Chapter 5Dimension Theory The geometric notion of the dimension of an affine algebraic variety V is closely related to algebraic properties of the coordinate ring of the variety, that is, the ring of polynomial functions on V . This relationship suggests that we look for various ways of defining the dimension of an arbitrary commutative ring. We will see that under appropriate hypotheses, several concepts of dimension are equivalent. Later, we will connect the algebraic and geometric ideas. 5.1 The Calculus of Finite Differences Regrettably, this charming subject is rarely taught these days, except in actuarial programs. It turns out to be needed in studying Hilbert and Hilbert-Samuel polynomials in the next section. 5.1.1 Lemma Let g and G be real-valued functions on the nonnegative integers, and assume that ∆G = g , that is, G(k + 1) − G(k ) = g (k ) for all k ≥ 0. (We call ∆G the difference of G.) Then b +1 g (k ) = G(k )|b = G(b + 1) − G(a). a k=a Proof. Add the equations G(a + 1) − G(a) = g (a), G(a + 2) − G(a + 1) = g (a + 1), . . . , G(b + 1) − G(b) = g (b). ♣ 5.1.2 Lemma If r is a positive integer, define k (r) = k (k − 1)(k − 2) · · · (k − r + 1). Then ∆k (r) = rk (r−1) . Proof. Just compute: ∆k (r) = (k + 1)(r) − k (r) = (k + 1)k (k − 1) · · · (k − r + 2) − k (k − 1) · · · (k − r + 1) = k (k − 1) · · · (k − r + 2)[k + 1 − (k − r + 1)] = rk (r−1) . ♣ 1 1. In a similar fashion we can find n k=1 k s for any positive integer s.2).2 Hilbert and Hilbert-Samuel Polynomials There will be two polynomial-like functions of interest.1. (We can allow r = 0 if we take the degree of the zero polynomial to be -1. 1 k=1 The first term on the right is (n + 1)n(n − 1)(n − 2)/4. The degree of f is taken to be the degree of g .3 Examples n k=1 k (2) +1 = [k (2) /2]|n = (n + 1)n/2.1. so k 3 = k (3) + 3k 2 − 2k . (The analogous result from differential calculus that a function with zero derivative is constant is harder to prove. Therefore n +1 k 3 = [k (4) /4]|n + 3n(n + 1)(2n + 1)/6 − 2n(n + 1)/2.1. along with the observation that a function whose difference is zero is constant. .) ♣ 5.) Proof.2 CHAPTER 5.1) and (5. In other words. DIMENSION THEORY 5. This follows from (5.5 Lemma Let f : N → Q. 5.4 Definitions and Comments A polynomial-like function is a function f from the natural numbers (nonnegative integers) N to the rational numbers Q. and we begin preparing for their arrival. Then f is a polynomial-like function of degree r if and only if ∆f is a polynomial-like function of degree r − 1. such that f eventually agrees with a polynomial g ∈ Q[X ]. 1 ∆k (2) = 2k (1) . 5. k (3) = k (k − 1)(k − 2) = k 3 − 3k 2 + 2k .1. f (n) = g (n) for all sufficiently large n (abbreviated n >> 0). so +1 k 2 = [k (3) /3]|n + (n + 1)n/2 = (n + 1)n(n − 1)/3 + (n + 1)n/2 1 k=1 = n(n + 1)(2n + 1)/6. so the sum of the first n cubes is [n(n + 1)/4][n2 − 3n + 2 + 2(2n + 1) − 4] which simplifies to [n(n + 1)/2]2 . so k 2 = k (k − 1) + k = k n + k (1) . and is usually accomplished via the mean value theorem. that is.1) and (1. As above.) The general result is accomplished by decomposing a long exact sequence into short exact sequences. HILBERT AND HILBERT-SAMUEL POLYNOMIALS 3 5. . then each Mn is a finitely generated R0 -module. Assume that R0 is Artinian and R is finitely generated as an algebra over R0 . then y is of the form i=1 ai xi with ai ∈ R. . all with finite length.6.5. Since M is Noetherian. If M = ⊕n≥0 Mn is a finitely generated graded R-module. so our second short exact sequence is 0 → im g → C → coker g → 0. m ≥ n. the length lRo (Mn ) of the R0 -module Mn is finite for all n ≥ 0.2.2. Apply (5.2.) To see how the process works.2. 0 → A → B → coker f → 0.3) and (1. hence M is a Noetherian Rmodule. xt . Let Nn be the direct sum of the Mm . It suffices to prove that y1 . Proof. consider an exact sequence GA f GB 0 Our first short exact sequence is g GC h GD i GE G 0. By (4.6. where the additivity property can be expressed as l(M ) = l(N ) + l(M/N ). This is probably familiar for a short exact sequence 0 → N → M → M/N → 0. we can write xi = yi + zi with yi ∈ Mn and zi ∈ ⊕m>n Mm . . . But just as we decomposed xi above.2. ♣ 5.3 Additivity of Length Suppose we have an exact sequence of R-modules 0 → A1 → A2 → · · · → An → 0. Section 7. so the third short exact sequence is 0 → im h → D → coker h → 0. Nn is finitely generated over R. Then we have additivity of length. . l(A1 ) − l(A2 ) + · · · + (−1)n−1 l(An ) = 0. 5. we can write ai = bi + ci where bi ∈ R0 and ci ∈ ⊕j>0 Rj .1.1 Proposition Let R = ⊕n≥0 Rn be a graded ring.14). yt generate t Mn over R0 .5. . (“Long” means longer than short. . say by x1 . If y ∈ Mn .2 Corollary In (5.13). coker g ∼ = im h (= ker i). Now coker f = B/ im f = B/ ker g ∼ = im g (= ker h). (See TBGY. . R is a Noetherian ring. . Thus t t y= i=1 (bi + ci )(yi + zi ) = i=1 bi y i because the elements bi zi .1). Since Nn = Mn ⊕ ⊕m>n Mm . Proof. Problem 5. ci yi and ci zi must belong to ⊕m>n Mm . ♣ We will need the following basic property of composition length.2. Applying additivity for short exact sequences. n) is polynomial-like of degree at most r − 1. . hence so is ∆h(M. so λr (Mn ) ⊆ Mn+1 . n) are defined and finite. After cancellation. .3). . n ≥ 0. By (5. n ∈ N. Then h.1) and (5. By hypothesis. ar belonging to R1 . is polynomial-like of degree at most r − 1.2. Problem 6. Assume that R0 is Artinian and R is finitely generated as an algebra over R0 . An ideal I of R is said to be an ideal of definition if Mn ⊆ I ⊆ M for some n ≥ 1. n) and h(C. then R = R0 . We argue by induction on r. Then K is a submodule of M and C a quotient of M . If Kn is the kernel.6. h(K.11) applies. ar ∈ R1 . . so K and C are finitely generated T -modules. h(M. h(K. n) are polynomial-like of degree at most r − 2. we call h the Hilbert polynomial of M .2). so we may replace the third short exact sequence by 0 → im h → D → E → 0. If r = 0. (If an ideal I annihilates an R-module M . . n) and h(C. define h(M. and note that I = M if and only if every prime ideal containing I is maximal. then Mn = 0 for n > d.2. with all generators a1 . Equivalently. so (1. and therefore h(M. of λr : Mn → Mn+1 . Thus K and C are finitely generated Noetherian graded R-modules. we have l(A) − l(B ) + l(coker f ) − l(im g ) + l(C ) − l(coker g ) + l(im h) − l(D) + l(E ) = 0. then M is an R/I -module. h(K.) By induction hypothesis.2. n) = h(C.4 CHAPTER 5. Section 4. n + 1) − h(C. n) + h(M. see TBGY. . n). as a function of n with M fixed. n). n) = 0 for n >> 0. ♣ 5.4 Theorem Let R = ⊕n≥0 Rn be a graded ring. Now assume r > 0. . and let λr be the endomorphism of M given by multiplication by ar .1. n) − h(K. If M is a finitely generated graded R-module. n) = 0 hence ∆h(M. where T is the graded subring of R generated over R0 by a1 .] .2. R/I is an Artinian √ ring. .2. Choose a finite set of homogeneous generators for M over R. 0 Let K be the direct sum of the Kn and C the direct sum of the Cn . n) − h(M.10). DIMENSION THEORY But coker h ∼ = im i = E . By (5. this becomes l(A) − l(B ) + l(C ) − l(D) + l(E ) = 0 as desired.3. [See (3. If d is the maximum of the degrees of the generators. and Cn the cokernel.5). Using slightly loose language. Now ar annihilates K and C . n) = lR0 (Mn ). ar−1 . Proof. we have the exact sequence G Kn G Mn λr G Mn+1 G Cn G 0. so by (5. 5.5 Definitions and Comments Let R be any Noetherian local ring with maximal ideal M.2. An induction argument using additivity of length shows that sI (M.2. n) is the same for all possible choices. . (Note that by definition of a graded ring. n) is finite.2.6 The Hilbert-Samuel Polynomial Let I be an ideal of definition of the Noetherian local ring R.2.2)] are given by grI (R) = ⊕n≥0 (I n /I n+1 ). . n + 1) − sI (M. Consequently ∆sI (M. sI (M.4). If M is a finitely generated R-module. n) + r(n) . which allows us to produce elements in Rt for arbitrarily large t. n) depends on the particular ideal of definition I . tn) ≥ sI (M. then M/IM is a finitely generated module over the Artinian ring R/I .5. If I is generated over R by a1 . from right n n to left. then the images a1 . By (5. the associated graded ring and the associated graded module [see (4. We define the Hilbert-Samuel polynomial by sI (M. To see this. n) = lR/I (I n M/I n+1 M ) < ∞. . . d2 and d3 . HILBERT AND HILBERT-SAMUEL POLYNOMIALS 5 5. n) ≥ sM (M. The Hilbert-Samuel polynomial sI (M. Now the sequence 0 → I n M/I n+1 M → M/I n+1 M → M/I n M → 0 is exact by the third isomorphism theorem. . Therefore all three degrees coincide. Then for every n ≥ 1 we have Mtn ⊆ I n ⊆ Mn . .4). With the I adic filtrations on R and M .) By (5. n) + sI (M . we have O(dn 1 ) ≤ O (d2 ) ≤ O (d3 ). so by (5. If the degrees of these polynomial are. 5. the additivity of length. n). with d3 = d1 . n) = h(grI (M ).1. ar in I/I 2 generate grI (R) over R/I . then N becomes an R/I -module via (a + I )x = ax. h(grI (M ). Then sI (M . . Again recall that if N is an R-module and the ideal I annihilates N . let t be a positive integer such that Mt ⊆ I ⊆ M. n) = lR (M/I n M ). so sM (M. n) is polynomial-like of degree at most r. and suppose we have an exact sequence 0 → M → M → M → 0 of finitely generated R-modules. h(grI (M ). ar . n) = sI (M.3).2. d1 . hence has finite length over R/I . n) is polynomial-like of degree at most r − 1.7 Theorem Let I be an ideal of definition of the Noetherian local ring R. grI (M ) = ⊕n≥0 (I n M/I n+1 M ).2. n). The Hilbert-Samuel polynomial satisfies a property analogous to (5. n) = sI (M.5). Ri Rj ⊆ Ri+j . It follows that we may replace lR/I by lR in the above formula. but the degree d(M ) of sI (M. . Thus M/IM is Artinian (as well as Noetherian).2.1. Thus for every n ≥ 0 we have Mn+m = M ∩ I n+m M ⊇ I n+m M . and consequently I n+m M ⊆ Mn+m = I n Mm ⊆ I n M . ♣ 5. n) and lR (M /Mn ) have the same degree and the same leading coefficient. n) − lR (M /Mn ). . as well as nonnegative for n >> 0.1 The Dimension Theorem Definitions and Comments The dimension of a ring R.2. n) − sI (M . sI (M .3 5.3. and the leading coefficient of r(n) is nonnegative. Set Mn = M ∩ I n M . Proof. sI (M. By the Artin-Rees lemma (4. An example of an infinitedimensional ring is the non-Noetherian ring k [X1 . the filtration {Mn } is I -stable. Proof. Then d(M ) ≤ d(M ). n). so IMn = Mn+1 for sufficiently large n.7). n) − sI (M . will be taken as its Krull dimension. which implies that lR (M /I n+m M ) ≥ lR (M /Mn+m ) ≥ lR (M /I n M ). the maximum length n of a chain P0 ⊂ P1 ⊂ · · · ⊂ Pn of prime ideals of R.6 CHAPTER 5. X2 . n) respectively. The left and right hand terms of this inequality are sI (M . Then by additivity of length.2. ]. . n). If there is no upper bound on the length of such a chain. Apply (5. n) = lR (M /Mn ) hence lR (M /Mn ) is polynomial-like. say. n + m) and sI (M . and it follows that sI (M . denoted by dim R. n) − lR (M /Mn ) = r(n) is polynomial-like of degree less than deg lR (M /Mn ) ≤ deg sI (M. we take n = ∞. . where M is a finitely generated module over the Noetherian local ring R. where k is a field. DIMENSION THEORY where r(n) is polynomial-like of degree less than d(M ). n) = lR (M /Mn ) and r(n) = sI (M . The following sequence is exact: 0 → M /(M ∩ I n M ) → M/I n M → M /I n M → 0.1. n ≥ m. Moreover.8 Corollary Let M be a submodule of M . We have . noting that we can ignore r(n) because it is of lower degree than sI (M.7). The result now follows upon adding the equations sI (M. ♣ 5. . .5. .3. and J/I is a prime ideal of R/I iff J is a prime ideal of R.3). provided that the prime ideals in the chain contribute to M in the sense that they belong to the support of M . and we take the dimension of the zero module to be −1.5. the height of P is the dimension of the localized ring RP . Then by (1. By (1. There are several other ideas that arise from the study of chains of prime ideals. ar belonging to M we have lR (M/(a1 . we define dim M = dim(R/ ann M ) if M = 0. Since M is finitely generated. . and the coheight of I as the supremum of the coheights of prime ideals P ⊇ I . denoted by dim M . By (0. We now assume that M is nonzero and finitely generated over the Noetherian ring R. 1. and S = (R/I )/(J/I ). we can assert that there is a smallest positive integer r. At the other extreme. By (1. 3.3. has dimension 0 by (1. will be measured by length of chains of prime ideals. P ⊇ ann M iff P ∈ Supp M . A Dedekind domain that is not a field has dimension 1. Formally. 2. ar )M < ∞. If M = 0 we take δ (M ) = −1. .9). called the Chevalley dimension δ (M ). The coheight of the prime ideal P (notation coht P ) is the maximum length n of a chain of prime ideals P = P0 ⊂ P1 ⊂ · · · ⊂ Pn . integrally closed integral domain in which every nonzero prime ideal is maximal. a field. .5). THE DIMENSION THEOREM 7 the infinite chain of prime ideals (X1 ) ⊂ (X1 .2 The Dimension of a Module Intuitively. X3 ) ⊂ · · · . X2 . A Dedekind domain is a Noetherian. Supp(M/MM ) = V (ann(M/MM )) which coincides with {M} by Problem 2. Algebraic number theory provides many examples.5). Every associated prime ideal of M is maximal. Every prime ideal in Supp M is maximal. By the above equivalent conditions. We make the additional assumption that R is a local ring with maximal ideal M. 5. The length of M as an R-module is finite. the minimal elements of AP(M ) and Supp M are the same. .4). we define the height of I as the infimum of the heights of prime ideals P ⊇ I .6.) If I is an arbitrary ideal of R.6. so S is an integral domain iff R/J is an integral domain. . and by (1. . It follows from the correspondence theorem and the third isomorphism theorem that the coheight of P is the dimension of the quotient ring R/P . dim M = 0. and more generally an Artinian ring.9). the dimension of an R-module M . then S ∼ = R/J . 4.5. We define the height of a prime ideal P (notation ht P ) as the maximum length n of a chain of prime ideals P0 ⊂ P1 ⊂ · · · ⊂ Pn = P . X2 ) ⊂ (X1 .3. the following conditions are equivalent. By (1. such that for some elements a1 . (If I and J are ideals of R with I ⊆ J . lR (M/MM ) is finite.6). Thus dim M = sup{coht P : P ∈ Supp M } = sup{coht P : P ∈ AP(M )}. because the ring of algebraic integers of a number field is a Dedekind domain.5.4. M has at least one associated prime. 3. and apply (1.2) there is an injective homomorphism from R/Q to R/(Ra + P ). .5.3.1) that for some associated prime P we have dim M = coht P = dim R/P . If we can show that dim R/P ≤ d(R/P ). and let a1 . we have (R/(Ra + P ))Q = 0. . . To prove this. . n) + sM (R/(Ra + P ). The dimension dim M of the module M .) But the support .5) is V (P ) ∩ V (I ) = V (Q) = Supp R/Q. Thus assume d(M ) ≥ 0.5.3. set Q = I + P . [Since Ra + P ⊆ Q. n) + r(n) where r(n) is polynomial-like of degree less than d(R/P ).3. and assume that the result holds up to t − 1. . the maximal ideal of R. the induction hypothesis implies that t − 1 ≤ d(R/Q). Since the chain Q ⊂ P2 ⊂ · · · ⊂ Pt is of length t − 1. then sM (M. ar ) and let P be the annihilator of M .7).8) we have d(R/Q) ≤ d(R/(Ra + P )). If d(M ) = −1. DIMENSION THEORY 5. 2.3. ar )M has finite length. . Assume M = 0 and r = δ (M ) ≥ 0. By NAK. M = 0 so dim M = −1. Proof. (For convenience we take I = M. .] By (1.3. (TBGY.8) we have d(R/P ) ≤ d(M ). see Problem 3. first note that M/IM ∼ = M ⊗R R/I . which by (1. so by (5. 2. sM (R/P. The following quantities are equal: 1. hence d(R/P ) = −1 and we are finished. (Note that the annihilator of R/I is I and the annihilator of R/Q is Q.2. Therefore t ≤ d(R/P ).) By (5. Choose a ∈ P1 \ P . the length t of the chain is at most d(R/P ). where the map from R/P to itself is multiplication by a. Now the sequence 0 → R/P → R/P → R/(Ra + P ) → 0 is exact. We divide the proof into three parts. M has only finitely many associated primes. Then choose Q to be a minimal element in the support of R/(Ra + P ). hence dim M is finite. so by (5. . We claim that the support of R/Q is {M}. Thus d(R/(Ra + P )) < d(R/P ). n).2. then R/P = 0 (because P is prime). as desired. n) = sM (R/P.10).2. and we specify that the degree of the zero polynomial is -1. and consequently t − 1 < d(R/P ). . 1. (The image of the map is Ra + P .9). Supp M/IM = Supp M ∩ Supp R/I . By (1. ar be elements of M such that M/(a1 . Let I be the ideal (a1 .). We can pick a Q belonging to AP(R/(Ra + P )).2) there is an injective homomorphism from R/P to M .2) and (5. . If δ (M ) = −1. Thus assume t ≥ 1. By (1. The degree d(M ) of the Hilbert-Samuel polynomial sI (M. it will follow that dim M = dim R/P ≤ d(R/P ) ≤ d(M ). n) = lR (M/Mn M ) = 0 for n >> 0. and consider prime ideals Q such that Ra + P ⊆ Q ⊆ P1 . It follows from (5. 3. where I is any ideal of definition of R. d(M ) ≤ δ (M ).3 Dimension Theorem Let M be a finitely generated module over the Noetherian local ring R. . subsection S7. hence t − 1 ≤ d(R/(Ra + P )).) By Problem 9 of Chapter 1. then M = 0 and d(M ) = −1.5.9) or (1. .8 CHAPTER 5. The Chevalley dimension δ (M ).1 of the supplement. If t = 0. It suffices to show that for any chain of prime ideals P = P0 ⊂ · · · ⊂ Pt in R. dim M ≤ d(M ). 5)] I is an ideal of definition.) Again by (1. (We have already noted in part 1 that dim M is finite. But by n the correspondence theorem. n). Finiteness of dimension follows from (5. M ⊆ ∪1≤i≤t Pi . ar )N (apply the first isomorphism theorem) . t. . then M = 0 by NAK. so assume M = 0.4. . . in particular. . Moreover.3. Pt }.3. . Q is an ideal of definition of R. n) is at most r. also.11). . dim R < ∞.4 Consequences of the Dimension Theorem In this section we will see many applications of the dimension theorem (5. and consider M as an R-module. . . By the induction hypothesis. then dim M < ∞. Now assume dim M > 0.3). To see this. of the number of generators of I . 3.10). coht Pi = dim M for all i = 1. 5.2). ar )M ∼ = N/(a1 . Q is M-primary. so δ (M ) = 0. . so M/(a. (If M/IM = 0. Then Supp N ⊆ Supp M \ {P1 . NPi = 0 for all i because a ∈ / Pi .3. and let a1 . note that if NP = 0. Then R is a Noetherian local ring and Q is an ideal of definition of R generated by a1 .5) and (3. .6. hence sQ (M.9). where ai = ai + P . ar be elements of M such that N/(a1 . Let R = R/P . By (5. AP(R/Q) = {M}.4. . . because dim M = coht Pi and we are removing all the Pi . then by (5. But M/(a. .3). ar . .9). . Q = Q/P . . then MP = 0. The last assertion follows from the definition of Chevalley dimension in (5. δ (N ) ≤ dim N . If M is a finitely generated R-module. over all ideals I of definition of R. a1 . Let r = δ (N ). . . In summary.5.2). ♣ 5. hence division by a is allowed. Choose an element a in M such that a belongs to none of the Pi .3. ♣ .1 Proposition Let R be a Noetherian local ring with maximal ideal M.3. . . Thus dim N < dim M . δ (M ) ≤ dim M . M has finite length. lR (M/Q M ) = lR (M/Qn M ). . . . Proof. ar )N has finite length. n) = sQ (M. a1 . .6). R/I has finite length iff (by the Noetherian hypothesis) R/I is Artinian iff [by (5. Pi ⊂ M for every i. ar )M also has finite length. .1. CONSEQUENCES OF THE DIMENSION THEOREM 9 of M/IM is {M} by (1. It follows that d(M ) ≤ r. .2. . δ (M ) ≤ r + 1 = δ (N ) + 1 ≤ dim N + 1 ≤ dim M. By (5. so by (1.3. Since the dimension of M is greater than 0. the degree of the Hilbert-Samuel polynomial sQ (M. . If dim M = −1 then M = 0 and δ (M ) = −1.6. . Pt be the associated primes of M whose coheight is as large as it can be. In more detail. If dim M = 0. . so by the prime avoidance lemma (0.2. that is. proving the claim. . . the dimension of R is the minimum.1). and let N = M/aM . . . Thus δ (M ) ≤ r + 1. .2. . contradicting our assumption.) Let P1 . Then dim R ≤ dimk (M/M2 ). . Proof. . ar form a k -basis of M/M2 . But gr(R) = ˆ ) is Noetherian. so by (5. sM (R. . let A be the localized ring RP0 . P is minimal subject to P ⊇ I . .2.3. . By (5. The elements ai must belong to P . we have dim R ≤ r by (5. . which in turn implies that R ˆ is Noetherian. (See (3. .10). Therefore d(R) = d(R the result follows from (5. such that P is a minimal prime ideal over I . . and note that if P is minimal over I iff I = P . . In particular. . . of prime ideals of R will stabilize if and only if the chain AP0 ⊃ AP1 ⊃ AP2 ⊃ · · · of prime ideals of A stabilizes. then so is gr(R).) ♣ 5. a1 .1). (Sketch: [Note that R ∼ gr(R ˆ ). Then the chain P0 ⊃ P1 ⊃ P2 ⊃ .3). Then dim R = dim R ˆ M ˆ n .2.2. . We may assume without loss of generality that R is a local ring. Conversely. and R ˆ is also Noetherian. the result is immediate because dim R < ∞.4. Let R be a Noetherian local ring with maximal ideal M.2 Proposition Let R be a Noetherian local ring with maximal ideal M and residue field k = R/M. if P0 is a prime ideal of R. ♣ ˆ is local by Problem 16 of Chapter 4. and two Hilbert-Samuel polynomials must have the same degree.4 Theorem If R is a Noetherian ring. which is a contradiction because J must be proper. . where ai = ai + M. ar be elements of M such that a1 . . by (5. DIMENSION THEORY 5.3 Proposition ˆ its M-adic completion.6). then the prime ideals of R satisfy the descending chain condition. Then by (0.4. .10 CHAPTER 5. The following conditions are equivalent: (a) ht P ≤ n. Invoking (3.10) as in the first part of the proof. .1) there is an ideal of definition J of RP generated by n elements a1 /s.2.4.3.1) and (5.3. R/Mn ∼ = R/ M ˆ ). the height of M is at most r. (Alternatively. and R ˆ. (b) There is an ideal I of R that is generated by n elements.9). Since M itself is an ideal of definition (see (5. ar generate M. else the ai /s will generate RP . . Proof. we conclude that I satisfies (b). if (a) holds then dim RP ≤ n. . ht P = dim RP ≤ n.4. the associated graded ring of R.5).) By (5. ♣ . then IRP is an ideal of definition of RP√that is generated by n elements.3.)] so gr(R 5.5 Generalization of Krull’s Principal Ideal Theorem Let P be a prime ideal of the Noetherian ring R. ˆ n). If (b) holds. By (4.4.4.4. n) = s ˆ (R. (In other words.4. an .5).) Proof. But if R is local as well as Noetherian. Take I to be the ideal of R generated by a1 . If R is Noetherian. an /s with s ∈ R \ P . see (5.4). Let a1 . . Explicitly. Since ht M = dim R.1).5)). .3. ♣ 5. . the result follows. the Proof. . 5. CONSEQUENCES OF THE DIMENSION THEOREM 11 5. ♣ The hypothesis that a is not a unit is never used. .4. This allows us to find elements a1 . We may assume that P is minimal in the support of R. . Then a.4. ♣ 5. Xn ]] be a formal power series ring in n variables over the field k . then a cannot belong to the prime ideal P and the theorem is vacuously true.3).1) and (3. .8 Corollary Let a be a non zero-divisor belonging to the prime ideal P of the Noetherian ring R. . .10). as ∈ M. we appeal to part 2 of the proof of (5. such that the images ai in R/(a) generate an M/(a)-primary ideal of R/(a). which contradicts the hypothesis that a is not a zero-divisor. .4. then for some s ∈ R \ P we have sa = 0.3. . with s = dim R/(a). a contradiction. . Thus assume ht P = dim RP = 0.4. In the proof of part 3 of the dimension theorem (5.3. It follows from (5.3. P consists entirely of zero-divisors. If a is neither a unit nor a zerodivisor. so by (5. so the dimension of R is at most n. P is an associated prime of R.6 Krull’s Principal Ideal Theorem Let a be a nonzero element of the Noetherian ring R. hence P ∈ Supp R. Proof. take M = R and N = R/(a) to conclude that dim R/(a) < dim R. the dimension is at least n because of the chain (0) ⊂ (X1 ) ⊂ (X1 . Xn ) of prime ideals. Proof. By (1. and let a ∈ M be a non zerodivisor. as generate an M-primary ideal of R. hence dim R/(a) ≤ dim R − 1. . because otherwise P has height 1 and we are finished. .5) that ht P ≤ 1.6).9). The unique maximal ideal is (X1 . . dim R ≤ 1 + s = 1 + dim R/(a). for if dim R = 0.6). If a is a unit.4. ♣ 5. so by (1.4. In (5. . X2 ) ⊂ · · · ⊂ (X1 . We have dim R > 0. To prove equality. a1 . and as in the proof of (5. For if a/1 = 0. On the other hand. . 5. Xn ). . . .4. . Proof. ♣ .3. Proof. . .9 Theorem Let R = k [[X1 . replace R by RP and R/(a) by (RP )Q .3). a contradiction.4. then M is the only prime ideal. .7 Theorem Let R be a Noetherian local ring with maximal ideal M.4. then every minimal prime ideal P over (a) has height 1. where Q is a minimal prime ideal over (a). but nothing is gained by dropping it. Then dim R = n. M consists entirely of zero-divisors. We claim that RP = 0.7). Then ht P/(a) = ht P − 1. Then dim R/(a) = dim R − 1.5. . Now Y1 satisfies the equation x1 = g (Y1 . so g (Y1 . .12 CHAPTER 5.) 2. we have x1 = g (Y1 . and we find elements x0 . . . For each i = 1. say α = β . To study them we will need a stronger version of Noether’s normalization lemma. (This is the standard normalization lemma. Then there will be a unique α that maximizes f .2 Theorem Let A be a finitely generated k -algebra. . xn ]. It suffices to let A be a polynomial ring k [Y1 . . . . . and therefore A is integral over B .5. Proof. α = (a1 . . xn ∈ A . satisfy the required conditions. . . Ym ). . . hence are integral over B ) all Yi are integral over B . Ym ]. . . relative to the ideals I0 ⊂ I1 ⊂ · · · ⊂ Ir . . m) such that A is integral over B = k [x1 . Ym ) for some nonconstant polynomial g with coefficients in k . where s is greater than the maximum of the aj .5 5. If Ii is the preimage of Ii under the canonical map A → A /I0 . . If we write the polynomial g as a sum of monomials the above equation becomes cα Y α . . where xi = Yi − Y1ri . . . . . and pick the ri so that all the f (α) are distinct. 5. The proof is by induction on r. A is integral over B = k [x1 . . xm )Y1j = 0 . then the images of xi−h(0) in A. . . . . . DIMENSION THEORY 5. xh(i) . xm + Y1rm ) − x1 = 0. cα = 0. . We first consider the case in which I1 is a principal ideal (x1 ) = x1 A with x1 ∈ / k . There exists a nonnegative integer n and elements x1 . . xn ∈ A algebraically independent over k such that the following conditions are satisfied. By our assumption that A is a polynomial ring. . i = 2. . . . . Affine algebras are of great interest in algebraic geometry because they are the coordinate rings of affine algebraic varieties. let f (α) = a1 + a2 r2 + · · · + am rm . In this section we will give the statement and proof. If we can show that Y1 is integral over B . then (since the xi belong to B .1 Strengthening of Noether’s Normalization Lemma Definition An affine k -algebra is an integral domain that is also a finite-dimensional algebra over a field k . . For we may write A = A /I0 where A = k [Y1 . cα Y1a1 (x2 + Y1r2 )a2 · · · (xm + Y1rm )am − x1 = 0. x2 + Y1r2 . . . . . . m.5. . Ym ]. . . . . We claim that there are positive integers ri (i = 2. . . . and I1 ⊂ · · · ⊂ Ir a chain of nonzero proper ideals of A. . r. . . page 42. For example. . To produce the desired ri . there is a positive integer h(i) such that Ii ∩ B is generated (as an ideal of B ) by x1 . following Serre’s Local Algebra. . and we have cβ Y1 f (β ) + j<f (β ) pj (x1 . Assume r = 1. 1. i > h(0). . take ri = si . . . am ). xm ]. . . and x1 A ∩ C = x1 C . . . hence. xm ) = B = k [x1 . t2 . . there are elements xs+1 . Ym ] is integral over B = k [x1 . . . . . Since A has transcendence degree m over k and an integral extension must be algebraic. . . If Q ⊂ Q . A = k [Y1 . we may assume that R is a field. and as we noted above. . Finally. (A .6 Properties of Affine k -algebras We will look at height. there are elements t2 . . . Note that we have also shown that A ∩ k (x1 .1 Proposition Let S = R[X ] where R is an arbitrary ring. But then every nonzero prime ideal of S is maximal. u ∈ A ∩ k (x1 . where Q and Q are prime ideals of S both lying above the same prime ideal P of R. we now consider the general case by induction on m. we may assume without loss of generality that P = 0. an integral domain. . and take J = IS and Q = P S . Proof. . we may assume that I = 0. . . dividing by x1 . Still assuming r = 1. . . Then Q1 ∩ R ⊆ Q ∩ R = P S ∩ R = P . Suppose that the prime ideal Q1 of S is properly contained in Q. . and let P be a prime ideal of R with I ⊆ P . tm satisfy the conditions of the theorem for the chain of ideals I1 ⊂ · · · ⊂ Ir−1 . . where I1 is principal). . Then x1 . tm ]. . If P is a minimal prime ideal over I . By localizing with respect to the multiplicative set R \ {0}. . then Q is a minimal prime ideal over J . A is integral over the polynomial ring C = k [x1 . Since A is integral over B . xm ]. let t1 . . xm are algebraically independent over k . tm ] and the ideal I1 ∩ k [t2 . . . . . To verify that Q is prime. . Since P S is also 0. Take xi = ti for 1 ≤ i ≤ s.6. then Q = P S . xm satisfy the desired conditions. . it is a unique factorization domain and therefore integrally closed. . tm ∈ A such that x1 . note that R[X ]/P R[X ] ∼ = R[X ]/P [X ] ∼ = (R/P )[X ]. and note that x1 ∈ / k because I1 is proper. . Thus x1 A ∩ B = x1 B . . ♣ 5. . xm ]. tm ]. By what we have just proved. Let x1 be a nonzero element of I1 . . . If m = 0 there is nothing to prove. . tm ].6. 5. and the proof of the first case is complete. ♣ 5. . . .2 Corollary Let I be an ideal of the Noetherian ring R. . . the result follows. . . tm ] satisfying the conditions for the ideal Ir ∩ k [ts+1 . . . it follows that x1 . xm ). . . Since R/P can be regarded as a subring of S/Q. Proof. Let S be the polynomial ring R[X ]. . there are elements x2 . Then t = x1 u with u ∈ A. . The right-to-left inclusion follows from our assumptions about x1 . hence Q = 0. we have u ∈ B . and let s = h(r − 1).5. xm ∈ k [ts+1 . . . . coheight and dimension of affine algebras. . . the second assertion will also hold. If we can show that I1 ∩ B = (x1 ) = x1 B . . we take the inductive step from r − 1 to r. By the induction hypothesis. . xm satisfying the conditions of the theorem for k [t2 . .6. PROPERTIES OF AFFINE K -ALGEBRAS 13 so Y1 is integral over B . . . By the argument of the previous paragraph. and we have already taken care of m = 1 (because A is then a PID). . . . . By modding out I . Thus the first assertion of the theorem holds (in this first case. . . . t2 . tm are algebraically independent over k . Since B is isomorphic to a polynomial ring. so let t belong to I1 ∩ B . . the dimension of R.6. Since the height of Pj is at least j . 1 ≤ i ≤ n. Proof. so ht Q ≥ ht P . P a prime ideal of R.4. . By the generalized Krull principal ideal theorem (5.6.6. 1. . Then ht(X1 . and let Pi = Qi ∩ R for every i = 0. . X2 ) ⊂ · · · ⊂ (X1 . ♣ 5. and by (5. If Qn = Pn S . . . . DIMENSION THEORY polynomial belonging to R coincides with its constant term. so again by (5. where R is a Noetherian ring. then Q0 ⊂ Q1 ⊂ · · · ⊂ Qn = Q. ♣ 5.3). (Note that X cannot belong to Qn because 1 ∈ / Pn .4. . . . Xn ]. . Pj = Pj +1 ⊂ Pj +2 ⊂ · · · ⊂ Pn so ht Pj + n − j − 1 ≤ dim R. if K is a field then dim K [X1 . .) It follows that dim S ≥ 1 + dim R.3). then dim R[X1 . ♣ If X is an affine algebraic variety over the field k . . Proof. This follows from (5. a contradiction. The height of (X1 .6. . . Then ht P = ht Q.3 Proposition As above.) By minimality. . . Conversely. Now consider a chain Q0 ⊂ Q1 ⊂ · · · ⊂ Qn of prime ideals of S .1).1). . Proof. .6 Corollary Let R = K [X1 .4) by induction.2). there is an ideal I of R generated by n elements such that P is a minimal prime ideal over I . ♣ 5.6. By (5. . 5. ht Qn = ht Pn . Xi ) = i.5 Corollary If R is a Noetherian ring. . then by (5. But J is also generated over S by n elements. Q1 ∩ R = P .6. Proof.5). . Xn ). in view of the chain (X1 ) ⊂ (X1 . let S = R[X ]. We now show that the geometric dimension coincides with the algebraic (Krull) dimension.6. By (5.14 CHAPTER 5. . The general result now follows by induction. Xn ) is at most n. ht Q ≤ ht P . if P0 ⊂ P1 ⊂ · · · ⊂ Pn = P ⊂ R and Qi = Pi [X ]. using (5. hence dim S ≤ 1 + dim R. Q = P S is a minimal prime ideal over J = IS . . Q1 = P S = Q. Xn ] = n + dim R. We may assume that the Pi are not all distinct (otherwise dim R ≥ dim S ≥ dim S − 1). .8). We abbreviate transcendence degree by tr deg. where K is a field.6. In particular. Let P0 ⊂ P1 ⊂ · · · ⊂ Pn be a chain of prime ideals of R. R Noetherian. . Let n be the height of P . Then dim S = 1 + dim R.5). ♣ We may now prove a major result on the dimension of a polynomial ring. . Xn ] = n. First consider i = n. Q = P S . we have n − 1 ≤ dim R. Let j be the largest index i such that Pi = Pi+1 . But the Q sequence can be extended via Qn ⊂ Qn+1 = Qn + (X ). But by choice of j . ht Pj = ht Qj ≥ j .4 Theorem Let S = R[X ]. and in fact the height is n. Qj = Pj S . By (5.6. . n.4.6. . . its (geometric) dimension is the transcendence degree over k of the function field (the fraction field of the coordinate ring). and does not increase height.5). along with (5. . . . . . . .3. . yn algebraically independent over k such that R is integral over k [y1 . . .5. yn ].) By the strong form (5. . then dim R = tr degk Frac R. xn ]. it follows from (5. . . . . Let F = Frac R and L = Frac k [x1 . . . . .6). . 5. algebraically independent over k .6. . . We can assume that R = k [X1 .6. Proof.7 Theorem If R is an affine k -algebra. there are elements y1 . . . . R is integral over a polynomial algebra. .5) that coht Q = dim k [yh+1 . .6. By Noether’s normalization lemma. PROPERTIES OF AFFINE K -ALGEBRAS 15 5. . yh ) Since k [y1 . .6. Xn ] with ht P = h. (See Problems 4. . we therefore have tr degk F = tr degk L = n = dim R. . .6. so ht P + coht P = h + (n − h) = n = dim R.2) of Noether’s normalization lemma.6. Proof.8 Theorem If P is a prime ideal of the affine k -algebra R. dim R = dim k [x1 . . . . . .5 and 6.1) and (5. . . then ht P + coht P = dim R. xn ∈ R. . yn ] and Q = P ∩ k [y1 . ♣ . Since an integral extension cannot increase dimension (see Problem 4). xn ]. then ht P + coht P ≤ dim R. Then F is an algebraic extension of L. and since an algebraic extension cannot increase transcendence degree. So if height plus coheight equals dimension in the smaller ring. there is equality. . yn ]/Q ∼ = k [yh+1 . yn ] = (y1 . But coht Q = coht P (Problem 5). . the same must be true in the larger ring. such that R is integral over k [x1 . yn ] = n − h. . . An integral extension preserves dimension and coheight. . ♣ It follows from the definitions that if P is a prime ideal of R. there are elements x1 . By Noether’s normalization lemma. xn ] = n by (5. . .5. If R is an affine k -algebra. .