) Prove that rx is nilpotent for all r ∈ R. Prove that [O : Of ] = f (index as additive abelian groups). A discrete valuation on K is a function ν : K × → Z satisfying (i) ν(ab) = ν(a) + ν(b) (ii) ν is surjective. Thus v(u + x) is a unit and hence u + x is a unit. (1 + x)(1 − x + x2 − x3 + · · · xm−1 ) = 1 Solution (d). The ring √ f is called the order O √ of conductor f in the field Q( D). Thus a + bω + Of and a + b ω + Of are the same coset if and only if b ≡ b mod f hence [O : Of ] = f . Proof (c). and (c) compute p(x) + q(x) and p(x)q(x) under the assumption that the coefficients of the two given polynomials are taken from the specified ring (where the integer coefficients are taken mod n in parts (b) and (c)): . then a + bf ω ∈ O so Of ⊆ O. So b − b = b f thus b ≡ b mod f . Suppose x = 0 and since xm = 0. Thus rx is nilpotent for all r ∈ R.1. (a) Prove that R is a subring of K which contains the identity.23. Since |Of | = |S|. q(x) be nonzero elements of R[x].1. The ring of integers O is called the maximal order in Q( D).1. Since bf ∈ Z.2. Let a + bf ω ∈ Of . Suppose that a subring. 7. a such that a + bω + Of = a + b ω + Of . In each of parts (a). In general. Proof (b).26. Then part (b) shows vx is nilpotent and part (c) shows 1 + vx is a unit. and (iii) ν(x + y) ≥ min {ν(x). Prove conversely that a subring of O containing the identity and having finite index f in O (as additive abelian group) is equal to Of . (3) R[x] is an integral domain. Now (a + bf ω)(c + df ω) = ac + (ad + bd)f ω + bdf 2 ω 2 ∈ Of and (a + bf ω) − (c + df ω) = (a − c) + (b − d)f ω ∈ Of . Let D be a squarefree integer. Since |O/S| = f . y ∈ K × with x + y = 0. b ∈ Z so Of ⊆ S. Then (1) degree p(x)q(x) = degree p(x) + degree q(x). Thus Of is a subring of O. Since R is commutative. p(x) + q(x) = x3 − x2 + x − 1 p(x)q(x) = x5 − x Solution (c). s entry is aqr and all other entries are zero. α + β = 6(1) + 3(1 2) − 3(2 3) + 14(1 2 3) − 7(1 3 2) . Let S be a ring with identity 1 = 0. (c) R = Z/3Z.35 (a) R = Z. Solution (a). 7. Solution (c). Compute the following elements: (a) α + β. for all A ∈ Mn (R). (e) α2 . (d) βα. then the (j. p(x) + q(x) = x p(x)q(x) = 2x6 + x4 + x3 + 2x 7. Proof. i)th element of ZEji is zjj . Now we observe that the (j.2.7).2. (c) αβ. (c) Deduce that Epq AErs is the matrix whose p. Let n ∈ Z+ and let A be an n × n matrix with entries from S whose i. Consider the following elements of the integral group ring ZS3 : α = 3(1 2) − 5(2 3) + 14(1 2 3) and β = 6(1) + 2(2 3) − 7(1 3 2) where (1) is the identity of S3 . Proof (a).2. (b) 2α − 3β. (b) R = Z/2Z. i)th element of Eji Z is zii and the (j. (a) Prove that Eij A is the matrix whose ith row equals the j th row of A and all other rows are zero. then AZ = ArI = rAI = rIA = ZA.6. j)th element of ZEji is 0. Let Eij be the element of Mn (S) whose i. If i = j. (b) Prove that AEij is the matrix whose j th column equals the ith column of A and all other columns are zero. Thus Z must be a scalar matrix. If Z = rI. j entry is 1 and whose other entries are all 0. Use Exercise 7.10. Proof (b). so zii = zjj .1. p(x) + q(x) = 9x3 − 3x2 + 37x − 9 p(x)q(x) = 14x6 − 21x5 + 94x4 − 142x3 + 144x2 − 181x + 20 Solution (b). so zij = 0. Exercise 7. Let [zij ] = Z ∈ Z(Mn (R)) so ZEij = Eij Z for all Eij . 7. j entry is aij . j)th element of Eji Z is zij and the (j.6.7.2. Solution (a). Prove that the center of the ring Mn (R) is the set of scalar matrices (cf. . 2α − 3β = −18(1) + 6(1 2) − 16(2 3) + 28(1 2 3) + 21(1 3 2) Solution (c). α2 = 34(1) − 70(1 2) − 28(1 3) + 42(2 3) − 15(1 2 3) + 196(1 3 2) 7.7). Proposition 7. then the map R → R/I defined by r →r+I (r + I) × (s + I) = (rs) + I is a surjective ring homomorphism with kernel I (this homomorphism is called the natural projection of R onto R/I). Conversely. . Ring Homomorphisms and Quotient Rings. Let G = {g1 . if I is any subgroup such that the above operations are well defined. (1) (The First Isomorphism Theorem for Rings) If ϕ : R → S is a homomorphism of rings.6. . the image of ϕ is a subring of S and R/ ker ϕ is isomorphic as a ring to ϕ(R). Then A + B = {a + b | a ∈ A. Let X = x1 g1 + · · · + xn gn for xi ∈ R. Furthermore. 7.2.36 DAVID S. βα = −108(1) + 18(1 2) + 63(1 3) − 51(2 3) + 84(1 2 3) + 6(1 3 2) Solution (e). Thus every ideal is the kernel of a ring homomorphism and vice versa. ker ϕ is closed under multiplication by elements from R. Let R and S be rings and let ϕ : R → S be a homomorphism. (2) The kernel of ϕ is a subring of R. αβ = −108(1) + 81(1 2) − 21(1 3) − 30(2 3) + 90(1 2 3) Solution (d). s ∈ R. (2) If I is any ideal of R.12. Proof. (1) The image of ϕ is a subring of S. gn } be a finite group. A ∩ B is an ideal of A and A + B/B ∼ A/(A ∩ B).7. = .e.8.1.3. Exercise 7. Let R be a ring. b ∈ B} is a subring of R. Then n   gi gj =gk  xj  gk n NX = k=1 n    xj  gk = k=1 n j=1   gi gj =gk  xi  gk = k=1 = XN. Theorem 7. Then the (additive) quotient group R/I is a ring under the binary operations: (r + I) + (s + I) = (r + s) + I and for all r. Proposition 7. then the kernel of ϕ is an ideal of R.5. Prove that the element N = g1 + g2 + · · · + gn is in the center of the group ring RG (cf. FOOTE Solution (b). then I is an ideal of R. . Let R be a ring and let I be an ideal of R. if α ∈ ker ϕ then rα and αr ∈ ker ϕ for every r ∈ R. Theorem 7. i.. (1) (The Second Isomorphism Theorem for Rings) Let A b ea subring and let B be an ideal of R. DUMMIT AND RICHARD M. Similarly to the proof in part (b). Furthermore. Let R be the ring of all functions from X into Z/2Z. Let X be a nonempty set and let P(X) be the Boolean ring of all subsets of X defined in exercise 21 of section 1. a1 Db1 b1 a1 a2 Db2 b2 a2 a (D − 1)b/4 b a+b | a. so we show it is closed under multiplication. Proof (c). For each A ∈ P(X) define the function χA : X → Z/2Z by χA (x) = 1. Now let √ √ a1 + b1 D = a2 + b2 D. then ϕ maps a + b D to this element. / χA is called the characteristic function of A with values in Z/2Z. We calculate √ √ √ ϕ((a1 + b1 D) + (a2 + b2 D)) = ϕ((a1 + a2 ) + (b1 + b2 ) D) = = a1 + a2 b1 + b2 D(b1 + b2 ) a1 + a2 a1 Db1 b1 a1 + a2 Db2 b2 a2 √ √ √ ϕ((a1 + b1 D)(a2 + b2 D)) = ϕ((a1 a2 + b1 b2 D) + (a1 b2 + a2 b1 ) D) = = so ϕ is a homomorphism of rings. .3. b ∈ Z .3. This is clearly a group under addition. so we show that S is closed under multiplication. 7. 0.37 (2) (The Third Isomorphism Theorem for Rings) Let I and J be ideal of R with I ⊆ J. prove that the set and is isomorphic to the quadratic integer ring O. a b Db a is √ √ (b) If D is not a perfect square in Z prove that the map ϕ : Z[ D] → S defined by ϕ(a + b D) = a ring isomorphism. Now if isomorphism. Let D be an integer that is not a perfect square in Z and let S = (a) Prove that S is a subring of M2 (Z). S is clearly a group under addition. The correspondence A ↔ A/I is an inclusion preserving bijection between the wset of subrings A of R that contain I and the set of subrings of R/I. (c) If D ≡ 1 mod 4 is squarefree. But a1 (D − 1)b1 /4 b1 a 1 + b1 a2 (D − 1)b2 /4 = b2 a2 + b2 a1 a2 + (D − 1)a1 b2 /4 (D − 1)(a1 b2 + b1 b2 )/4 a1 b2 + a2 b1 + b1 b2 (D − 1)b1 b2 /4 + a1 a2 + a1 b2 + a2 b1 + b1 b2 a Db b a a1 a2 + b1 b2 D a1 b2 + a2 b1 D(a1 b2 + a2 b1 ) a1 a2 + b1 b2 D a1 Db1 a1 Db1 b1 a1 b1 a1 = a2 Db2 a2 Db2 b2 a2 b2 . = (3) (The Fourth Isomorphism Theorem for Rings) Let I be an ideal of R. so a1 = a1 .15. A (a subring containing I) is an ideal of R if and only if A/I is an ideal of R/I. Prove that the map P(X) → R defined by A → χA is a ring isomorphism. b ∈ Z is a subring of M2 (Z) Proof (a). this is isomorphic to the quadratic integer ring O.12. Proof (b). But = a1 a2 + Db1 b2 a1 b2 + a2 b1 D(a2 b1 + a1 b2 ) Db1 b2 + a1 a2 so S is a subring of M2 (Z). a Db b a | a. x ∈ A. 7. Then J/I is an ideal of R/I and (R/I)/(J/I) ∼ R/J. b1 = b2 and thus a2 √ ∈ S. x ∈ A. Therefore ϕ is an so this is a subring of M2 (Z). So ϕ(A) = χA = f and ϕ is surjective. B ⊆ X so that ϕ(A + B) = ϕ((A\B) ∪ (B\A)) = χ(A\B)∪(B\A) . . exercise 15. 7. .3. a2 .3. So χA = χB . (a) Prove tha tp(x) is a unit in R[x] if and only if a0 is a unit and a1 .  0.   1. x ∈ A ∩ B. Deduce that if S is an integral domain then every ring homomorphism from R to S sends the identity of R to the identity of S. an are nilpotent in R. there exists v ∈ R such that uv = 1R . DUMMIT AND RICHARD M. x ∈ X\(A ∩ B). with ϕ(u)−1 = ϕ(v) = ϕ(u−1 ). But then 0 = xn = (r + N (R))n = rn + N (R). Prove that if R is a commutative ring and N (R) is its nilradical (cf. If S is an integral domain.. Let A = A so that ϕ(A) = χA = χA = ϕ(A ). x ∈ B\A. section 1). Let R and S be nonzero rings with identity and denote their respective identities by 1R and 1S . =   1. 7.13) that an element x ∈ R is nilpotent if xn = 0 for some n ∈ Z+ . If ϕ(1R ) = 1S .3. Then there exists r ∈ R such that x = r + N (R).30. Let a = −1 so that Since 1 = 0. So ϕ(1R ) is a zero divisor in S. prove that N (R/N (R)) = 0.29) then zero is the only nilpotent element of R/N (R) i. Therefore ϕ is a ring isomorphism. 7. Let ϕ : R → S be a nonzero homomorphism of rings. Proof (b). Therefore ϕ(1R ) = 1S . (b) Prove that if ϕ(1R ) = 1S then ϕ(u) is a unit in S and that ϕ(u−1 ) = ϕ(u)−1 for reach unit u of R. thus A = B and ϕ is injective. x ∈ A\B. then (ϕ(1R ) − 1S )ϕ(1R ) = ϕ(1R )ϕ(1R ) − 1S ϕ(1R ) = ϕ(1R ) − ϕ(1R ) = 0. This implies rn ∈ N (R) so r ∈ N (R) and thus x = 0. Since an integral domain has a 1 = 0 and ϕ(1R ) = 0 if ϕ(R) = 0. Now let f = g ∈ R so there exist A. 7. x ∈ B\A. exercise 7. Similarly. So ϕ(u)ϕ(v) = ϕ(uv) = ϕ(1R ) = 1S . then R has characteristic 2. But   1. = ϕ(A)ϕ(B). Prove that the set of nilpotent element form an ideal – called the nilradical of R and denoted by N (R). x ∈ X\(A ∩ B) But simple calculations show this is ϕ(A) + ϕ(B).e. Proof (a).38 DAVID S. 7.   0.1.33. then ϕ(1R ) − 1S = 0. Proof.1. Now let A.   0. [See exercise 7. Proof. If u ∈ R is a unit.] 0 = (−1)2 + 1 = 1 + 1. x ∈ A\B. exercise 7. . Prove that a nonzero Boolean ring has characteristic 2 (cf. Let x ∈ R/N (R) with xn = 0 for some n ∈ N. Proof. Let R be a commutative ring. ϕ(A × B) = ϕ(A ∩ B) = χA∩B   0. so ϕ is well defined. then there are no zero divisors. Thus ϕ(u) is a unit.17. . Recall (cf.3. .14. χ(A\B)∪(B\A) =  0 x ∈ A ∩ B. B such that ϕ(A) = f = g = ϕ(B).3. Let p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 be an element of the polynomial ring R[x]. Now let f ∈ R so let A ⊆ X be the set such that f (A) = 1 and f (X\A) = 0. Assume R is commutative.3. (a) Prove that if ϕ(1R ) = 1S then ϕ(1R ) is a zero divisor in S. FOOTE Proof.29.27. 4. then ra = r a = 0. Assume R is commutative. But then (gi − 1)(g1 + · · · + gn ) = gi g1 + · · · gi gn − g1 − · · · − gn = g1 + · · · + gn − g1 − · · · − gn = 0.13. Proof (a). . Assume R is commutative.4. Proof (b). Thus ra = 0 and R is an integral domain. then pn pn 0 = ϕ( i=1 ai gi ) m =( i=1 a i )m pn .7. 7. Thus i=1 ai gi ∈ ker ϕ. . (a) Let p be a prime and let G be an abelian group of order pn .2. (x) is prime if and only if R is an integral domain. Since P is prime. Therefore ker ϕ = N (Fp G). we know that R = R/P is an integral domain. 7. proposition 7. If ϕ is the augmentation map.4.9.14. . a1 . 7. Assume R is commutative. . (2) Assume R is commutative. we must have ker ϕ ⊆ N (Fp G).3. a ∈ R\P . Suppose there exists m such that ( pn i=1 ai gi )m = 0. Proof (b). we must have ai = 0.4. 7. n n Now let g − 1 ∈ ker ϕ. Let ϕ : R → S be a homomorphism of commutative rings.10. exercise 7. Assume R is commutative. Similarly.13. .13.12 shows (x) is maximal if and only if R is a field. gn } be a finite group and assume R is commutative. an are nilpotenet elements of R.39 (b) Prove that p(x) is nilpotent in R[x] if and only if a0 . So if r. Prove that the augmentation ideal in the group ring RG is generated by {g − 1 | g ∈ G}. Since exercise 7. Since Fp \ {0} is a multiplicative group. .4. (b) Let G = {g1 . Proof.2 we may assume r = gi − 1. In a ring with identity every proper ideal is contained in a maximal ideal. .4. . Proof.12. . Corollary 7. . Prove that (x) is a maximal ideal if and only if R is a field.4. Corollary 7. pn i=1 7.3. Assume R is commutative.2 states ker ϕ is generated by {g − 1 | g ∈ G}.29).) − 1 = 0 in Fp G. Then R is a field if and only if its only ideals are 0 and R. Proposition 7. By exercise 7. Let ϕ : R[x] → R be given by ϕ(p(x)) = p(0) and notice ker ϕ = (x). The ideal M is a maximal ideal if and only if the quotient ring R/M is a field. Proof (a). . Let R be a commutative ring with 1. Thus R[x]/(x) ∼ R so by proposi= tion 7.10. Prove that if r is any element of the augmentation ideal of RG then r(g1 + · · · + gn ) = 0. Then (g − 1)p = g p + p(. Every maximal ideal of R is a prime ideal.11. (1) I = R if and only if I contains a unit. Then the ideal P is a prime ideal in R if and only if the quotient ring R/P is an integral domain. 7. Prove that the principal ideal generated by x in the polynomial ring R[x] is a prime ideal if and only if R is an integral domain.4. Proposition 7. Proposition 7. Properties of Ideals. Let R be a ring with identity 1 = 0. Prove that if G = σ is cyclic then the augmentation ideal is generated by σ − 1. Let I be an ideal of R. Prove that the nilradical of the group ring Fp G is the augmentation ideal (cf. Prove that if P is a prime ideal of R and P contains no zero divisors then R is an integral domain. If R is a field then any nonzero ring homomorphism from R into another ring is an injection. Proposition 7. Since π1 ◦ ϕ is surjective. then ϕ−1 (P ) is either R or a prime ideal of R. Now let rs ∈ ϕ−1 (P ). = 7. c1 ∈ {0. Apply this to the special case when R is a subring of S and ϕ is the inclusion homomorphism to deduce that if P is a prime ideal of S then P ∩ R is either R or a prime ideal of R. Since R is commutative. c1 such that n i=0 ai xi = c0 + c1 x for all n.4. n n−1 ai xi = i=0 i=0 n−1 ai xi + an xn = i=0 n−1 ai xi + an x(x + 1)(n−1)/2 = i=0 bi x i = c0 + c1 x. exercise 7. Let x2 + x + 1 be an element of the polynomial ring E = F2 [x] and use the bar notation to denote passage to the quotient ring F2 [x]/(x2 + x + 1).29). ϕ−1 (P ) is an ideal. (b) Prove that if M is a maximal ideal of S and ϕ is surjective then ϕ−1 (M ) is a maximal ideal of R. (b) Write out the 4 × 4 addition table for E and deduce that the additive group E is isomorphic to the Klein 4-group. 1}. Proof (a). So r or s is in ϕ−1 (P ) so ϕ−1 (P ) is prime or ϕ−1 (P ) = R. 1. The multiplication table in table 7. Prove that rad I is an ideal containing I and that (rad I)/I = N (R/I) (cf. Let r ∈ ϕ−1 (P ).40 DAVID S. R/ ker(π1 ◦ ϕ) ∼ S/M . F2 [x]/(x2 +x+1) ⊆ 0. so ϕ(r)ϕ(s) = ϕ(rs) ∈ P so ϕ(r) or ϕ(s) is in P because P is prime. rad I = r ∈ R | rn ∈ I for some n ∈ Z+ .2 shows E × ∼ Z3 .3. × (c) Write out the 4 × 4 multiplication table for E and prove that E is isomorphic to the cyclic group of order 3. x + 1 . Solution (b). = Solution (c). If n = 1 then n n−1 1 i=0 = a0 + a1 x. FOOTE (a) Prove that if P is a prime ideal of S then either ϕ−1 (P ) = R or ϕ−1 (P ) is a prime ideal of R. Now if n > 2 is odd. Give an example to show that this need not be the case if ϕ is surjective.1 shows F2 [x]/(x2 + x + 1) ∼ V4 . Let that n−1 i=0 n i=1 ai xi be an element of F2 [x]. Proof (a). Specifically if R ⊂ S and ϕ is the inclusion homomorphism. The addition table in table 7. Let π1 : S → S/M and π2 : R/ ker(π1 ◦ ϕ) be the canonical projection maps. So there exist c0 . x and x + 1. Suppose there exists c0 . ai xi = i=0 i=0 n−1 ai xi + an xn = i=0 n−1 ai xi + an (x + 1)n/2 = i=0 ci xi = c0 + c1 x. Let I be an ideal of the commutative ring R and define called the radical of I. c1 such ai xi = c0 + c1 x.15.4. Since c0 . (a) Prove that E has 4 elements: 0. But ker(π1 ◦ ϕ) = ϕ−1 (M ). Therefore E is a field. = 7. Proof (b). DUMMIT AND RICHARD M. Then if n is even. 1. But −1 ϕ (P ) = P ∩ R. Deduce that E is a field.30. So R/ϕ−1 (M ) is a field and ϕ−1 (M ) is maximal. x. so ϕ(rs) = ϕ(r)ϕ(s) ∈ P because ϕ(r) ∈ P and P is an ideal. let K be a field. then I = Ak for some k ≥ 0. 7.31. ν(−b)} = min {ν(a). Multiplication Table 0 0 1 x 1+x 0 0 0 0 1 x 1+x 0 1+x 1 x 0 0 1 x x 1+x 1+x 1 Table 7. Let k = min {ν(I)} with ν(r) = k. Thus b = ba−1 a ∈ (a). Deduce that (n) is a radical of Z if and only if n is a product of distinct primes in Z.4. Now suppose n is a product of distinct primes to the first power. (b) Prove that if I is any nonzero ideal of R. Every prime ideal of R is a primary ideal. A proper ideal Q of the commutative ring R is called primary if whenever ab ∈ Q and a ∈ Q then bn ∈ Q / for some positive integer n. so rad P = P . For each integer k ≥ 0 define Ak = {r ∈ R | ν(r) ≥ k} ∪ {0}. 7. either r or rn−1 is in P . (a) Prove that Ak is a principal ideal and that A0 ⊇ A1 ⊇ A2 ⊇ · · ·.4.4.e. Ak is nonempty. Following the notation of exercise 7. Then Ak = (r) = I. Proof (a). Thus α must contain all prime factors of n so α = 0 mod n. Since −1 ∈ R× .2. then αm contains all of the prime factors of n. so Ak ⊆ (a). If r ∈ R and a ∈ Ak then ν(ra) = ν(r) + ν(a) ≥ k and similarly ν(ar) ≥ k. An ideal Q of R is primary if and only if every zero divisor in R/Q is a nilpotent element of R/Q. So ν(a − b) ≥ min {ν(a).1. exercise 7. Note P ⊆ rad P . . Now let ν(a) = k by surjectivity of ν.1. ν(−1) = 0.. If Q is a primary ideal then rad (Q) is a prime ideal (cf. n is square free).41. An ideal I of the commutative ring R is called a radical ideal if rad I = I.41 Addition Table 0 0 1 x 1+x 1 x x 1+x 0 1 1+x 1+x x 1 0 0 1 1 0 x 1+x 1+x x Table 7. 7. (a) (b) (c) (d) The primary ideal of Z are 0 and (pn ).4. Also ν(ab) = ν(a) + ν(b) ≥ k so Ak is a subring. Then mp = 0 mod n and (mp)2 = 0 mod n which is a contradiction. Prove that 0 is a radical ideal in Z/nZ if and only if n is a product of distinct primes to the first power (i. ν(b)} ≥ k. If b ∈ Ak then ν(ba−1 ) = ν(b) − ν(a) ≥ 0 so ba−1 ∈ R. F2 [x]/(x2 + x + 1). Deduce that R is a local ring with unique maximal ideal A1 . Notice that 0 = ν(1) = ν(aa−1 ) = ν(a) + ν(a−1 ) so ν(a) = −ν(a−1 ). Establish the following facts about primary ideals. (b) Let n > 1 be an integer. Thus n is a product of distinct primes to the first power. From part (a) any proper ideal is contained in A1 so A1 is the unique maximal ideal of R. Since ν is surjective.26. Let r ∈ R so if rn ∈ P .30). where p is a prime and n is a positive integer. Therefore rad 0 = 0. Therefore Ak is a principal ideal and it is clear that A0 ⊇ A1 ⊇ A2 ⊃ · · ·. Suppose rad 0 = 0 and n = mp2 . let ν be a discrete valuation on K and let R be the valuation ring of ν.39. F2 [x]/(x2 + x + 1). So Ak is an ideal. If αm = 0 mod n. Proof (a). Induction shows r ∈ P . Proof (b). (a) Prove that every prime ideal of R is a radical ideal. Proof (b). does not contain any zero divisors and is closed under multiplication (i.16. Let S be any commutative ring with idenitty and let ϕ : R → S be any injective ring homomorhpism such that ϕ(d) is a unit in S for every d ∈ D. Rings of Fractions. Then re − sd = 0 and sf − te = 0. Let I ⊂ Z be a primary ideal so I = (n) for some n ∈ Z since every ideal of Z is principal. 0 d for any d ∈ D and the additive . ab ∈ D for all a. (2) (uniqueness of Q) The ring Q is the “smallest” ring containing R in which all elements of D become units. Note that d = de in Q for all e ∈ D. e) if and only if re = sd. / Proof (c).e. Since e ∈ D is neither zero nor a zero divisor we must have rf − td = 0. b ∈ D).. if D = R\ {0} then Q is a field. Let ab ∈ rad (Q) and a ∈ rad Q. It is immediate that this relation is reflexive and symmetric. DUMMIT AND RICHARD M. Let n = pa for p some prime. (1) every element of Q is of the form rd−1 for some r ∈ R and d ∈ D. If a field F contains a subring R isomorphic to R then the subfield of F generated by R is isomorphic to Q. l > 1. theorem 15. hence an equivalence relation. Then ab ∈ Q with ak ∈ Q and bl ∈ Q for k. Let ab ∈ Q with a ∈ Q. Let R be a commutative ring with identity 1 = 0. Then b is a zero divisor in / k k R so b = 0. since D is closed under multiplication. Proof. So ak = 0 thus every zero divisor is nilpotent in R. Done the equivalence class of (r. Let R be a commutative ring.e. d) by d : r = {(a. Thus b ∈ Q and Q is a primary ideal of R. d) ∼ (t. d ∈ D} and define the relation ∼ on F by (r. d) | r ∈ R.. Let P be a prime ideal of R with ab ∈ P .1.15 (cf. e) and (s. Therefore rad (Q) is a prime ideal.36). 7. f ). d) ∼ (s. d) ∼ (s.5. FOOTE Proof (a). Let ab ∈ (pk ) with a ∈ (pk ). Then there is an injective ring homomorphism Φ : Q → S such that Φ|R = ϕ.5. In particular. So n|pi shows that n = pk for / / some k ≥ 1. in the following sense. where the additive identity is inverse of a is −a . d r re Let Q be the set of equivalence classes under ∼. b ∈ D and rb = ad} .42 DAVID S. any ring containing an isomorphic copy of R in which all the elements of D become units must also contain an isomorphic copy of Q. Let D be any nonempty subset of R that does not contain 0. e) ∼ (t. (r. Now suppose every zero divisor in R is a nilpotent element. In other words.15. If a ∈ P then b ∈ P so P is a primary ideal. Corollary 7. Proof (b). Thus (n) = (pk ). so p|b. d d (3) multiplication is associative. Now consider (pk ) ⊆ Z. distributive and commutative. Prove theorem 7. The ring Q has the following additional properties. Let F = {(r. We now define an additive and multiplicative structure on Q: a c ad + bc a c ac + = and × = . 7. a ∈ (n) with (n) a primary ideal implies that pi ∈ (n). Since Q is primary then / / bnk ∈ Q so b ∈ rad (Q). Then there is a commutative ring Q with 1 such that Q contains R as a subring and every element of D is a unit in Q. Suppose (r. Proof (d). Thus pk |bk . b) | a ∈ R. so p ∈ (n). and (4) Q has an identity. Let a b = 0 in R = R/Q. r This proves ∼ is transitive. f ). Then pk |ab and pk a. Theorem 7. let R be an integral domain and let Q be the field of fractions of R. Multiplying the first of these equations by f and the second by d and adding them gives rf − td)e = 0. i. so (ab)n ∈ Q and am ∈ Q for all m ∈ Z+ . so bk ∈ (pk ) and / k (p ) is a primary ideal. b d bd b d bd In order to prove that Q is a commutative ring with identity there are a number of things to check: (1) these operations are well defined (2) Q is an abelian group under addition. We must show that (ad + bc)(b d ) = (a d + b c )(bd). d1 d2 d1 d3 . Hence multiplication of fractions is well defined. d3 ∈ Q so 1 2 3 ( r1 r2 r3 r1 r2 r3 × )× = × d1 d2 d3 d1 d2 d3 r1 r2 r3 = d1 d2 d3 r1 r2 r3 = × d1 d2 d3 r2 r3 r1 × ( × ). d r 0 r r 0 To show Q is an abelian group under addition note that d ∈ Q with d1 + d2 = d11d2 = d2 so there exists an r1 r2 r3 additive identity in Q. we see Q is an abelian group under addition. Since addition is clearly commutative in Q. Hence addition of fractions is well defined. then ( r2 r3 r1 d2 + r2 d1 r3 r1 + )+ = + d1 d2 d3 d1 d2 d3 r1 d2 d3 + r2 d1 d3 + r3 d1 d2 = d1 d2 d3 r1 r2 d3 + d1 d2 = + d1 d2 d3 r1 r2 r3 = + ( + ). d2 . which is the right hand side.. d2 . The left hand side of this equation is (ab )(cd ) substituting a b for ab and c d for cd gives (a b)(c d) which is the right hand side.e. 2 d d d d So Q is closed under inverse. bd d a b = a b (i. then −r d ∈ Q with r −r rd − rd 0 + = = . = d1 d2 d3 r1 r2 r3 d1 . d2 . i. The left hand side of this equation is ab dd + cd bb substituting a b for ab and c d for cd gives a bdd + c dbb . d1 d2 d3 r d Thus addition is associative. ab = a b) and c d = c d . r r r To show multiplication is associative. Now if ∈ Q. let d1 .e. let ( ∈ Q so r1 r2 r3 r1 d2 + r2 d1 r3 + )× = × d1 d2 d3 d1 d2 d3 r1 r3 d2 + r2 r3 d1 = d1 d2 d3 r1 r3 d2 r2 r3 d1 = × + × d1 d2 d3 d2 d1 d3 r1 r3 r2 r3 = × + × d1 d3 d2 d3 r1 r2 r3 r1 r2 d3 + r3 d2 ×( + )= × d1 d2 d3 d1 d2 d3 r1 r2 d3 + r1 r3 d2 = d1 d2 d3 r1 r2 d3 r1 r3 d2 = × + × d1 d2 d3 d1 d2 d3 r1 r2 r1 r3 = × + × . d3 ∈ Q. c c To check that multiplication is well defined assume a = a and d = d . Now if d1 .. We must show b b (ac)(b d ) = (a c )(bd).43 To check that addition is well defined assume ad+bc = a db +b c . d3 To show multiplication is distributive. Extend ϕ to a map Φ : Q → S by defining Φ(rd−1 ) = ϕ(r)ϕ(d)−1 for all r ∈ R. This map is well defined. let ∈ Q so r1 r2 r1 r2 × = d1 d2 d1 d2 r2 r1 = d2 d1 r2 r1 = × . if D = R\ {0}. e ∈ D. if d is represented by the fraction de then e e its multiplicative inverse is de . One then sees that every element of Q may be written as r · d−1 for some r ∈ R and some d ∈ D. since rd−1 = se−1 implies re = sd. Finally. d1 d d1 d d1 ∈ Q so Next we embed R into Q by defining rd where d is any element of D. The subring ι(R) of Q is therefore isomorphic to R. d for all d. d ∈ D. In particular. FOOTE r1 r2 d1 . r2 ∈ R so ι : R → Q by ι: r → ι(r1 + r2 ) = (r1 + r2 )d d r1 d r2 d = + d d = ι(r1 ) + ι(r2 ) Since rd d = re e ι(r1 × r2 ) = (r1 r2 )d d r1 d r2 d × = d d = ι(r1 )ι(r2 ). and then Φ(rd−1 ) = ϕ(r)ϕ(d)−1 = ϕ(s)ϕ(e)−1 = Φ(se−1 ). r2 d−1 ∈ Q. 2 So Φ is a ring homomorphism. notice that d d ∈ Q for any d ∈ D. Then 1 2 Φ(r1 d−1 + r2 d−1 ) = Φ((r1 d2 + r2 d1 )(d1 d2 )−1 ) 1 2 = ϕ(r1 d2 + r2 d1 )ϕ(d1 d2 )−1 = ϕ(r1 )ϕ(d2 )ϕ(d1 )−1 ϕ(d2 )−1 + ϕ(r2 )ϕ(d1 )ϕ(d1 )−1 ϕ(d2 )−1 = ϕ(r1 )ϕ(d1 )−1 + ϕ(r2 )ϕ(d2 )−1 −1 = Φ(r1 d1 ) + Φ(r2 d−1 ) 2 Φ(r1 d−1 r2 d−1 ) = Φ((r1 r2 )(d1 d2 )−1 ) 1 2 = ϕ(r1 r2 )ϕ(d1 d2 )−1 = ϕ(r1 )ϕ(d1 )−1 ϕ(r2 )ϕ(d2 )−1 −1 = Φ(r1 d1 )Φ(r2 d−1 ). Next note that each d ∈ D has a multiplicative inverse in Q: namely. It remains to establish the uniqueness property of Q.44 DAVID S. Assume ϕ : R → S is an injective ring homomorphism such that ϕ(d) is a unit in S for all d ∈ D. Now let r1 . Now let r1 d−1 . d2 d1 r d1 To show Q has an identity. DUMMIT AND RICHARD M. Now let r d rd r × = = . Φ is injective because rd−1 ∈ ker Φ implies r ∈ ker Φ ∩ R = ker ϕ. Furthermore. every nonzero element of Q has a multiplicative inverse and Q is a field. so ϕ(r)ϕ(e) = ϕ(s)ϕ(d). ι is injective because rd 0 ι(r) = 0 ⇔ = ⇔ rd2 = 0 ⇔ r = 0 d d because d is neither zero nor a zero divisor. ι(r) does not depend on the choice of d ∈ D. so ι is a ring homomorphism. . since ϕ is injective this forces r and hence also rd−1 to be zero. We henceforth identify each r ∈ R with ι(r) and so consider R as a subring of Q. d2 To show multiplication is commutative. Let F = Q( D) be a quadratic field with associated quadratic integer ring O and field norm N as in section 7. Thus a is a unit. −67 or −163. 8. Proposition 8. 8. Prove that O is not a Euclidean Domain with respect to any norm.5. In particular. theorem 7. Since x ∈ (α). If R is a Euclidean Domain then there are universal side divisors in R. Let m be the minimum integer in the set of norms of nonzero elements of R. then d = ud for some unit u. Let R be a Euclidean Domain and let a and b be nonzero elements of R.8(a). Proposition 8. Let R be a Euclidean Domain. If two elements d and d of R generate the same principal ideal.1.2. (a) Suppose D is −1. Proof. −3.e.1. and (2) the principal ideal (d) is the ideal generated by a and b.. then d is a greatest common divisor of a and b. where d is any nonzero element of I of minimum norm. and Unique Factorization Domains 8. Proof. i. r = 0.3. But now x + I = r + I. If a and B are nonzero element sin the commutative ring R such that the ideal generated by a and b is a principal ideal (d).4. (b) Suppose that D = −43. Let R be an integral domain and let D be a nonempty subset of R that is closed under multiplication. Let x + I be a coset of I. Euclidean. Let Q be the quotient field of R. Proposition 8. Then (1) d is a greatest common divisor of a and b. −2. 2 2 Let F = {f ∈ R | N (f ) < N (α)}. if d and d are both greatest common divisors of a and b. 8.15 states there exists an injective ring homomorhpism ψ : D−1 R → Q with ψ|R = ι. .8.1. Let R be an integral domain that is not a field. Let a ∈ R with N (a) = m.15 there exists an injective ring homomorhpism ι : R → Q with ι(R\ {0}) ⊆ Q× . D\ {0} is also a multiplicative set of R. Prove that the quotient ring Z[i]/I is finite for any nonzero ideal I of Z[i]. there exists q. Prove that O is a Euclidean Domain with respect to N . y = −217 − m · 527. Prove that the ring of fractions D−1 R is isomorphic to a subring of the quotient field of R (hence is also an integral domain). r ∈ R such that 1 = qa + r with N (r) < N (a) or r = 0. 8.1.10.45 7. Determine all integer solutions of the following equations: (a) (b) 17x + 29y = 31 (c) Proof (b). By exercise 8. Since R is a domain. In particular.1.5.2.1. √ 8. Proof. Since R is a Euclidean Domain. Principal Ideal. then r = 0 so N (r) < N (α).e. Deduce that a nonzero element of norm zero (if such an element exists) is a unit.. −7 or −11. Then if f = f1 + f2 i and α = a1 + a2 i we have f1 + f2 < a2 + a2 . Let R be an integral domain. Then there exists q.1. x = 372 + m · 899. So any solution to this equation has form for any m ∈ Z. Euclidean Domains.5. there are elements x and y in R such that d = ax + by. Since D\ {0} ⊆ R\ {0}. Z[i] is a Euclidean Domain. Every ideal in a Euclidean Domain is principal. r such that x + α = qα + r. if I is any nonzero ideal in the Euclidean Domain R then I = (d). d can be written as an R-linear combination of a and b. Since there 1 2 are only a finite number of (f1 .3. Since a is chosen to be minimum norm. Theorem 8. Let d = rn be the last nonzero remainder in the Euclidean Algorithm for a and b described at the beginning of this chapter. More precisely. So I = (α) for some α ∈ Z[i]. f2 ) ∈ Z × Z with this property. By theorem 7. Z[i]/I is finite for any nonzero ideal I of Z[i]. Proposition 8. (d) = (d ). then d = ud for some u ∈ R× . so any coset of I has a representative with norm / less than N (α). i. Notice that 372 · 17 − 217 · 29 = 31. Prove that every nonzero element of R of norm m is a unit.1. Since a and b are nonzero then e is also nonzero so e = ste and e(1 − st) = 0 implies 1 = st because R is a domain. . eb) and (a. In particular. 9. Prove that the rings F [x. Proof (a).b) | e. e be two least common multiples of a and b. Since R is a Euclidean Domain. there exists e ∈ R with (e) = (a) ∩ (b). Now suppose there is a largest principal ideal (e) contained in (a) ∩ (b). Then ∼ R[x]/(I) = (R/I)[x]. First note that ( (a. Since a | e and b | e we must have (e) ⊆ (a) and (e) ⊆ (b). Thus d M1 ≡ M dd mod N mod N mod N. Since (e) is the largest such principal ideal. ϕ(p(x. then (e ) ⊆ (a) ∩ (b). Proof (b). if I is a prime ideal of R then (I) is a prime ideal of R[x]. y]/(y 2 − x2 ) are not isomorphic for any field F . y] → R[y] be given by Let p(x. Proof. If a | e and b | e . q(x) are nonzero (2) the units of R[x] are just the units of R (3) R[x] is an integral domain. Since y 2 − x2 = (y − x)(y + x) then F [x. b) is the greatest common divisor of a and b. (a) Prove that a least common multiple of a and b (if such exists) is a generator for the unique largest principal ideal contained in (a) ∩ (b). Then e | e and e | e. Let R be an integral domain. Let M be an integer relatively prime to N and let d be an integer relatively prime to ϕ(N ). A least common multiple of a and b is an element e of R such that (i) a | e and b | e. and (ii) if a | e and b | e then e | e . y]/(y 2 − x2 ) is not a domain.1. Now let e. So we have (e) ⊆ ( (a.1. If a and b are nonzero elements of R then (a) ∩ (b) is the largest ideal contained in (a) ∩ (b).13. y)(y 2 − x)) = p(y 2 . Thus ab | (ea. But then e | e . Thus least common multiples are unique up to multiplication by a unit. Proof. Proposition 9.1.b) ). Then abx = eb and aby = ea ab ab ab so ab | eb and ab | ea. Prove that if M1 ≡ M d mod N then d M ≡ M1 mod N where d is the inverse of d mod ϕ(N ) : dd ≡ 1 mod ϕ(N ). y]/(y 2 − x) and F [x. Then (1) degree p(x)q(x) = degree p(x) + degree q(x) if p(x). Polynomial Rings 9. Not let ϕ : R[x.46 DAVID S. y)(y 2 − y 2 ) = 0.b) ) ⊆ (e).11.12 (A Public Key Code). Since (M.1.b) is a least common multiple of a and b. y)(y 2 − x) ∈ (y 2 − x) so that ϕ(p(x. Therefore (a. Then e | a and e | b.b) ) ⊆ (a) ∩ (b) so ( (a.1. Let R be a commutative ring with 1 and let a and b be nonzero elements of R. y). So part (a) shows e is a least common multiple of a and b. N ) = 1 we know that M ϕ(N ) ≡ 1 mod N . So e = se and et = e . If (e ) is a principal ideal such that (e ) ⊆ (a) ∩ (b) then e | a and e | b. where (a. ≡M ≡ M αϕ(N )+1 9. Let N be a positive integer. First suppose there exists a least common multiple e of a and b. Definitions and Basic Properties. then (e ) ⊆ (e). Let I be an ideal of the ring R and let (I) = I[x] denote the ideal of R[x] generated by I (the set of polynomials with coefficients in I). 8. where ϕ denotes Euler’s ϕ-function.2. Thus e | e and (e ) ⊆ (e) so (e) is the largest principal ideal contained in (a) ∩ (b). Proposition 9. ab ab Proof (c). Now let ax = e = by.b) . y)) = p(y 2 . DUMMIT AND RICHARD M. ab (c) Prove that in a Euclidean Domain the least common multiple of a and b is (a. FOOTE 8. (b) Deduce that any two nonzero elements in a Euclidean Domain have a least common multiple which is unique up to multiplication by a unit. . . . Let p(x1 . xn ) be a homogeneous polynomial of degree k in R[x1 . .14. . j be relatively prime integers. First we write m n p(x1 . If F is a field. This is a contradiction because a prime is not a unit. 9.4. Since F [x] is a unique factorization domain we can write m n i=1 pi (x) and 1 + p(x) = i=1 pαi (x) where each pαi (x) is irreducible. Then m n p(λx1 . Note that we can write p(x. Polynomial Rings Over Fields I. Proof. . Let p(x) = notice 1 + p(x) ∈ F [x]. .15. .47 Thus (y 2 − x) ⊆ ker ϕ. Proof.1. . xn ) = i=0 ai j=1 xj j . x2 . . . Prove that F [x] contains infinitely many primes. x2 . 9.2. .2. Corollary 9. Theorem 9. λxn ) = i=0 ai m (λxj )kj j=1 n = λk i=0 k ai j=1 xj j k = λ p(x1 . 9. y) = q(x. . . y). Let F be a field. . . . 9. So r = 0 and s = 0. then F [x] is a Principal Ideal Domain and a Unique Factorization Domain. Now let p(x. Proof. x2 . . then there are unique q(x) and r(x) in F [x] such that a(x) = q(x)b(x) + r(x) with r(x) = 0 or deg r(x) < deg b(x).5. . . 9. x2 . . .1. y]/(y 2 −x). y) ∈ ker ϕ so that 0 = ϕ(p(x. . . xn ]. . . λx2 . pn (x) are all of the primes in F [x]. Suppose on the other hand that p1 (x). xn ). The polynomial ring F [x] is a Euclidean Domain. xn ). . λxn ) = λk p(x1 . . Proof. y) = r(x)+ys(x)+q(x. λx2 . .2. . . if a(x) and b(x) are two polynomials in F [x] with b(x) nonzero. But now m 1= i=1 pαi (x) − p(x) m = pα1 (x) i=2 pαi (x) − p(x) pα1 (x) . Exhibit all the ideals in the ring F [x]/(p(x)). k where n j=1 kj = k. .17. Prove that the ideal (xi − y j ) is a prime ideal in R[x. . Therefore p(x. y].1. . y)) = p(y 2 . An ideal I in R[x1 .3. Specifically. where F is a field and p(x) is a polynomial in F [x].4. Prove that an ideal is a homogeneous ideal if and only if it may be generated by homogeneous polynomials. Prove that for all λ ∈ R we have p(λx1 . . xn ] is called a homogeneous ideal if whenever p ∈ I then each homogeneous component of p is also in I. y)(y 2 −x) by considering cosets of F [x. 9. y)(y 2 − x) so ker ϕ ⊆ (y 2 − x). Let F be a finite field. . . Let R be an integral domain and let i. But then 0 = ϕ(r(x)+ys(x)) = r(y 2 )+ys(y 2 ). qi (x) irreducible.48 DAVID S. If p(x) is reducible in F [x] then p(x) is reducible in R[x].6. In particular. Then p(x) is irreducible in R[x] if and only if it is irreducible in F [x]. y 2 . y]/(x2 . Let R be a Unique Factorization Domain with field of fractions F and let p(x) ∈ R[x]. (p(x)) ⊆ I. y 2 . y 2 . . Corollary 9. Let p(x) = an xn + an−1 xn−1 + · · · + a0 be a polynomial of degree n with integer coefficients. Theorem 9. for some α ∈ F . then p(x) has no roots in Q. FOOTE Proof. y))2 + (x2 .2. Proposition 9. Then α i=1 ri (x) i=1 qi (x) = i=1 pi (x). Using division of polynomials we calculate x3 + 4x2 + x − 6 = (50x2 + 45x − 95)(x/50 + 31/500) + (31/10 x2 + 1395/500 x − 2945/500) So we write the greatest common divisor as 31/10 x2 + 1395/500 x − 2945/500 = 50x2 + 45x − 95 = (31/10 x2 + 1395/500 x − 2945/500)(500/31).7.10.. 1 because if there is no constant term then every term has even degree and if there is a constant term. By the lattice isomorphism theorem. Solution.8. it is also a principal ideal domain so let I = s t (q(x)). y) + 1 so this is in 1. s ∈ F such that rA(x) = a(x) and sB(x) = b(x) both lie in R[x] and p(x) = a(x)b(x) is a factorization in R[x]. Therefore (q(x)) = ( i=1 qi (x)) = ( i=1 pσ(i) (x)). So α(x.11. Since (p(x)) ⊆ (q(x)). Proof (d). Proposition 9.5 (Gauss’ Lemma). if p(x) = A(x)B(x) for some nonconstant polynomials A(x). let F be its field of fractions and let p(x) ∈ R[x].e. 9. The highest power of x and y is 1 along with the largest coefficient. Let r(x) = i=1 ri (x) and q(x) = i=1 qi (x) s t n for ri (x). Proposition 9. if p(x) is a monic polynomial with integer coefficients and p(d) = 0 for all itnegers d dividing the constant tterm of p(x). there is an α ∈ F with p(α) = 0.B(x) ∈ F [x]. then there are nonzero elements r. i. More precisely. where each pi (x) is irreducible. R is a Unique Factorization Domain if and only if R[x] is a Unique Factorization Domain. Corollary 9. p(x) = r(x)q(x) for some r(x) ∈ F [x]. Suppose the greatest common divisor of the coefficients of p(x) is 1. y) has no constant term. Proposition 9. If r/x ∈ Q is in lowest terms (i. y) + (x2 . If R is a Unique Factorization Domain. Then p(x) has a facto of degree one if and only if p(x) has a roote in F .4. y) = β(x. The elements with α = 0 are the elements with no constant term. Let R be a Unique Factorization Domain. 9. 9. let α(x. Since F is a field. 9. if p(x) is a moni polynomial that is irreducible in R[x]. x5 − 6x + 5 = (x3 + 4x2 − x − 6)(x2 − 4x + 15) − (50x2 + 45x − 95) = (x3 + 4x + x − 6)[(x2 − 4x + 15)(x/50 + 31/500) − 1] − (x5 − 6x + 5)[x/50 + 31/500]. DUMMIT AND RICHARD M.e. F [x] is a Euclidean domain and in particular a unique factorization domain. In particular. Determine the characteristics of each of these rings.2. Determine the greatest common divisor of a(x) = x3 + 4x2 + x − 6 and b(x) = x5 − 6x + 5 in Q[x] and write it as a linear combination in Q[x] of a(x) and b(x). then a polynomial ring in an arbitrary number of variable with coefficients in R is also a Unique Factorization Domain. 2) ∈ 0. r and s are relatively prime integers) and r/s is a root of p(x). Since F [x] is a Euclidean domain. y) + 1)2 = β(x.3. So let n p(x) = i=1 pi (x). 2). then p(x) is irreducible in F [x]. y) + 1 where β(x. Let α = α(x. y)2 + 2β(x. A polynomial of degree two or three over a field F is reducible if and only if it has a root in F . then r divides the constant term and s divides the leading coefficient of p(x). Irreducibility Criteria.10. and I is an ideal of F [x]. Let I/(p(x)) ⊆ F [x]/(p(x)) be an ideal.. Describe the ring structure of the following rings: (d) Z[x. y)2 = (β(x. Polynomial Rings That are Unique Factorization Domains.9. So there exists σ ∈ Sn t t such that αi qi (x) = pσ(i) (x) for all 1 ≤ i ≤ t. 2) so α2 = (α(x.6. αi ∈ F . Let F be a field and let p(x) ∈ F [x]. Show that α2 = 0 or 1 for every α in the last ring and determine those elements with α2 = 0. 1. Proof. Let I be a proper ideal in the integral domain R and let p(x) be a nonconstant monic polynomial in R[x].14 (Eisenstein’s Criterion for Polynomial Ring over Z). Proof. as usual. We shall prove in section 13. For those that are reducible. . where p is an odd prime.16.2. 9. Suppose p divides ai for all i ∈ {0. 9. the polynomial f (x) = (x−1)2 (x−2) ∈ Q[x] has α = 1 as a multiple root and the polynomial f (x) = x4 + 2x2 + 1 = (x2 + 1)2 ∈ R[x] has α = ±i ∈ C as multiple roots. . Dx (f (x)). Construct fields of each of the following orders: (a) 9.3. 9. .13 (Eisenstein’s Criterion).4. (b) If f (0) = 0 prove that f is irreducible if and only if g is irreducible. Prove that K1 = F11 [x]/(x2 + 1) and K2 = F11 [y]/(y 2 + 2y + 2) are both fields with 121 elements. (a) Describe the coefficients of g in terms of the coefficients of f . p a prime. Show that the polynomial (x − 1)(x − 2) · · · (x − n) − 1 is irreducible over Z for all n ≥ 1. a0 are all elements of P and suppose a0 is not an element of P 2 .4.4.1. . n ≥ 1. Solution (b). If the image of p(x) in (R/I)[x] cannot be factored in (R/I)[x] into two polynomials of smaller degree.6. Then f (x) is irreducible in R[x].4. The derivative. Corollary 9. Suppose an−1 . 9. Solution (a). The polynomial g(x) = xn f (1/x) is called the reverse of f (x).19. Let F be a field and let f (x) = an x + an−1 xn−1 + · · · + a0 ∈ F [x]. (b) x3 + x + 1 in F3 [x]. x Proof (c). Let p be a prime in Z and let f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 ∈ Z[x]. Note that Dx (f (x)) is again a polynomial with coefficients in F . The polynomial f (x) is said to have a multiple root if there is some field E containing F and some α ∈ E such that x−α)2 divides f (x) in E[x]. 9. For example. determine their factorization into irreducibles. (b) 49. 9.8. (c) x4 + 1 in F5 [x].4.5 that a nonconstant polynomial f (x) has a multiple root if and only if f (x) is not relatively prime to . 9. (c) 8. then p(x) is irreducible in R[x]. Prove that the map which sends the element p(x) of K1 to the element p(y + 1) of K2 (where p is any polynomial with coefficients in F11 ) is well defined and gives a ring isomorphism from K1 to K2 . n − 1} but that p2 does not divide a0 . Solution (d). Proposition 9.12. Let P be a prime ideal of the integral domain R and let f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 be a polynomial in R[x] (here n ≥ 1). Solution (c). na = a + · · · + a (n times). Proof (d). Let F be a field and let f (x) be a polynomial of degree n in F [x]. Proof (b). a1 . Then f (x) is irreducible in both Z[x] and Q[x]. . (d) 81. .4. The notaion Fp denotes the finite field Z/pZ. Determine whether the following polynomials are irreducible in the rings indicated. Prove that the following polynomials are irreducible in Z[x]: (c) x4 + 4x3 + 6x2 + 2x + 1 p p (d) (x+2) −2 . Solution (b). of f (x) is defined by Dx (f (x)) = nan xn−1 + (n − 1)an−1 xn−2 + · · · + a1 where. . . Solution (a).4.49 Proposition 9. Solution (c). Corollary 9. . DUMMIT AND RICHARD M. fm ∈ F [x1 . . The maximal ideals in F [x] are the ideal (f (x)) generated by irreducible polynomials f (x).23.22. Then m h= i=2 bi S(fi−1 . . . . In particular. a polynomial of degree n in one variable over a field F has at most n roots in F .20. . Let g(x) be a nonconstant monic element of F [x] and let g(x) = f1 (x)n1 f2 (x)n2 · · · fk (x)nk be its factorization into irreducibles. Then (1) Every polynomial f ∈ R can be written uniquely in the form f = fI + r where fI ∈ I and no nonzero monomial term of the remainder r is divisible by any of the leading terms LT (g1 ). . LT (gm ). for all α ≥ 2 (3) (Z/pα Z)× is a cyclic group of order pα−1 (p − 1). . . then {g1 . Suppose f1 . 9. (2) Both fI and r can be computed by general polynomial division by g1 . xn ] are polynomials with the same multidegree α and that the linear combination h = a1 f1 + · · · + am fm with constants ai ∈ F has strictly smaller multidegree. . gm and are independent of the order in which these polynomials are used in the division. . . where the fi (x) are distinct. gm are any elements of I such that LT (I) = (LT (g1 ). We have the following isomorphisms of multiplicative groups: (1) (Z/nZ)× ∼ (Z/pαi Z)× × · · · × (Z/pαr Z)× = r i (2) (Z/2α Z)× is the direct product of a cyclic group of order 2 and a cyclic group of order 2α−2 . . where p1 . Let p be a prime. then f (x) has (x − α1 ) · · · (x − αk ) as a factor.24. F [x]/(f (x)) is a field if and only if f (x) is irreducible. . Then we have the following isomorphism of rings: F [x]/(g(x)) ∼ F [x]/(f1 (x)n1 ) × F [x]/(f2 (x)n2 ) × · · · × F [x]/(fk (x)nk ). . .21 (Hilbert’s Basis Theorem). .6. Polynomial Rings Over Fields II. . pr are distinct r primes. Proposition 9. The multiplicative group (Z/pZ)× of nozero residue classes pα1 1 mod p is cyclic. xn ] and let I be a nonzero ideal in R. Fix a monomial ordering on R = F [x1 . o Theorem 9. 9. gm } is a Gr¨bner o basis for I. x2 . α2 . then the multiplicative group F × of nonzero elements of F is a cyclic group. . . for all odd primes p. A finite subgroup of the multiplicative group of a field is cyclic. gm } is a Gr¨bner basis for o the nozero ideal I in R.25. Let n ≥ 2 be an integer with factorization n = · · · pαr in Z. . .19. . Proposition 9.15. In particular. . . . fi ). Proposition 9. . (3) The remainder r provides a unique representative for the coset of f in the quotient ring F [x1 . Polynomials in Several Variables Over a Field and Gr¨bner Bases.50 DAVID S. LT (gm )). . . . xn ] and suppose {g1 . even counted with multiplicity. . . . . . Fix a monomial ordering on R = F [x1 . . for some constants bi ∈ F.17. In particular. . xn ] with coefficients from a field F is finitely generated. . If R is a Noetherian ring then so is the polynomial ring R[x]. . (1) If g1 . . xn ]/I. .5. Use this criterion to determine whether the following polynomials have multiple roots: (a) x3 − 3x − 2 ∈ Q[x] (b) x3 + 3x + 2 ∈ Q[x] (c) x6 − 4x4 + 6x3 + 4x2 − 12x + 9 ∈ Q[x] (d) Show for any prime p and any a ∈ Fp that the polynomial xp − a has a multiple root. αk in F (not necessarily distinct). Corollary 9. .18. In particular. Corollary 9. . . . . (2) The ideal I has a Gr¨bner basis. f ∈ I if and only if r = 0. FOOTE its derivative (which can be detected by the Euclidean Algorithm in F [x]). . = Proposition 9. if F is a finite field. . . . Every ideal in the polynomial ring F [x1 . Theorem 9. If the polynomial f (x) has roots α1 . o Lemma 9. . .16. . Proposition 9. . . . . . If I and J are any two ideals in F [x1 . Fix a monomial ordering on R = F [x1 . Proposition 9. . Then I = J if and only if I and J have the same reduced Gr¨bner basis with respect to any fixed monomial ordering on F [x1 . . xn ] = ∅.29 (Elimination). . . . xn ] and I ∩ J = (tI + (1 − t)J) ∩ F [x1 . . . . gm } is a Gr¨bner basis for the nonzero ideal I in F [x1 . . . .27. . xn ] o with respect to the lexicographic monomial ordering x1 > · · · > xn . gj ) ≡ 0 o mod G for 1 ≤ i < j ≤ m. x1 . xn ]. gm ) is a nonzero ideal in R. Suppose G = {g1 . . I ∩ F [xi+1 . . o Proposition 9. . . . . xn ] and fix a monomial ordering on R. . . . .28. . In particular. . . . . . . Theorem 9. . If I = (g1 . . . . . . xn ] is a Gr¨bner basis o of the ith elimination ideal Ii = I ∩ F [xi+1 . Let I and J be two ideal in F [x1 . . Then there is a unique reduced Gr¨bner basis for o every nonzero ideal I in R. Let R = F [x1 . . . xn ] then tI + (1 − t)J is an ideal in F [t. gm } is a Gr¨bner basis for I if and only if S(gi . . .26 (Buchberger’s Criterion). xn ] of I. Then G ∩ F [xi+1 .30. Corollary 9. then G = {g1 . . I ∩ J is the first elimination ideal of tI + (1 − t)J with respect to the ordering t > x1 > · · · > xn . . . . . . . xn ] = 0 if and only if G ∩ F [xi+1 . . xn ]. . . . . . xn ]. . . xn ]. . .51 Proposition 9. . In particular. we calculate o S(f1 . y. x − yz). x − yz) in F [x. o Now to find the reduced Gr¨bner basis for J. y.32. y 2 z. y] are equal. z]. 9. corollary 9. o S(f2 . x − yz) and finally I ∩ J = (y 2 z. tx − tyz − x + yz. So fix the lexicographical ordering x > y > z for F [x. G = x3 + y. g3 ) = −yg1 − x g3 = −x y + x y ≡ 0 S(g2 . z].28 two ideals in F [x. yz + y) and the ideal J = (x3 z + x3 . f2 } . x3 + y) o in F [x. then our reduced Gr¨bner basis for I is G = x3 + y. yz + y is the reduced Gr¨bner o o basis for J. xy 2 . mod {g1 . z] are equal if and only if I. yz + y . fj ) we find is given by a Gr¨bner basis. tyz + x − yz.30 states By calculating the S(fi . x2 − xyz. Then proposition 9. f2 . tz. g2 . g2 ) = g1 − zg2 = x3 − yz ≡ −yz − y 3 3 3 S(g1 . xz − yz 2 . f2 ) = yzf1 − x3 f2 = −y 2 z 2 − x3 y ≡ 0 mod {f1 . To find a Gr¨bner basis for o I. Let I = (x. g2 .24. Use reduced Gr¨bner bases to show that the ideal I = (x3 − yz. Since the o o reduced Gr¨bner basis is the same for both ideals. Use Gr¨bner bases to show that (x.6. By corollary 9. . x − yz. Since LT (g2 ) mod LT (g1 ). g3 } tI + (1 − t)J = (tx. o S(g1 . z) and J = (y 2 − yz). x − yz). g2 } Thus {g1 .6. y 2 z is a Gr¨bner basis so x − yz. Now to find a Gr¨bner basis for (xy. x − yz) = (xy. J have the same reduced Gr¨bner basis o for any monomial ordering. z) ∩ (y 2 . Thus o I ∩ J = (tI + (1 − t)J) ∩ F [x. f3 } . g2 . y. y 2 z is the reduced Gr¨bner basis because x | xy. ty 2 − y 2 . mod {g1 . x − yz) tI + (1 − t)J = (tz. y 2 z. Proof.52 DAVID S. z]. y.28 states the ideals are equal. f2 ) = f1 − yf2 = y 2 z S(f1 . S(f1 . g3 } . o Proof. g3 } is a Gr¨bner basis for J. ty 2 − y 2 . g3 ) = x g2 + yzg3 = x − y z ≡ 0 3 3 2 mod {g1 . DUMMIT AND RICHARD M. 3 Since x − yz = −1(yz + y) + (x3 + y). f3 ) = y 2 zf2 − xf3 = −y 3 z 2 ≡ 0 mod {f1 . f3 ) = yzf1 − xf3 = 0 Thus xy. tyz + x − yz. FOOTE 9. then if a component contains a nonzero element then that component of the submodule must contain all of F . y ∈ N . Introduction to Module Theory 10. y = zm2 . The set of torsion elements is denoted Tor(M ) = {m ∈ M | rm = 0 for some nonzero r ∈ R} . Now f (x + y) = (x + y) · 1A = f (x) + f (y).1. Show that if R is the ring of 2 × 2 matrices over a field and e is the matrix with a 1 in position 1. Example (b). If ym = 0. Since (r · 1A ) · a = (r · a) · 1A = a · (r · 1A ). then this is such an example. Proof (a). 10. then r1 r2 (x + ry) = r2 r1 x + rr1 r2 y = 0 so Tor(M ) is a submodule of M because r1 r2 = 0. Thus by definition A is an R-algebra and the R-module structure on A induced by its algebra structure is precisely the original R-module structure.1. where zM = {zm | m ∈ M }. zM is clearly nonempty. Tor(M ) = ∅ because r0 = r(m − m) = rm − rm = 0. Basic Definitions and Examples. then it would not be T -stable. Prove that the submodules Uk described in the example of F [x]-modules are all of the F [x]-submodules for the shift operator. 1 and zeros elsewhere then eR is not a left R-submodule (where M = R is considered as a left R-module as in example 1) – in this case the matrix e is not in the center of R. r ∈ R. Conclude that A is an R-algebra and that the R-module structure on A induced by its algebra structure is precisely the original R-module structure. (b) Give an example of a ring R and an R-module M such that Tor(M ) is not a submodule. Suppose A is a ring with identity 1A that is a (unital) left R-module satisfying r · (ab) = (r · a)b = a(r · b) for all r ∈ R and a. (a) Prove that if R is an integral domain then Tor(M ) is a submodule of M called the torsion submodule of M.. f (xy) = (xy) · 1A = f (x)f (y) and f (1R ) = 1R · 1A = 1A .1. Thus it is clear that the Uk s are all of the F [x]-submodules by the bjiection of F [x]-submodules and subspaces invariant under T . If f (x) = f (y) then x = x · 1A = y · 1A = y so f is well defined.8. and (2) x + ry ∈ N for all r ∈ R and for all x. f1 0 f2 0 = f1 f1 f2 f2 . Note since F is a field.53 Part III – Modules and Vector Spaces 10. Prove that the map f : R → A defined by f (r) = r · 1A is a ring homomorphism mapping 1R to 1A and that f (R) is contained in the center of A. Proposition 10.14. If r1 x = r2 y = 0. 10. If R = 0. Let z be an element of the center of R. Let x = zm1 .16. zr = rz for all r ∈ R. i. b ∈ A. f1 f2 If a ∈ eR then a = 0 0 but 1 1 1 1 so eR is not a left R-submodule. But if a component is zero to the left of some nonzero component. An element m of the R-module M is called a torsion element if rm = 0 for some nonzero element r ∈ R. Proof. A subset N of M is a submodule of M if and only if (1) N = ∅. Proof.1. 10.1 (The Submodule Criterion). Proof. 10. Proof (c). Let R be a ring and let M be an R-module. Let x. (c) If R has zero divisors show that every nonzero R-module has nonzero torsion elements.1.e. y ∈ R\ {0} with xy = 0.22. then xym = 0 so either way there are nonzero torsion elements. r ∈ R so x + ry = zm1 + rzm2 = zm1 + zrm2 = z(m1 + rm2 ) so zM is a submodule of M . Prove that zM is a submodule of M . L (x) divides mα. the degree of α over F is the degree of the extension it generates over F . Then F (α) = a0 + a1 α + a2 α2 + · · · + an−1 αn−1 | a0 . Basic Theory of Field Extensions. such that σ restricted to F is the isomorphism ϕ. so that the image of ϕ is either 0 or isomorphic to F . then it is algebraic. Theorem 13. Theorem 13. Proposition 13.10.2. . i. Theorem 13. Let α be algebraic over the field F and let F (α) be the field generated by α over F . an−1 ∈ F Corollary 13.12. a1 . θn−1 are a basis for K as a vector space over F .4. a1 . 13. Basic Theory of Field Extensions.1. θ2 . Then ϕ is either identically 0 or is injective. Then F (α) ∼ F [x]/(mα (x)) = so that in particular [F (α) : F ] = deg mα (x) = deg α. Proposition 13. K = a0 + a1 θ + a2 θ2 + · · · + an−1 θn−1 | a0 .1.F (x) divides f (x) in F [x]. More precisely. . . Let F be a field and let p(x) ∈ F [x] be an irreducible polynomial. so the degree of the extension is n. Suppose K is an extension field of F containing a root α of p(x): p(α) = 0. Let θ = x mod (p(x)) ∈ K. .. is either 0 or a prime p.4. Let α be algebraic over F . . Hence consists of all polynomials of degree < n in θ.11. Corollary 13. Proposition 13.F (x) ∈ F [x] which has α as a root. Field Theory 13.e. . . Corollary 13. an−1 ∈ F ⊆ K.5. .8. ch (F ). A polynomial f (x) ∈ F [x] has α as a root if and only if mα.9. Then there is an isomorphism σ : F (α)→F (β) ˜ mapping α to β and extending ϕ.e.3. .13. Let α be a root of p(x) (in some extension of F ) and let β be a root of p (x) (in some extension of F ). . If ch (F ) = p then for any α ∈ F. and let a(θ). p · α = α + α + · · · + α = 0. Theorem 13. Let ϕ : F → F be a homomorphism of fields.7.. if α is an element of an extension of degree n over F then α satisfies a polynomial of degree at most n over F and if α satisfies a polynomial of degree n over F then the degree of F (α) over F is at most n. Proposition 13.F (x) in L[x]. Let F be a field and let p(x) ∈ F [x] be an irreducible polynomial. Then addition in K is defined simply by usual polynomial addition and multiplication in K defined by a(θ)b(θ) = r(θ) where r(x) is the remainder (of degree < n) obtained after dividing the polynomial a(x)b(x) by p(x) in F [x]. The element α is algebraic over F if and only if the simple extension F ⊂ F (α) is finite. Then ∼ F (α) = F [x]/(p(x)). Let F (α) denote the subfield of K generated over F by α.6. Suppose in theorem 13.6 that p(x) is of degree n. then mα. If the extension F ⊂ K is finite. Corollary 13.. FOOTE 13. The characteristic of a field F . Let K be as in theorem 13. b(θ) ∈ K be two polynomials of degree < n in θ. Identifying F with this isomorphic copy shows that there exists an extension of F which p(x) has a root. DUMMIT AND RICHARD M. Then there exists a field K containing an isomorphic copy of F in which p(x) has a root. Then there is a unique monic irreducible polynomial mα. i. Then the elements 1.e. [K : F ] = n. . Let p(x) ∈ F [x] be an irreducible polynomial and let ˜ p (x) ∈ F (x) be the irreducible polynomial obtained by applying the map ϕ to the coefficients of p(x). Proposition 13. i.2. If L ⊂ F is an extension of fields and α is algebraic over both F and L. Let ϕ : F →F be an isomorphism of fields. . . Let p(x) ∈ F [x] be an irreducible polynomial of degree n over the field F and let K be the field F [x]/(p(x)).54 DAVID S. θ. / Proof. F ⊆ K ⊆ L. . 3) : Q] = 4. Prove that 3 2 ∈ F . Then the collection of elements of L that are algebraic over F form a subfield K of L. .2. Let K1 and K2 be two finite extensions of a field F contained in K. . then L is algebraic over F . . 3 are √ √ contained in Q( 2. Conclude that [Q( 2+ 3) : Q] = 4.7. extension degrees are multiplicative. . the field generated over F by α and β is the field generated by β over the field F (α) generated by α. β) = (F (α))(β). n. F (α. . αn ) where αi ∈ Q for i = 1. 3). i. 3) √ √ √ √ √ √ Since Q( 2. the other side is also infinite. Thus 2 ∈ Q( 2 + 3) thus √ √ √ √ √ √ √ 3 ∈ Q( 2√ √ By definition we have Q( √2. . 3). . If α1 . . More precisely. Hence by √ definition Q( 2 + 3) ⊆ Q( 2.18. 3) = Q( 2 + 3) then [Q( 2 + 3) : Q] = [Q( 2. a field generated over F by a finite4 number of algebraic elements of degrees n1 . . . If p were to be reducible. . .15.e. First note that [Q( 3 + 2 2) : Q] = [Q( 3 + 2 2 : Q( 2)][Q( 2) : Q]. Then with equality if and only if an F -basis for one of the fields remains linearly independent over the other field. Notice that p( 2 + 3) = 0 where p(x) = x4 − 10x2 + 1. . .16.. . Then [K1 K2 : F ] = [K1 : F ][K2 : F ] = nm. αn and β1 . Corollary 13. Find an irreducible polynomial √ √ satisfied by 2 + 3. Prove that |F| = pn for some positive integer n. Lemma 13. α2 . . But x2 − 3 + 2 2 = (x − 1 + √ √ √ √ √ 2 2)(x + 1 + 2 2) so [Q( 3 + 2 2) : Q( 2)] = 1. . Proof. . Since F is finite then for some n ∈ Z+ . Since pZ ⊆ ker ϕ then ϕ : Fp → F is well defined. Corollary 13.17. 3). .22. √ √ √ √ √ √ √ √ Proof. Notice ϕ(1Fp ) = 1F so ϕ is injective. Let ϕ : Z → F be the ring map given by ϕ(n) = n · 1F . Since √ Q. . . αn ) : Q] = 2l for some l ∈ Z+ . β2 . . . / . Therefore p is irreducible in Q.10.2. (in particular α−1 for α = 0) are all algebraic. Prove that Q( 2+ 3) = Q( 2. α/β (for β = 0). .21. βm are bases for K1 and K2 over F .14. Corollary 13. Then α ± β.13. Suppose α and β are algebraic over F . Then [K : F ] divides [L : F ]. √ √ √ √ √ √ √ √ To prove the other inclusion notice that ( 2 + 3)3 − 9( 2 + 3) /2 = 2. Thus [Q( 3 + 2 2) : Q] = 2. Corollary 13. Since 3 does not divide 2l for l ∈ Z+ then 3 2 ∈ F .e. then Q. respectively. αβ.2.20. n and j = 1. . √ 13. . Suppose F = Q(α1 . If K is algebraic over F and L is algebraic over K. But ¯ ϕ is a ring map of fields so it is trivial or injective. 3). . Determine the degree of the extension Q( 3 + 2 2) over Q. 13. √ √ √ √ √ Solution. . Then [L : F ] = [L : K][K : F ]. n2 . αk ) : Q(α1 . 2}.. 3) ⊆ Q( 2 + √ and thus equality. Suppose F ⊂ L is a finite extension and let K be any subfield of L containing F . then the elements αi βj for i = 1. Now if α √ √ √ 3 2 ∈ F then Q ⊂ Q( 3 2) ⊆ F then [Q( 3 2 : Q)] | [F : Q]. . then there would be an irreducible polynomial with degree less than 4. . .55 Theorem 13. . √k−1 )] ∈ {1. 2 + 3 are contained in Q( 2. [K2 : F ] = m in proposition 13. |F| = |Fp | [F : Fp ] [K1 K2 : F ] ≤ [K1 : F ][K2 : F ] = pn √ √ √ √ √ √ 13. nk is algebraic of degree ≤ n1 n2 · · · nk .1. Thus we can consider F as ¯ ¯ ¯ a vector space over Fp . where n and m are relatively prime: (n. 2.19. m span K1 K2 over F . Let F ⊂ L be an arbitrary extension. Theorem 13. . then [Q(α1 . .2. This contradicts the minimality of the degree of m√2+√3 = 4. + 3). . i. Let F be a finite field of characteristic p. . .21. . The extension F ⊂ K is finite if and only if K is generated by a finite number of algebraic elements over F . m) = 1. Let F ⊆ K ⊆ L be fields. Since [Q(α1 . Theorem 13. Proposition 13. . Suppose that [K1 : F ] = n. √ 2 13. where if one side of the equation is infinite. Let F be a field of characteristic p.37. Corollary 13. The field C is algebraically closed. Q. For any field F there exists an algebraically closed field K containing F . None of the classical Greek problems: (I) Doubling the Cube. Let E be a splitting field for f (x) over F and let E be a splitting field for f (x) over F .31. b ∈ F . and notice since p is irreducible. Since p(α) = 0 we have Since ai ∈ F ⊆ R and α ∈ R then α−1 ∈ R and R is a field. Theorem (Fundamental Theorem of Algebra). Proof.. Every irreducible polynomial over a field of characteristic 0 is separable.36. Let f (x) ∈ F [x] be a polynomial and let f (x) ∈ F [x] ˜ be the polynomial obtained by applying ϕ to the coefficients of f (x).34. Proposition 13. Proposition 13.25. A polynomial f (x) has a multpile root α if and only if α is also a root of Dx f (x). Then the isomorphism ϕ extends to an isomorphism σ : E →E . (a + p)p = ap + bp . Classical Straightedge and Compass Constructions. An algebraic closure of F is unique up to isomorphism. Let ϕ : F →F be an isomorphism of fields.2. Corollary 13.e. and (ab)p = ap bp . The field C contains an algebraic closure for any of its subfields. Theorem 13. Let F be an algebraic closure of F .e. a0 = 0. Proposition 13. Corollary 13.29. Any two splitting fields for a polynomial f (x) ∈ F [x] over a field F are isomorphic. f (x) is separable if and only if it is relatively prime to its derivative: (f (x).16. the collection of complex numbers algebraic over Q. Suppose that F is a finite field of characteristic p. If the element α ∈ R is obtained from a field F ⊆ R by a series of compass and straightedge constructions then [F (α) : F ] = 2k for some integer k ≥ 0. For any field F .26.3. A splitting field of a polynomial of degree n over F is of degree at most n! over F . Proposition 13.e. σ ˜ restricted to F is the isomorphism ϕ. Then every element of F is a pth power in F. Proposition 13. and (III) Squaring the Circle.33. A polynomial in F[x] is separable if and only if it is the product of distinct irreducible polynomials in F[x]. F = Fp . is possible.56 DAVID S. i. Let p(x) be given by p(x) = an xn + · · · + a0 . α−1 = −a−1 (an αn−1 + · · · + a1 ). Put another way. A polynomial over such a field is separable if and only if it is the product of distinct irreducible polynomials. .. Let K be an algebraically closed field and let F be a subfield of K. Proposition 13.30.28 (Uniqueness of Splitting Fields). i.4. FOOTE 13. Every irreducible polynomial over a finite field F is separable. Corollary 13. Separable and Inseparable Extensions.27. Then F is algebraically closed. Proposition 13. 13. In particular. (II) Trisecting an Angle. In particular. the pth -power map defined by ϕ(a) = ap is an injective field homomorphism from F to F .5. Splitting Fields and Algebraic Closures. Then the collection of elements F of K that are algebraic over F is an algebraic closure of F . Proposition 13. Then for any a. 13. is an algebraic closure of Q.24. Let F ⊂ K be an algebraic extension and let R be a ring contained in K and containing F .. 0 13. Theorem 13. Let α ∈ R\ {0} so since α ∈ F then there is some irreducible p ∈ R[x] such that p(α) = 0. if f (x) ∈ F [x] then there exists an extension K of F which is a splitting field for f (x). f (x) and Dx f (x) are both divisible by the minimal polynomial for α. Show that R is a subfield of K containing F . DUMMIT AND RICHARD M.32. Dx f (x)) = 1.35. Theorem 13. i.23. If G1 = G2 are distinct finite subgroups of automorphisms of a field K then their fixed fields are also distinct. .11.10. In particular.12. If σ1 . Corollary 13. σ2 . Corollary 14. K/F is Galois if and only if F is the fixed field of Aut(K/F ). The cyclotomic polynomial Φn (x) is a monic polynomial in Z[x] of degree ϕ(n).13. any polynomial with coefficients in F having α as a root also has σα as a root. Proposition 14. The extension K/F is Galois if and only if K is the splitting field of some separable polynomial over F . Let p(x) be an irreducible polynomial over a field F of characteristic p. Proposition 14. Let K/F be a field extension and let α ∈ K be algebraic over F .2. with Galois group G. Aut(K/F ) permutes the roots of irreducible polynomials. Then [K : F ] = n = |G| . In particular distinct automorphisms of a field K are linearly independent as functions on K. σn are distinct embeddings of a field K into a field L. Let E be the splitting field over F of the polynomial f (x) ∈ F [x]. Aut(K) is a group under composition and Aut(K/F ) is a subgroup.5. Theorem 14. .4.8. Theorem 14. Every finite extension of a perfect field is separable. Furthermore. Corollary 13. 14. The degree over Q of the cyclotomic field of nth roots of unity is ϕ(n): [Q(ζn ) : Q] = ϕ(n). χn are distinct characters of G with values in L then they are linearly independent over L.9.41. Proposition 14.40. .2. Galois Theory 14. The cyclotomic polynomial Φn (x) is an irreducible monic polynomial in Z[x] of degree ϕ(n). Basic Definitions. Cyclotomic Polynomials and Extensions.7 (Linear Independence of Characters). .38.42. Then |Aut(K/F )| ≤ [K : F ] with equality if and only if F is the fixed field of Aut(K/F ).e. i. Then with equality if f (x) is separable over F . and (2) if H1 ≤ H2 ≤ Aut(K) are two subgroups of automorphisms with associated fixed fields F1 and F2 .39. σn } be a subgroup of automorphisms of a field K and let F be the fixed field. Then every automorphism of K fixing F is contained in G.6. . . then F2 ⊆ F1 . Then there is a unique integer k ≥ 0 and a unique irreducible separable polynomial psep (x) ∈ F [x] such that p(x) = psep (xp ). Corollary 14. Equivalently.14 (Fundamental Theorem of Galois Theory). Then the collection F of elements of K fixed by all the elements of H is a subfield of K.3. .e. If K is the splitting field over F of a separable polynomial f (x) then K/F is Galois. Let H ≤ Aut(K) be a subgroup of the group of automorphisms of K. Let G = {σ1 = 1. respectively. Aut(K/F ) = G. . If χ1 . if this is the case then every irreducible polynomial with coeffiecients in F which has a root in K is separable and has all its roots in K (so in particular K/F is a separable extension). The Fundamental Theorem of Galois Theory. Proposition 14.. The association of groups to fields and fields to groups defined above is inclusion reversing.1. Theorem 13. . 14. Proposition 14. then they are linearly independent as functions on K.57 Proposition 13. Theorem 14. every finite extension of either Q or a finite field is separable. .. Then for any σ ∈ Aut(K/F ).6. . . Corollary 14. Lemma 13. |Aut(E/F )| ≤ [E : F ] k Corollary 14. Theorem 14. .1. σα is a root of the minimal polynomial for α over F i. Put another way. namely (1) if F1 ⊆ F2 ⊆ K are two subfields of K then Aut(K/F2 ) ≤ Aut(K/F1 ). so that K/F is Galois. Let K/F be any finite extension. Let G be a finite subgroup of automorphisms of a field K and let F be the fixed field. 13. Corollary 14. 37. if f is not nilpotent in R. element of R.2. π : R → D−1 R is an injetion if and only if D contains neither zero nor any zero divisors of R. Localization. 15. Prove that Rf ring in the variable x. r r Let d1 = d2 .58 DAVID S. d2 ∈ D with ϕ(d1 d2 ) = ϕ(d1 )ϕ(d2 ) so d1 d2 ∈ ϕ−1 (D ) because D is a multiplicative set. Let R be any commutative ring with 1 and let f be any {f n | n ≥ 0} of nonnegative powers of f in R. Proof. 15. ∼ R[x]/(f x − 1) where R[x] is the polynomial = Proof.15). hence if and only if D contains nilpotent elements. Let d1 .4. Radicals and Affine Varieties. Commutative Rings and Algebraic Geometry 15. Let D = ϕ−1 (D ). DUMMIT AND RICHARD M. (1) ker π = {r ∈ R | xr = 0 for some x ∈ D}. theorem 7. Theorem 15.18.1.21. and (2) D−1 R = 0 if and only if 0 ∈ D. Define Rf = D−1 R. in particular. 15. Prove D is a multiplicatively closed subset of R and the map ϕ : D−1 R → D −1 S given by ϕ (r/d) = ϕ(r)/ϕ(d) is a ring homomorphism. Let D be the multiplicative set Note that Rf = 0 if and only if f is nilpotent. Integral Extensions and Hilbert’s Nullstellensatz. then f becomes a unit in Rf . let R be a commutative ring with 1 = 0 and let D be a multiplicatively closed subset of R containing 1.36. there is a unique homomorphism Ψ : D−1 R → S such that Ψ ◦ π = ψ. Suppose ϕ : R → S is a ring homomorphism with ϕ(1R ) = 1S and D is a multiplicatively closed subset of S. If f is not nilpotent. Noetherian Rings and Affine Algebraic Sets.36 (cf. 15. Then there is a commutative ring D−1 R and a ring homomorphism π : R → D−1 R satisfying the following universal property: for any homomorphism ψ : R → S of commutative rings that sends 1 to 1 such that ψ(d) is a unit in S for every d ∈ D.4. 15. FOOTE 15. Corollary 15.3. Then r1 d2 = r2 d1 so 1 2 r1 r2 d2 r1 )=ϕ( ) d1 r2 d1 d2 ϕ(r1 r2 d2 ) = ϕ(r2 d1 d2 ) ϕ(r2 d1 )ϕ(r2 ) = ϕ(r2 d1 )ϕ(d2 ) r2 =ϕ( ) d2 ϕ( .4. In the notation of theorem 15. 59 So ϕ is well defined. d2 ∈ D−1 R.4. Then ϕ( r1 r2 r1 d2 + r2 d1 + )=ϕ( ) d1 d2 d1 d2 ϕ(r1 d2 + r2 d1 ) = ϕ(d1 d2 ) ϕ(r1 ) ϕ(r2 ) = + ϕ(d1 ) ϕ(d2 ) r1 r2 =ϕ( )+ϕ( ) d1 d2 r1 r2 r1 r2 ϕ( )=ϕ( ) d1 d2 d1 d2 ϕ(r1 r2 ) = ϕ(d1 d2 ) ϕ(r1 ) ϕ(r2 ) = ϕ(d1 ) ϕ(d2 ) r1 r2 = ϕ ( )ϕ ( ).22. .4. Prove that the localization RP is isomorphic to the localization of RQ at the prime ideal P RQ (cf.21). d1 d2 15. exercise 15. Proof. Suppose P ⊆ Q are prime ideals in R and let RQ be the localization of R at Q. Now let r1 r2 d1 .