Supplementary Notes on Commutative Algebra1

April 11, 2017

1These notes are meant as a supplement to the Lectures on Commutative Algebra, that are avail- able online at http://www.math.iitb.ac.in/∼srg/Lecnotes/AfsPuneLecNotes.pdf and are primarily meant for the students of MA 526 at IIT Bombay in Spring 2017. The contents in these notes are mainly drawn from parts of Chapters 2 and 3 of Combinatorial Topology and Algebra, RMS Lecture Notes Series No. 18, Ramanujan Math. Soc., 2012. Contents

1 Graded Rings and Modules 3 1.1 Graded Rings ...... 3 1.2 Graded Modules ...... 5 1.3 Primary Decomposition in Graded Modules ...... 7

2 Hilbert Functions 10 2.1 Hilbert Function of a graded module ...... 10 2.2 Hilbert-Samuel polynomial of a local ring ...... 12

3 Dimension Theory 14 3.1 The dimension of a local ring ...... 14 3.2 The Dimension of an Afﬁne Algebra ...... 18 3.3 The Dimension of a Graded Ring ...... 20

References 23

2 Chapter 1

Graded Rings and Modules

This chapter gives an exposition of some rudimentary aspects of graded rings and modules. It may be noted that exercises are embedded within the text and not listed separately at the end of the chapter.

1.1 Graded Rings

In this section, we shall study some basic facts about graded rings. The study extends to graded modules, which are discussed in the next section. The notions and results in the first two sections will allow us to discuss, in Section 1.3, some special properties of associated primes and primary decomposition in the graded situation. The prototype of a graded ring is the polynomial ring A[X1,...,Xn] over a ring A. Before giving the general definition, we point out that the set of all nonnegative integers is denoted by N. Definition: A ring R is said to be N-graded, or simply graded, if it has additive subgroups Rd, for d ∈ N, such that R = ⊕d∈NRd and RdRe ⊆ Rd+e, for all d, e ∈ N. The family {Rd}d∈N th is called an N-grading of R, and the subgroup Rd is called the d graded component of R. Let R = ⊕d∈NRd be a graded ring. Elements of Rd are said to be homogeneous of degree d a ∈ R a = P a a ∈ R . Every can be uniquely written as d∈N d with d d such that all except finitely many ad’s are 0; we call ad’s to be the homogeneous components of a. Given a homogeneous element b ∈ R, we sometimes write deg(b) to denote its degree. An ideal of R generated by homogeneous elements is called a homogeneous ideal. Note that an ideal is homogeneous iff it contains the homogeneous components of each of its elements. If I is a homogeneous ideal of R, then R/I has the induced N-grading given by (R/I)d = Rd/(I ∩ Rd). A subring S of R is said to be graded if S = ⊕d∈N(S ∩ Rd). Note that R0 is 0 0 a graded subring of R and each Rd is an R0-module. If R = ⊕d∈NRd is any graded ring, then a homomorphism φ : R → R0 is called a graded ring homomorphism, or a homomorphism 0 of graded rings, if φ(Rd) ⊆ Rd for all d ∈ N; note that in this case the kernel of φ is a homogeneous ideal of R and the image of φ is a graded subring of R0.

Examples: 1. Let A be a ring and I be a homogeneous ideal of A[X1,...,Xn]. Then R = th A[X1,...,Xn]/I is a graded ring, with its d graded component being given by Rd =

3 A[X1,...,Xn]d/Id, where Id = I ∩ A[X1,...,Xn]d. Note that in this case R as well as Rd are f. g.A-algebras, and in particular, A-modules. Graded rings of this type are usually called graded A-algebras. def d d+1 2. Let A be a ring and I be any ideal of A such that I 6= A. Then grI (A) = ⊕d∈NI /I is a graded ring. It is called the associated graded ring of A w.r.t. I.

(3.1) Exercise: Let R = ⊕d∈NRd be a graded√ ring, and I,J be homogeneous ideals of R. Then show that I + J, IJ, I ∩ J, (I : J) and I are homogeneous ideals.

(3.2) Exercise: Let R = ⊕d∈NRd be a graded ring, and I be a homogeneous ideal of R. Show that I is prime iff I is prime in the graded sense, that is, I 6= R and

a, b homogeneous elements of R and ab ∈ I ⇒ a ∈ I or b ∈ I.

∗ Given an ideal I of a graded ring R = ⊕d∈NRd, we shall denote by I the largest homogeneous ideal contained in I. Note that I∗ is precisely the ideal generated by the homogeneous elements in I. ∗ (3.3) Lemma. If R = ⊕d∈NRd is a graded ring and p is a prime ideal of R, then p is a prime ideal of R. Proof: Follows from (3.2). 2

(3.4) Lemma. Let R = ⊕d∈NRd be a graded ring. Then 1 ∈ R0. 1 = P a a ∈ R b ∈ R b = P ba Proof: Write d∈N d with d d. Then for any e, d∈N d. Equating the degree e components, we ﬁnd b = ba0. This implies that ca0 = 1 for all c ∈ R. Hence 1 = a0 ∈ R0. 2

Given a graded ring R = ⊕d∈NRd, we deﬁne R+ = ⊕d>0Rd. Note that R+ is a homogeneous ideal of R and R/R+ ' R0. Thus if R0 is a ﬁeld, then R+ is a homogeneous maximal ideal of R; moreover, R+ is also maximal among all homogeneous ideals of R other than R, and so it is the maximal homogeneous ideal of R.

(3.5) Lemma. Let R = ⊕d∈NRd be a graded ring and x1, . . . , xn be any homogeneous elements of positive degree in R. Then R+ = (x1, . . . , xn) iff R = R0[x1, . . . , xn]. In particular, R is noetherian iff R0 is noetherian and R is a f. g.R0-algebra.

Proof: Suppose R+ = (x1, . . . , xn). We show by induction on d that each x ∈ Rd is in R0[x1, . . . , xn]. The case of d = 0 is clear. Suppose d > 0. Write x = a1x1 + ··· + anxn, where a1, . . . , an ∈ R. Since x is homogeneous, we may assume, without loss of generality, that each ai is homogeneous. Then for 1 ≤ i ≤ n, we must have deg(ai) = d − deg(xi) < d, and so, by induction hypothesis, ai ∈ R0[x1, . . . , xn]. It follows that R = R0[x1, . . . , xn]. Conversely, if R = R0[x1, . . . , xn], then it is evident that R+ = (x1, . . . , xn). The second assertion in the Lemma follows from the ﬁrst one by noting that f. g. algebras over noetherian rings are noetherian. 2

(3.6) Exercise: Let R = ⊕d∈NRd be a graded ring. Show that R is noetherian iff it satisﬁes a.c.c. on homogeneous ideals.

4 (3.7) Exercise: Let A be a ring and I = (a1, . . . , ar). If a¯1,..., a¯r denote the images of 2 a1, . . . , ar mod I , then show that grI (A) = (A/I)[¯a1,..., a¯r] ' (A/I)[X1,...,Xr]/J, for some homogeneous ideal J of (A/I)[X1,...,Xr]. Deduce that if (A/I) is noetherian and I is f. g., then grI (A) is noetherian.

(3.8) Graded Noether Normalisation Lemma. Let k be an infinite field and R = k[x1, . . . , xn] be a graded k-algebra such that R0 = k and deg(xi) = m for 1 ≤ i ≤ n. Then there exist homogeneous elements θ1, . . . , θd of degree m in R such that θ1, . . . , θd are algebraically independent over k and R is integral over the graded subring k[θ1, . . . , θd]. In particular, R is a finite k[θ1, . . . , θd]-module.

Proof: If k is an inﬁnite ﬁeld, then in (4.11) and (4.13) of Chapter 1, the elements θ1, . . . , θd can be chosen to be k-linear combinations of x1, . . . , xn. The result follows.

1.2 Graded Modules

The notion of a graded ring extends readily to that of graded modules. For simplicity, we initially consider N-graded modules, but point out toward the end of this section that many of the notions and results below extend easily to Γ-graded modules where Γ is an s s arbitrary additive monoid (in particular, Γ can be Z, Z , or N , where s is any positive integer).

Deﬁnition: Let R = ⊕d∈NRd be a graded ring, and M be an R-module. Then M is said to be N-graded or simply, graded, if it contains additive subgroups Md, for d ∈ N, such that M = ⊕d∈NMd, and RdMe ⊆ Md+e, for all d, e ∈ N. Such a family {Md}d∈N is called an th N-grading of M, and Md the d graded component of M.

Let R = ⊕d∈NRd be a graded ring and M = ⊕d∈NMd be a graded R-module. By a graded submodule of M we mean a submodule N of M such that N = ⊕d∈N(N ∩ Md). Note that if N is a graded submodule of M, then M/N is a graded R-module with the induced 0 0 grading given by (M/N)d = Md/(N ∩Md). If M = ⊕d∈NMd is any graded R-module, then by graded module homomorphism, or a homomorphism of graded modules, we mean a R-module 0 0 homomorphism φ : M → M such that φ(Md) ⊆ Md, for all d ∈ N. For any such φ, the kernel of φ and the image of φ are graded submodules of M and M 0 respectively.

Examples: 1. Let R = A[X1,...,Xn] and I be a homogeneous ideal of R. Then M = R/I is a graded R-module. In this example, the graded components Md are ﬁnite A-modules. 2. Let A be a ring and I be an ideal of A. If M is any A-module, then grI (M) = d d+1 ⊕d∈NI M/I M is a graded grI (A)-module. It is called the associated graded module of M w.r.t. I. (3.9) Exercise: Let the notation be as in (3.7). Let M be an A-module. Suppose M = Ax1 + ··· + Axm, and x¯1,..., x¯m denote the images of x1, . . . , xm in M/IM, then show that grI (M) = grI (A)¯x1 + ··· + grI (A)¯xm. Deduce that if A is noetherian and M is f. g., then grI (M) is a noetherian grI (A)-module.

(3.10) Graded Nakayama’s Lemma. Let R = ⊕d∈NRd be a graded ring and M = ⊕d∈NMd

5 be a graded R-module. If R+M = M, then M = 0. Proof: If M 6= 0, we can find a nonzero element x of least degree in M. Now the assumption x ∈ R+M leads to a contradiction. 2 (3.11) Artin-Rees Lemma. Let A be a noetherian ring, I an ideal of A and M a f.g. A- module. Suppose M = M0 ⊇ M1 ⊇ M2 ⊇ · · · is a chain of submodules of M such that 0 IMn ⊆ Mn+1 for all n ≥ 0 with equality for n ≥ r (for some r). Then for any submodule M 0 0 0 0 of M, there exists s ≥ 0 such that IMn = Mn+1 for all n ≥ s, where Mi denotes M ∩ Mi. Proof: Let t be an indeterminate over A and let R be the subring of A[t] defined by R = n n 2 2 ⊕n≥0I t = A⊕It⊕I t ⊕· · · . If I = (a1, . . . , am), then R = A[a1t, . . . , amt] and so R is a f.g. graded A-algebra [called the Rees algebra of I]. Thus by (3.5), R is noetherian. Moreover, n 0 0 0 L = ⊕n≥0Mnt is naturally a graded R-module. Now clearly IMn ⊆ M ∩ IMn ⊆ Mn+1, 0 0 n and hence L = ⊕n≥0Mnt is an R-submodule of L. By (3.1) of Chapter 1, each Mn is a f.g. n A-module, and hence so is Mfn = M0 ⊕ M1t ⊕ · · · ⊕ Mnt . Consequently, the R-module ˜ n n+1 2 n+2 Ln = M0 ⊕ M1t ⊕ · · · ⊕ Mnt ⊕ IMnt ⊕ I Mnt ⊕ · · · , generated by Mfn, is f.g. Since IMn = Mn+1 for n ≥ r, it follows that L = Lñ for n ≥ r, and so L is a noetherian R- ˜0 ˜0 ˜0 module. Hence the chain L0 ⊆ L1 ⊆ L2 ⊆ · · · of R-submodules of L terminates, i.e., there ˜0 ˜0 0 0 exists s ≥ 0 such that Ln = Ls for n ≥ s. It follows that IMn = Mn+1 for n ≥ s. 2 0 n 0 n n 0 n−s Remark: By taking Mn = I M in (3.11), we get M ∩ I M = I (M ∩ I M) for n ≥ s. This, or the particular case when M = A and M 0 = J, an ideal of A, is often the version 0 n of Artin-Rees Lemma used in practise. For example, putting M = ∩n≥0I M and using Nakayama’s Lemma, we get a proof of Krull’s Intersection Theorem [viz., (3.3) of Ch. 1]. Artin-Rees Lemma is most useful in the theory of completions. For more on completions, which may be regarded as the fourth fundamental process, one may refer to [AM, Ch. 10].

Sometimes, it is useful to consider gradings which are more general than N-grading. s Definition: Let R be a ring and s be a positive integer. By a Z -grading on R we mean a family {Rα}α∈Zs of additive subgroups of R such that R = ⊕α∈Zs Rα and RαRβ ⊆ Rα+β, s for all α, β ∈ Z . Elements of Rα are said to be homogeneous of degree α. A ring with a s s Z -grading is called a Z -graded ring. The definitions of graded subring, homogeneous ideal, graded ring homomorphism, s in the context of Z -graded rings, are exactly similar to those in the case of N-graded rings. s s If R = ⊕α∈Zs Rα is a Z -graded rings, and M is an R-module, then M is said to be Z - s graded if it contains additive subgroups Mα, for α ∈ Z , such that M = ⊕α∈Zs Mα and s RαMβ ⊆ Mα+β, for all α, β ∈ Z . Corresponding notions of homogeneous element, graded submodule, and graded module isomorphism, etc. are defined in a similar fashion. s s If {Rα}α∈Zs is a Z -grading on a ring R such that Rα = 0 for all α = (α1, . . . , αs) ∈ Z s with αi < 0 for some i, then it is called an N -grading; note that in this case R = ⊕α∈Ns Rα. s Similarly, one has the notion of N -gradings for modules. n Examples: 1. Let A be a ring and R = A[X1,...,Xn]. Then R has a Z -grading {Rα}α∈Zs α n α given by Rα = AX , where for α = (α1, . . . , αn) ∈ Z , X denotes the monomial α1 αn s α s X1 ··· Xn if α ∈ N and X = 0 if α 6∈ N . Note that if A = k, a field, then the ho-

6 mogeneous ideals of R, w.r.t. the above grading, are precisely the monomial ideals of k[X1,...,Xn]. 2. Let A be a ring and R = A[X1,...,Xn]. Let d1, . . . , dn be any integers. For d ∈ α1 αn Z, let Rd be the A-submodule of R generated by the monomials X1 ··· Xn for which d1α1 + ··· + dnαn = d. Then {Rd}d∈Z deﬁnes a Z-grading on R which is an N-grading iff di = deg(Xi) ≥ 0 for 1 ≤ i ≤ n. The corresponding homogeneous polynomials and homogeneous ideals are sometimes called weighted homogeneous polynomials and weighted homogeneous ideals respectively.

The notion of Z-gradings permits us to define the following simple but useful operation on graded rings. If R = ⊕d∈NRd is a graded ring and m is any integer, then we define R(m) to be the graded R-module obtained from R by shifting, or by twisting, the grading by m, i.e., R(m)d = Rm+d for d ∈ Z and R(m) = ⊕d∈ZR(m)d. Further, if M = ⊕d∈NMd is a graded R-module, then we define M(m) = ⊕d∈ZM(m)d, where M(m)d = Mm+d. Note that M(m) is a graded R-module.

s Remark: Most of the results proved earlier in this section extend to Z -graded rings and s Z -graded modules over them. The notion of grading can also be extended by replacing s s Z byN or more generally, a monoid; most results extend to this case as well provided the monoid is assumed to be torsionfree. See [B] or [N] for more on this.

1.3 Primary Decomposition in Graded Modules

Throughout this section, R = ⊕d∈NRd denotes a graded ring and M = ⊕d∈NMd a graded R-module. Given a submodule N of M, by N ∗ we denote the largest graded submodule of N, i.e., N ∗ is the submodule generated by the homogeneous elements in N. Note that ∗ N = ⊕d∈NN ∩ Md. (4.1) Lemma. Every associated prime of M is a homogeneous prime ideal and the annihi- lator of some homogeneous element of M. Proof: Suppose p ∈ Ass(M) and p = (0 : x) for some x ∈ M. Then x 6= 0, and we can write x = xe + xe+1 + ··· + xd with xj ∈ Mj and xe 6= 0. Now for any a ∈ p with a = ar + ar+1 + ··· + as and ai ∈ Ri, the equation ax = 0 yields the equations

arxe = 0, arxe+1 + ar+1xe = 0, arxe+2 + ar+1xe+1 + ar+2xe = 0,...

2 3 d−e+1 d−e+1 which imply that arxe = a xe+1 = a xe+2 = ··· = a xd = 0. Hence a x = 0, p r r r r i.e., ar ∈ (0 : x). Since p = (0 : x) is prime, we have ar ∈ p. Now (a − ar)x = 0 and using arguments similar to those above, we ﬁnd that ar+1 ∈ p. Proceeding in this manner, we see that ai ∈ p for r ≤ i ≤ s. This proves that p is homogeneous. Moreover, if we let d Ij = (0 : xj), then we have p ⊆ Ij for e ≤ j ≤ d and ∩j=eIj ⊆ p. Since p is prime, we have Ij ⊆ p for some j and thus p = (0 : xj). 2 (4.2) Corollary. Given any p ∈ Ass(M), there exists an integer m and a graded ring isomorphism of (R/p)(−m) onto a graded submodule of M.

7 Proof: Write p = (0 : x) for some homogeneous element x of M. Let m = deg(x). The map a 7→ ax of R → M is a homomorphism with p as its kernel and it maps Rd into Mm+d. Hence it induces a desired graded ring isomorphism of (R/p)(−m). 2 (4.3) Exercise: Show that if p ∈ Supp (M), then p∗ ∈ Supp (M).

(4.4) Theorem. If R is noetherian and M is f. g., then there exists a chain 0 = M0 ⊆ M1 ⊆ · · · ⊆ Mn = M of graded submodules of M, homogeneous prime ideals p1,..., pn of R and integers m1, . . . , mn such that Mi/Mi−1 ' (R/pi)(−mi), the isomorphism being that of graded modules. Proof: Similar to the proof of (1.7) in view of (4.1) and (4.2) above. 2 Now let us turn to primary submodules of M. (4.5) Lemma. Let N be a graded submodule of M such that N 6= M and for any homogeneous elements b ∈ R and y ∈ N we have

by ∈ N and y∈ / N =⇒ bnM ⊆ N for some n ≥ 1.

Then N is a primary submodule of M. Proof: Let a ∈ R and x ∈ M be any elements such that ax ∈ N and x∈ / N. Then 0 0 we can write x = x + xe + xe+1 + ··· + xd with x ∈ N, xj ∈ Mj and xe ∈/ N. Let a = ar + ar+1 + ··· + as where ai ∈ Ri. Now a(xe + xe+1 + ··· + xd) ∈ N, and since N is n1 graded, we find that arxe ∈ N. By assumption, there exists n1 ≥ 1 with ar M ⊆ N. Now n1 n1n2 (a − ar) (xe + xe+1 + ··· + xd) ∈ N, and hence there exists n2 ≥ 1 with ar+1 M ⊆ N. n0 Proceeding in this manner, we can find n0 ≥ 1 such that ai M ⊆ N for r ≤ i ≤ s. n Therefore, we can find n ≥ 1 [e.g., n = n0(s − r + 1)] such that a M ⊆ N. Thus N is primary. 2 (4.6) Lemma. If Q is a p-primary submodule of M, then Q∗ is p∗-primary. Proof: Since the homogeneous elements of Q (resp: p) are elements of Q∗ (resp: p∗), it follows from (4.5) that Q∗ is a primary submodule. Further, if a is any homogeneous el- p ement of p = Ann(M/Q), then anM ⊆ Q for some n ≥ 1, and since M is graded, p anM ⊆ Q∗. It follows that p∗ ⊆ Ann(M/Q∗). On the other hand, since M/Q∗ is graded p and Ann(M/Q∗) ⊆ Ann(M/Q), we see that Ann(M/Q∗) is a homogeneous ideal con- p tained in p, and hence Ann(M/Q∗) ⊆ p∗. Thus Q∗ is p∗-primary. 2 (4.7) Theorem. Suppose A is noetherian, M is f. g., and N is a graded submodule of M. Let N = Q1 ∩ · · · ∩ Qh be a primary decomposition of N, and p1,..., ph be the associated primes corresponding to Q1,...,Qh, respectively. Then we have

∗ ∗ ∗ ∗ (i) pi = pi , Qi is pi-primary for 1 ≤ i ≤ h and N = Q1 ∩ · · · ∩ Qh.

(ii) If the primary decomposition N = Q1 ∩· · ·∩Qh is irredundant, then so is the primary ∗ ∗ decomposition N = Q1 ∩ · · · ∩ Qh.

(iii) If pi is minimal among p1,..., ph, then Qi is a graded submodule of N.

8 ∗ ∗ Proof: Applying (4.1) to M/N, we get pi = pi . So by (4.6), we ﬁnd that Qi is pi-primary. ∗ ∗ ∗ Since N is graded and N ⊆ Qi, we have N ⊆ Qi . Thus N ⊆ Q1 ∩· · ·∩Qh ⊆ Q1 ∩· · ·∩Qh = N, which proves (i). If N = Q1 ∩ · · · ∩ Qh is an irredundant primary decomposition, then p1,..., ph are distinct, and Ass(M/N) = {p1,..., ph}. Thus the associated primes corre- ∗ ∗ ∗ ∗ sponding to Q1,...,Qh are distinct. Moreover, if the decomposition N = Q1 ∩ · · · ∩ Qh can be shortened, then Ass(M/N) would have less than h elements, which is a contradiction. This proves (ii). Finally, if pi is minimal among p1,..., ph, then the corresponding primary ∗ component Qi is unique. Hence from (i), we obtain Qi = Qi , i.e., Qi is graded. 2 Remark: As a special case of the results of this section, we obtain some useful results about ideals in graded rings. You may ﬁnd it instructive to write these down explicitly. (4.8) Exercise: Show that all the results of this section remain valid if R is replaced by s s a Z -graded ring and M by a Z -graded module over it. Deduce from this that if I is a monomial ideal in k[X1,...,Xn], then the associated primes of I are monomial ideals and that I has a primary decomposition such that each of the primary ideals occurring in it is a monomial ideal.

9 Chapter 2

Hilbert Functions

In this chapter, we discuss basic facts about Hilbert functions, and related objects such as Hilbert series and Hilbert polynomials, in the setting of f.g. graded modules over graded rings, and also of local rings and f.g. modules over local rings.

2.1 Hilbert Function of a graded module

Let R = ⊕α∈Ns Rα be a graded ring (s some non-negative integer). Assume that R0 is Artinian and R is finitely generated as an R0-algebra. Let N = ⊕α∈Zs Nα be a finitely generated graded R-module. Then each Nα is a finitely generated R0-module, hence of s finite length. The Hilbert function of N is the map H(N): Z → Z given by H(N)(α) =

`(Nα) = length R0 (Nα). Let T1,...,Ts be indeterminates and put X F (N,T ) = H(N)(α)T α, α∈Zs

α α1 αs where T = T1 ...Ts for α = (α1, . . . , αs). This is called the Hilbert series of N.

Proposition 2.1.1. Let y ∈ Rα and let K = (0 : y)N , C = N/yN. Then

(1 − T α)F (N,T ) = F (C,T ) − T αF (K,T ).

In particular, if y is a nonzerodivisor on N, then

(1 − T α)F (N,T ) = F (C,T ).

s Proof. For β ∈ Z we have an exact sequence y 0 → Kβ → Nβ → Nβ+α → Cβ+α → 0, where y is multiplication by y. This gives

H(N)(β + α) − H(N)(β) = H(C)(β + α) − H(K)(β).

β+α s Multiplying by T and adding over β ∈ Z , we get the desired equality.

10 Example 2.1.2. Let R = k[X1,...Xr] be the polynomial ring in r variables over a ﬁeld k. s Let α1, . . . , αr be nonzero elements of N and let R be graded by assigning degree αi to Xi. Qr αi Then R0 = k. We claim that F (R,T ) = 1/ i=1(1 − T ). For, letting y = Xr in the above αr proposition we have (1 − T )F (R,T ) = F (C,T ). Since C = K[X1,...,Xr−1], the claim follows by induction on r. Two special cases are noteworthy: (1) s = 1. In this case T is a single variable and α1, . . . , αr are positive integers. In particular, for the usual gradation of R we have α1 = ··· = αr = 1 and F (R,T ) = 1/(1 − T )r. (2) s = r and αi = (0,..., 1 ..., 0) with 1 in the i–th place. In this case F (R,T ) = Qr 1/ i=1(1 − Ti).

Theorem 2.1.3. Suppose R = R0[y1, . . . , yr] with yi ∈ Rαi for 1 ≤ i ≤ r. Then F (N,T ) = Qr αi −1 −1 G(T )/ i=1(1 − T ) for some G(T ) ∈ Z[T1,...,Ts,T1 ,...,Ts ].

Proof. Induction on r. If r = 0 then R = R0. Therefore Nα = 0 for all but ﬁnitely many −1 −1 α and so F (N,T ) ∈ Z[T1,...,Ts,T1 ,...,Ts ]. For r > 0, put K = (0 : yr)N and C = N/yrN. Then (1 − T αr )F (N,T ) = F (C,T ) − T αr F (K,T ) by 2.1.1. Since both K and C are annihilated by yr, these are ﬁnitely generated over R0[y1, . . . , yr−1]. Applying the induction hypothesis to K and C we get F (N,T ) expressed in the desired form.

For the purpose of later application we have proved the above results in a form slightly more general than immediately necessary. However, we now restrict ourselves to the following special situation: s = 1 and R is generated as an R0-algebra by ﬁnitely many elements of R1. Then we have

Corollary 2.1.4. If R is generated as an R0-algebra by r elements of R1 then F (N,T ) = G(T )/(1− r −1 T ) for some G(T ) ∈ Z[T,T ]. 2

Call a map f : Z → Q a polynomial function if there exists a (necessarily unique) polynomial g(X) ∈ Q[X] such that f(n) = g(n) for all n >> 0 [i.e., there exists n0 ∈ Z such that f(n) = g(n) for all n ≥ n0]. In this case g is said to be the polynomial associated to f and the degree (resp. the leading coefficient) of g is also called the degree (resp. leading coefficient) of f. We follow the convention that the zero polynomial has degree −1 and leading coefficient 0. For a map f : Z → Q we define Df : Z → Q by Df(n) = f(n) − f(n − 1).

Lemma 2.1.5. Let f : Z → Q be a map and let r be a non-negative integer. Then f is a polynomial function of degree ≤ r if and only if Df is a polynomial function of degree ≤ r − 1. Consequently, f is a polynomial function of degree r if and only if Df is a polynomial function of degree r − 1.

Proof. If f is a polynomial function of degree ≤ r then it is easily veriﬁed that Df is a polynomial function of degree ≤ r − 1 (check this). We prove the converse by induction on r. If r = 0, then Df(n) = 0 for n >> 0. This means f(n) is constant for n >> 0 and so it is a polynomial function of degree ≤ 0. Now, let r > 0 and suppose there is a polynomial

11 r−1 r−2 g(X) = a0X + a1X + ··· + ar−1 ∈ Q[X] such that Df(n) = g(n) for n >> 0. Then for n >> 0 we have

r r f(n) − f(n − 1) = g(n) = (a0/r)(n − (n − 1) ) + h(n), 0 0 where h(X) ∈ Q[X] is a polynomial of degree ≤ r − 2. Deﬁne f : Z → Q by f (n) = r 0 0 f(n) − (a0/r)n . Then Df (n) = h(n) for n >> 0. By induction f is a polynomial function of degree ≤ r − 1. Hence f is a polynomial function of degree ≤ r.

G(T )/(1 − T )r = P f(n)T n G(T ) ∈ [T,T −1] f Corollary 2.1.6. Let n∈Z with Z . Then is a polynomial function of degree ≤ r − 1. More precisely, f is a polynomial function of degree e − 1, where e is the order of the pole at T = 1 of G(T )/(1 − T )r−1. In particular, if G(1) 6= 0, then f is of degree r − 1.

r > 0 G(T )/(I − T )r−1 = P Df(n)T n Proof. If then n∈Z . Therefore the assertion follows from the above lemma by induction on r.

Theorem 2.1.7. Let R = ⊕n∈Z+ Rn be a graded ring with R0 Artinian and R generated as an R0-algebra by r elements of R1. Let N = ⊕n∈ZNn be a ﬁnitely generated graded R-module. Then H(N) is a polynomial function of degree ≤ r − 1.

Proof. Follows from 2.1.4 and 2.1.6.

The polynomial associated to H(N) is called the Hilbert polynomial of N.

2.2 Hilbert-Samuel polynomial of a local ring

Let A be a Noetherian local ring with maximal ideal m and let M be a finitely generated A- module. Let q be an m-primary ideal of A. Then qnM/qn+1M and M/qnM are A-modules n of finite length for every n ∈ Z, where we put q = A for all n ≤ 0. We define

n n+1 Hq(M): Z → Q by Hq(M)(n) = `(q M/q M), and n Pq(M): Z → Q by Pq(M)(n) = `(M/q M), where ` = length A. Clearly Hq(M) = DPq(M). For a ﬁnitely generated A-module L, let µ(L) denote the least possible number of gen- erators of L.

Theorem 2.2.1. Pq(M) is a polynomial function of degree ≤ µ(q).

Proof. Put R = gr q(A), N = gr q(M). By 2.1.7 and Exercise 3.7 of Chapter 2, H(N) is a polynomial function of degree ≤ µ(q) − 1. Since DPq = Hq(M) = H(N), by 2.1.5 it follows that Pq(M) is a polynomial function of degree ≤ µ(q).

Lemma 2.2.2. The degree of Pq(M) does not depend upon the m-primary ideal q of A.

12 Proof. There exists a positive integer r such that mr ⊆ q ⊆ m. Therefore mrn ⊆ qn ⊆ mn and so Pm(M)(rn) ≥ Pq(M)(n) ≥ Pm(M)(n), for every positive integer n. It follows that Pq and Pm have the same degree (check this).

The polynomial associated to Pq(M) is called the Hilbert-Samuel polynomial of M w.r.t. q. Its degree, which is independent of q, is denoted by d(M).

Proposition 2.2.3. Let 0 → M 0 → M → M 00 → 0 be an exact sequence of ﬁnitely generated 0 00 A-modules with M 6= 0. Then deg(Pq(M ) + Pq(M ) − Pq(M)) < deg Pq(M).

0 0 0 Proof. We may assume that M 6= 0 and that M is a submodule of M. Put Q = Pq(M ) + 00 PqM − Pq(M). Then the given sequence induces, for every n, an exact sequence

0 0 n+1 00 n+1 00 (∗) 0 → M /Mn+1 → M/q M → M /q M → 0

0 0 n 0 0 0 where Mn = M ∩ q M. This gives Q = Pq(M ) − f where f(n) = `(M /Mn+1). By Artin- 0 0 Rees Lemma, there exists an integer m such that qMn = Mn+1 for all n ≥ m. Therefore m+n 0 0 n 0 n 0 0 q M ⊆ Mm+n = q Mm ⊆ q M for n >> 0 and we get Pq(M )(m + n) ≥ f(m + n) ≥ 0 0 Pq(M )(n). for n >> 0. It follows that Pq(M ) and f have the same degree and the same leading coefﬁcient (check this). Therefore deg Q < deg f. Since f(n) ≤ Pq(M)(n) for all n by (∗) and these are nonnegative functions, we get deg f ≤ deg Pq(M). Therefore deg Q < deg Pq(M), and the proposition is proved.

Corollary 2.2.4. Let 0 → M 0 → M → M 00 → 0 be an exact sequence of ﬁnitely generated A-modules. Then d(M) = max(d(M 0), d(M 00)).

0 00 Proof. Since Pq(M ),Pq(M) and Pq(M ) are all nonnegative functions the assertion is immediate from the above proposition (check this).

13 Chapter 3

Dimension Theory

Recall that for a ring A its (Krull) dimension, denoted dim (A), is deﬁned by

dim (A) = sup{r | p ,..., p ∈ Spec (A) p ⊆ ...⊆ p }. there exist 0 r with 0 / / r It is immediate from the deﬁnition that

dim (A) = sup dim (Ap) = sup dim (Am) p∈Spec (A) m∈Max (A) where by Max (A) we denote the set of all maximal ideals of A. Our main aim in this chapter is to prove the Dimension Theorem, which characterises the dimension of a Noetherian local or graded ring as the degree of its “Hilbert Polyno- mial” and also as the length of a “system of parameters”. These characterisations allow one to make a more effective computation of the dimension. The treatment here will be fairly self-contained and Krull’s Principal Theorem (which was proved earlier using symbolic powers) will be derived as a consequence of the Dimension Theorem.

3.1 The dimension of a local ring

Let A be a Noetherian local ring with maximal ideal m, and let M be a finitely generated A-module. As done above for a ring, we define the Krull dimension of a module. By a chain of r Supp (M) p ⊆ ... ⊆ p p ,..., p ∈ Supp (M) length in we mean a sequence 0 / / r with 0 r . If M 6= 0 then we define

dim (M) = sup{r | there exists a chain of length r in Supp (M)}.

The dimension of the zero module is defined to be −1. Suppose M 6= 0. Since the minimal primes of Supp (M) belong to Ass(M), which is a finite set, there exists p ∈ Ass(M) such that dim (M) = dim (A/p) (check this). Moreover, if dim (M) < ∞ then the primes p satisfying this property belong to Ass(M) and hence are finite in number.

14 If M 6= 0, the Chevalley dimension of M, denoted by s(M), is deﬁned by

s(M) = inf{r | there exist a1, . . . , ar ∈ m with `(M/(a1, . . . , ar)M) < ∞}.

Since `(M/mM) < ∞ and the ideal m is ﬁnitely generated, we have s(M) < ∞. We deﬁne s(0) = −1.

Theorem 3.1.1. (Dimension Theorem). dim (M) = d(M) = s(M).

Proof. We show that dim (M) ≤ d(M) ≤ s(M) ≤ dim (M). n (1) dim (M) ≤ d(M). If d(M) = −1 then Pm(M)(n) = 0, i.e., M = m M, for n >> 0. Therefore M = 0 by Nakayama, and dim (M) = −1. Now, assume that d(M) ≥ 0. Then M 6= 0. Choose p ∈ Ass(M) such that dim (M) = dim (A/p). There exists an A-monomorphism A/p → M, so that d(A/p) ≤ d(M) by 2.2.4. Thus it is enough to dim (A/p) ≤ d(A/p) p ⊆ ... ⊆ p prove that . To do this we need to show that if 0 / / r is a chain in Spec (A) then r ≤ d(A/p0). We prove this by induction on r. There is nothing to 0 prove for r = 0, since d(A/p0) 6= −1. So assume that r ≥ 1. Choose a ∈ p1 \ p0 and p ∈ 0 0 Ass(A/(p0 +Aa)) with p ⊆ p1. Then there exists an A-monomorphism A/p → A/(p0 +Aa) d(A/p0) ≤ d(A/(p + Aa)) p0 ⊆ p ⊆ ... ⊆ p and so 0 by 2.2.4. Now, since / 2 / / r is a chain, we get r − 1 ≤ d(A/p0) by induction hypothesis. Applying 2.2.3 to the exact sequence

a 0 → A/p0 → A/p0 → A/p0 + Aa → 0

0 we get deg Pm(A/(p0 + Aa)) < deg Pm(A/p0) and hence r ≤ 1 + d(A/p ) ≤ 1 + d(A/(p0 + Aa)) ≤ d(A/p0). This completes the proof of (1). (2) d(M) ≤ s(M). We may assume that M 6= 0. Put s = s(M) and choose a1, . . . , as ∈ m such that `(M/(a1, . . . , as)M) < ∞. Let a = Ann(M) and q = a + (a1, . . . , as). Since `(M/(a1, . . . , as)M) < ∞ we have

{m} = Supp (M/(a1, . . . , as)M) = Supp (M) ∩ Supp (A/(a1, . . . , as)) (check this) = Supp (A/a) ∩ Supp (A/(a1, . . . , as)) = Supp (A/q). Therefore q is m-primary. Put A0 = A/a, m0 = m/a and q0 = q/a. Then q0 is m0-primary 0 and µ(q ) ≤ s. So by 2.2.1, deg Pq0 (M) ≤ s. Since Pq(M) = Pq0 (M) (check this), we get d(M) ≤ s. (3) s(M) ≤ dim (M). We prove this by the induction on dim (M), which is ﬁnite by (1). We may assume that M 6= 0. If dim (M) = 0 then Supp (M) = {m}, so that `(M) < ∞ and hence s(M) = 0. Assume now that dim (M) ≥ 1. Let p1,..., pr be all those primes in Supp (M) for which dim (M) = dim (A/pi), 1 ≤ i ≤ r. Then pi 6= m for every i and so by the Prime Avoidance Lemma, m 6⊆ p1 ∪ ... ∪ pr. Choose a ∈ m, a 6∈ p1 ∪ ... ∪ pr, 0 0 and let M = M/aM. Then Supp (M ) ⊂ Supp (M) \{p1,..., pr} (check this). Therefore dim (M 0) < dim (M). Let s = s(M 0). Then s ≤ dim (M 0) by induction hypothesis. Let 0 0 0 0 a1, . . . , as ∈ m be such that `(M /(a1, . . . , as)M ) < ∞. Then, since M /(a1, . . . , as)M = 0 M/(a, a1, . . . , as)M we get s(M) ≤ s + 1 ≤ dim (M ) + 1 ≤ dim (M).

15 Corollary 3.1.2. If q is an m-primary ideal of A, then dim (M) ≤ µ(q). In particular, dim (M) ≤ µ(m) < ∞.

Proof. `(M/qM) < ∞.

Let M 6= 0 and let d = dim (M). A set of d elements a1, . . . , ad ∈ m with the property that `(M/(a1, . . . , ad)M) < ∞ is called a system of parameters of M. By the above theorem, every nonzero M has a system of parameters.

Corollary 3.1.3. Let M 6= 0 and let a1, . . . , ar ∈ m. Then dim (M/(a1, . . . , ar)M) ≥ dim (M)− r. The equality holds if and only if a1, . . . , ar can be completed to a system of parameters of M.

0 0 0 Proof. Put M = M/(a1, . . . , ar)M. By Nakayama, M 6= 0. Let e = dim (M ) and let 0 b1, . . . , be be a system of parameters of M . Then

0 0 `(M/(a1, . . . , ar, b1, . . . , be)M) = `(M /(b1, . . . , be)M ) < ∞ and so dim (M) ≤ r + e. If the equality holds then a1, . . . , ar, b1, . . . , be is a system of parameters of M. Conversely, suppose a1, . . . , ar can be completed to a system of parameters a1, . . . , ar, c1, . . . , cs of M. Then

0 0 `(M /(c1, . . . , cs)M ) = `(M/(a1, . . . , ar, c1, . . . , cs)M) < ∞.

Therefore e ≤ s and we get r + e ≤ r + s = dim (M).

Corollary 3.1.4. Let M 6= 0 and let a ∈ m. Then dim (M/aM) ≥ dim (M) − 1. If a is a nonzerodivisor for M then the equality holds.

Proof. The ﬁrst part is a particular case of the above corollary. To prove the second part, let p1,..., pr be all those primes in Supp (M) for which dim (M) = dim (A/pi), 1 ≤ i ≤ r. Then pi ∈ Ass(M) for every i. If a is a nonzerodivisor for M then a 6∈ p1 ∪...∪pr. Therefore Supp (M/aM) ⊆ Supp (M) \{p1,..., pr}. It follows that dim (M/aM) ≤ dim (M) − 1.

Assume that M 6= 0. Let a1, . . . , ar be elements of a proper ideal of A. We say a1, . . . , ar is an M-regular sequence if ai is a nonzerodivisor on M/(a1, . . . , ai−1)M for every i, 1 ≤ i ≤ r.

Corollary 3.1.5. Let M 6= 0 and let a1, . . . , ar ∈ m be an M-regular sequence. Then we readily see that dim (M/(a1, . . . , ar)M) = dim (M) − r. In particular a1, . . . , ar can be completed to a system of parameters of M.

Proof. The ﬁrst part follows from 3.1.4 by induction on r, and the second part follows from 3.1.3.

In the remainder of this section, let A be any ring, not necessarily local. For a prime ideal p of A, the height of p, denoted ht (p), is deﬁned by ht (p) = dim (Ap). Thus ht (p) is the supremum of the lengths of all chains of prime ideals contained in p. For an arbitrary ideal a of A, we deﬁne ht (a) = min{ht (p): p ⊇ I}.

16 Theorem 3.1.6. (Principal Ideal Theorem). Suppose A is Noetherian. Let a be an ideal of A and let p be a prime ideal of A minimal among prime ideals containing a. Then ht (p) ≤ µ(a). In particular, ht (p) ≤ µ(p). Moreover, if a is generated by an A-regular sequence a1, . . . , ar then ht (p) = r. Conversely, if q is a prime ideal of A with ht (q) = r, then there exist r elements b1, . . . , br such that q is a minimal prime of (b1, . . . , br).

Proof. The ideal aAp is pAp–primary. Therefore ht (p) = dim (Ap) ≤ µ(aAp) ≤ µ(a) by 3.1.2. This proves the ﬁrst part. If a1, . . . , ar is A-regular then it is also Ap-regular (check this). Therefore by 3.1.5 a1, . . . , ar can be completed to a system of parameters of Ap. In particular, r ≤ dim (Ap) = ht (p) ≤ µ(a) ≤ r. For the converse, if ht (q) = r, then by 3.1.1, s(Aq) = r, and so there exist c1, . . . , cr ∈ qAq such that `(Aq/(c1, . . . , cr)Aq < ∞. Write bi c = where b ∈ q and s ∈ A \ q. Then (b , . . . , b )Aq is qAq–primary and so q is a i si i i 1 r minimal prime of (b1, . . . , br).

Exercise 1. Let A ⊆ B be rings, B integral over A. Recall then that the map Spec (B) → Spec (A) is surjective. Prove the following results: p⊆ q B p ∩ A⊆ q ∩ A (i) If / is a chain of prime ideals of then / .

(ii) Going-up theorem. Let p1 ⊆ ... ⊆ pn be a sequence of prime ideals of A, and let q1 ⊆ ... ⊆ qm, m < n, be a sequence of prime ideals of B such that qi ∩ A = pi for i = 1, . . . , m. Then the sequence q1 ⊆ ... ⊆ qm can be extended to a sequence q1 ⊆ ... ⊆ qn of prime ideals of B such that qi ∩ A = pi for i = 1, . . . , n. (iii) dim (A) = dim (B) (use (i) and (ii)).

Proposition 3.1.7. Let A be a ring and let X,X1,...,Xn be indeterminates over A. Then: (i) 1 + dim (A) ≤ dim (A[X]) ≤ 1 + 2dim (A).

(ii) If A is Noetherian then dim (A[X1,...,Xn]) = n + dim (A). p ⊆ ... ⊆ p A p [X] ⊆ ... ⊆ p [X] ⊆ Proof. (i) Let 0 / / n be a chain of prime ideals in . Then 0 / / n / (pn,X) is a chain of prime ideals in A[X]. Hence 1 + dim (A) ≤ dim (A[X]). p ⊆ q ⊆ r To prove the other inequality it is enough to show that if / / is a chain of prime ideals in A[X] then p ∩ A 6= r ∩ A. Assume the contrary. We may assume that p ∩ A = (0). Let S = A \ (0) and B = S−1(A[X]). Then B is a principal ideal domain and therefore dim (B) = 1 pB ⊆ qB ⊆ rB B . This is a contradiction as / / is a chain of prime ideals in . (ii) By induction we may assume that n = 1. Put X = X1. We ﬁrst show that if p is a prime ideal of A of height r then p[X] is also of height r. Since ht (p) = dim (Ap) = r, by 3.1.6, there exists an ideal I of A generated by r elements such that p is a minimal prime of I. Then p[X] is a minimal prime of I[X]. Therefore, by 3.1.6, ht (p[X]) ≤ r. On the other hand, ht (p[X]) ≥ r (check this). Hence ht (p[X]) = r. Now let q be a prime ideal of A[X] and let q ∩ A = p. Then by the earlier argument, ht (p) = ht (p[X]) ≤ ht (q). Let s = ht (q). By induction on s we show that s ≤ ht (p) + 1. s = 0 ht (p) = 0 s ≥ 1 q0 ⊆ q A[X] If then . Now suppose . Let / be a prime ideal of of height s − 1 and let q0 ∩ A = p0. Then by induction, s − 2 ≤ ht (p0) ≤ s − 1. Therefore either ht (p0) = s − 2 ht (p0) = s − 1 ht (p0) = s − 2 p0[X] ⊆ q0 ⊆ q p0 ⊆ p or . If then / / and hence / , as shown in (i). Hence s − 1 ≤ ht (p). If ht (p0) = s − 1 then, clearly, s − 1 ≤ ht (p).

17 Exercise 2. Let A be a Noetherian ring. Let A[[X1,...,Xn]] denote the ring of formal power series in X1,...,Xn with coefficients in A. Show that dim (A[[X1, . . . , xn]]) = n + dim (A). Exercise 3. This exercise provides an example of a Noetherian ring which is not finite di- mensional. Let A = k[X1,X2,...,Xi,...] be the polynomial ring in variables X1,X2,...,Xi,.... Let m1, m2, . . . , mi,... be an increasing sequence of positive integers such that mi+1 −mi > mi − mi−1 for all i ≥ 1. Let pi = (Xmi+1,...,Xmi+1 ), i ≥ 1, and let S = A \ ∪i≥1pi. Then check the following: (i) If I ⊆ ∪i≥1pi then I ⊆ pi for some i ≥ 1. (ii) If a ∈ A \ 0 then a ∈ pi for only finitely many values of i. Next, let B = S−1(A). Then check that B has the following properties: (i) Max (B) = {piB | i ≥ 1}. (ii) If b ∈ B \ 0, then B/(b) is semilocal [i.e., B/(b) has only finitely many maximal ideals]. (iii) If m is a maximal ideal of B, then Bm is Noetherian. From these properties it follows that B is Noetherian (check this). Since ht (piB) = mi+1 − mi, we have dim (B) = ∞.

3.2 The Dimension of an Afﬁne Algebra

In this section we prove some results on the dimension of an affine algebra. Recall that if k is a field then by an affine algebra over k we mean a ring which is a finitely generated k-algebra. We begin by recalling a version of

(NNL) Noether’s Normalization Lemma. Let A be an afﬁne algebra over a ﬁeld k. Let a1 ⊆ ... ⊆ a ⊆ A A x , . . . , x ∈ A r / be a sequence of ideals of . Then there exist 1 n such that

(i) x1, . . . , xn are algebraically independent over k;

(ii) A is integral over k[x1, . . . , xn];

(iii) ai ∩ k[x1, . . . , xn] = (x1, . . . , xh(i)) with 0 ≤ h(1) ≤ · · · ≤ h(r) ≤ n.

Let us deduce two corollaries from this: Corollary 3.2.1. With the notation and assumptions of NNL, assume that A is an integral domain. Then ht (ai) = h(i) for every i.

Corollary 3.2.2. If A = k[Y1,...,Ym] is the polynomial ring in m variables over a ﬁeld k then dim(A) = m.

To prove these corollaries, we ﬁrst recall the

Going-down Theorem. Let A ⊆ B be integral domains, A integrally closed, B integral over A. Let p1 ⊇ ... ⊇ pn be a sequence of prime ideals of A, and let q1 ⊇ ... ⊇ qm, m < n, be a sequence of prime ideals of B such that qi ∩ A = pi for i = 1, . . . , m. Then the sequence q1 ⊇ ... ⊇ qm can be extended to a sequence q1 ⊇ ... ⊇ qn of prime ideals of B such that qi ∩A = pi for i = 1, . . . , n.

18 Proof of Corollary 3.2.1: Put C = k[x1, . . . , xn]. By the surjectivity of the spec map corresponding to C/ai ∩ C,→ A/ai, there exists a prime ideal p of A such that ai ⊆ p and p ∩ C = ai ∩ C. Therefore ht (ai) ≤ ht (p) ≤ ht (p ∩ C) in view of Exercise 1. Further, ht (p ∩ C) = ht (ai ∩ C) = ht ((x1, . . . , xh(i))) ≤ h(i) by 3.1.6. Thus ht (ai) ≤ h(i). For the other inequality, let q be a prime ideal of A such that ai ⊆ q and ht (ai) = ht (q). As C is integrally closed and A is a domain, the going-down theorem implies that ht (q) ≥ ht (q ∩ C) ≥ ht (ai ∩ C) = ht ((x1, . . . , xh(i))) ≥ h(i), where the last inequality results from 0⊆ (x )⊆ · · ·⊆ (x , . . . , x ). 2 the chain / 1 / / 1 h(i)

Proof of Corollary 3.2.2: Let p be any prime ideal of A. Applying NNL to the sequence p⊆ A, k[x , . . . , x ] A A / we get a polynomial subalgebra 1 n of over which is integral and such that p ∩ k[x1, . . . , xn] = (x1, . . . , xh) for some h. By 3.2.1, ht p = h ≤ n. Thus ht p ≤ n for p A, dim(A) ≤ n. 0 ⊆ (Y ) ⊆ · · · ⊆ every prime ideal of so On the other hand, the chain / 1 / / (Y1,...,Ym) shows that dim(A) ≥ m. Finally, we have n = m because the quotient ﬁeld of A is algebraic over the quotient ﬁeld of k[x1, . . . , xn]. So dim(A) = m. 2

Theorem 3.2.3. Let A be an affine algebra over a field k. Assume that A is an integral domain. Then dim (A) = tr.degkK, where K is the quotient field of A. (0) ⊆ A, x , . . . , x ∈ A A Proof. Applying NNL to the sequence / there exist 1 n such that is integral over k[x1, . . . , xn] and x1, . . . , xn are algebraically independent over k. Hence tr.degkK = n. On the other hand, by Exercise 1 and 3.2.2, dim (A) = dim (k[x1, . . . , xn]) = n.

Theorem 3.2.4. Let A be an afﬁne algebra over a ﬁeld k. Assume that A is an integral domain. Let p be a prime ideal of A. Then ht (p) + dim (A/p) = dim (A). p ⊆ A, x , . . . , x ∈ A A Proof. Applying NNL to the sequence / there exist 1 n such that is integral over C = k[x1, . . . , xn] and p ∩ C = (x1, . . . , xh) for some h. By 3.2.1, we have h = ht (p). Now A/p is integral over C/p ∩ C = k[xh+1, . . . , xn]. Therefore, by Exercise 1 and 3.2.2, dim (A/p) = dim (C/p ∩ C) = n − h = dim (A) − ht (p).

Example 3.2.5. If A is not a domain then result of 3.2.4 need not hold. For example take A = k[X,Y,Z]/(XY,XZ) = k[x, y, z] and p = (y, z). Then dim (A) = 2, ht (p) = 0 and dim (A/p) = 1. Example 3.2.6. A ring A is said to be equidimensional if for all minimal primes p of A, dim (A) = dim (A/p). Let A be as in previous example. Let p = (y, z) and p0 = (x). Then p and p0 are minimal primes of A. Since dim (A/p) = 1 and dim (A/p0) = 2, A is not equidimensional. Example 3.2.7. A ring A is said to be puredimensional if A is equidimensional and Ass(A) consists of minimal primes of A. We give an example of an equidimensional ring which is not puredimensional. Let A = k[X,Y ]/(X2,XY ) = k[x, y], where k is a ﬁeld and X,Y are indeterminates over k. Then Ass(A) = {(x), (x, y)}, so A is not puredimensional. Since (x) is the only minimal prime of A, A is equidimensional.

19 Exercise 4. Let A be an afﬁne algebra over a ﬁeld k and let f be a non-zero divisor of A. Show that dim (A) = dim(Af ).

3.3 The Dimension of a Graded Ring

Let R = ⊕n∈Z+ Rn be a Noetherian graded ring. Then R0 is Noetherian and R is ﬁnitely generated as an R0-algebra. The graded dimension of R, denoted grdim (R), is deﬁned to be the supremum of lengths of chains of homogeneous prime ideals of R. The main result of this section is

Theorem 3.3.1. grdim (R) = dim (R).

The proof of the theorem is preceded by several lemmas. For a homogeneous prime ideal p of R we deﬁne its graded height, denoted grht (p), to be the supremum of lengths of chains of homogeneous prime ideals contained in p. Put R+ = ⊕n>0Rn. For an ideal a of R, let

∗ a = ⊕n≥0(a ∩ Rn).

This is the largest homogeneous ideal of R contained in a. Sometimes it is also denoted by agr

Lemma 3.3.2. (1) If m = m0 + R+ with m0 ∈ Max (R0) then m ∈ Max (R) and m is homogeneous. Any maximal ideal of R which is homogeneous is of this form.

(2) Every proper homogeneous ideal of R is contained in a homogeneous ideal of R which is maximal.

(3) If p is a prime ideal of R then so is p∗.

Proof. Exercise.

Lemma 3.3.3. Assume that R is a domain. Let S be the set of all nonzero homogeneous elements of R. Then either S−1R is a ﬁeld or S−1R =∼ K[X,X−1], where K is a ﬁeld and X is transcendental over K.

Proof. Let L be the quotient ﬁeld of R and let

K = {a/b | a, b ∈ Rn for some n, b 6= 0}.

−1 −1 Then K is a subﬁeld of L and K ⊆ S R. If R = R0 then K = S R. Otherwise, let

R = R0[y1, . . . , yr] with yi ∈ Rni , ni > 0, yi 6= 0. Let d = GCD(n1, . . . , nr). Then every homogeneous element of R has degree a multiple of d. Write d = m1n1 + ··· + mrnr with m1, . . . , mr ∈ Z. We may assume that mi ≥ 0 for 1 ≤ i ≤ t and mi < 0 for t + 1 ≤ i ≤ r, m1 mt −mt+1 −mr for some 1 ≤ t ≤ r. Put Y = y1 . . . yt , Z = yt+1 . . . yr . Then Y,Z are homogeneous elements of R and deg(Y ) = d + deg(Z). Put X = Y/Z. Then K[X,X−1] ⊆ S−1R. Let b be a nonzero homogeneous element of R. Then, as noted above, deg(b) = nd for some n ≥ 0. We have b = (bZn/Xn)Xn and b−1 = (Y n/bZn)X−n, so that b, b−1 ∈ K[X,X−1].

20 This shows that K[X,X−1] = S−1R. It remains to show that X is transcendental over p p−1 K. Suppose c0X + c1X + ··· + cp = 0 with ci ∈ K. We have to show that ci = 0 for every i. Multiplying by a common homogeneous denominator, we may assume that c0, . . . , cp are homogeneous elements of R of the same degree, say e. Since X = Y/Z, we p p−1 p p−i i get c0Y + c1Y Z + ··· + cpZ = 0. Now, ciY Z is homogeneous of degree e + (p − i) deg(Y ) + i deg(Z) = e + (p − i)(d + deg(Z)) + i deg(Z) = e + (p − i)d + p deg(Z). Since p−i i d 6= 0, these degrees are distinct for distinct i. So ciY Z = 0; hence ci = 0, for all i.

Corollary 3.3.4. With the notation of the above lemma, we have dim (S−1R) ≤ 1.

Proof. dim (K[X,X−1]) ≤ dim (K[X]) = 1 by 3.1.6, since K[X] is a PID.

Lemma 3.3.5. ht (p/p∗) ≤ 1 for every prime ideal p of R.

Proof. Going modulo p∗ we may assume that p∗ = 0. Then R is a domain by 3.3.2. Let S be the set of all nonzero homogeneous elements of R. Then S ∩ p = ∅. Therefore ht (p) = ht (S−1p) ≤ dim (S−1R) ≤ 1 by the above corollary.

Lemma 3.3.6. Let p ∈ Spec (R). If p∗ 6= p, then ht (p∗) = ht (p) − 1.

Proof. It is clearly enough to prove the following statement:

(∗) q⊆ p ht (q) ≤ ht (p∗ . If / is any chain of prime ideals, then ) (∗) q∗ = p∗ p∗ ⊆ q ⊆ p q = p∗ We prove ﬁrst in the special case . In this case / implies that by the above lemma and so ht (q) = ht (p∗). Now we prove (∗) in the general case by induction on ht (p∗). If ht (p∗) = 0 then q∗ = p∗ and we are in the special case. Assume ht (p∗) > 0 q∗ ⊆ p∗ q ⊆ q ht (q ) ≤ ht (q∗) now that and that / . If 1 / then 1 by induction. So ht (q) ≤ 1 + ht (q∗) ≤ ht (p∗). q⊆ p r q⊆ r⊆ p Call a chain / of prime ideals saturated if there exists no prime ideal with / / . q ⊆ p Lemma 3.3.7. Let / be chain of homogeneous prime ideals. If it is saturated as a chain of homogeneous prime ideals then it is also saturated as a chain of arbitrary prime ideals.

Proof. Going modulo q we may assume that q = 0. Then R is a domain, p is minimal among nonzero homogeneous prime ideals and we have to show that ht (p) = 1. Let a ∈ p be a nonzero homogeneous element and let r be a prime minimal with the property a ∈ r ⊆ p. Since a ∈ r∗, r∗ is a nonzero homogeneous prime contained in p. Therefore p = r∗ = r. So ht (p) = ht (r) = 1, by 3.1.6.

Lemma 3.3.8. grht (p) = ht (p) for every homogeneous prime ideal p of R.

Proof. We have only to prove that ht (p) ≤ grht (p). We do this by induction on ht (p). Put h = ht (p) g = grht (p) h = 0 h > 0 q ⊆ p , . If then there is nothing to prove. Let and let / be a chain such that ht (q) = h−1. If q is homogeneous then by induction h−1 ≤ grht (q) ≤ g−1 h ≤ g q q∗ ⊆ q ⊆ p and so . Suppose is not homogeneous. Then / / and so by 3.3.7 the chain q∗ ⊆ p grht (q∗) ≤ g −2 / is not saturated as a chain of homogeneous prime ideals. Therefore . Now, by induction and 3.3.6 we get g − 2 ≥ grht (q∗) ≥ ht (q∗) = ht (q) − 1 = h − 2.

21 Proof of Theorem 3.3.1: It is enough to prove that ht (p) ≤ grdim (R) for every p ∈ Spec (R). If p is homogeneous then by 3.3.8, ht (p) = grht (p) ≤ grdim (R). If p is not homogeneous then ht (p) = 1 + ht (p∗) = 1 + grht (p∗), by 3.3.6. Let m ∈ Max (R) be a p∗ p∗ ⊆ p p∗ homogeneous ideal containing (this exists by 3.3.2). Since / , is not maximal. So we get ht (p) = 1 + grht (p∗) ≤ grht (m) ≤ grdim (R). 2

From now on we assume that R0 = k is a ﬁeld. Note that for k-modules M, the length `(M) equals the rank over k or the vector space dimension over k; this will be denoted by rank kM. Also, when R0 is a ﬁeld, we see by 3.3.2 that R+ is the unique homogeneous maximal ideal of R. Put m = R+.

We say that R is standard if R = k[R1]. We assume that k is inﬁnite.

Corollary 3.3.9. dim (R) = dim (Rm). In particular, dim (R) is ﬁnite.

Proof. Every homogeneous prime ideal of R is contained in m by 3.3.2. Therefore, by 3.3.8, we have dim (R) = grdim (R) = grht (m) = ht (m) = dim(Rm).

Let us deﬁne graded Chevalley dimension of R to be

∗ s (R) = inf{r | there exist homogeneous elements a1, . . . ar of positive degree such that rank k(R/(a1, . . . , ar)R) < ∞}.

Sometimes s∗(R) it is also denoted by sgr (R).

Theorem 3.3.10. dim (R) = s∗(R).

We ﬁrst prove a lemma:

Lemma 3.3.11. Let p1,..., pr ∈ Spec (R).

(1) If ∪n>0Rn ⊆ p1 ∪ ... ∪ pr, then m ⊆ p1 ∪ ... ∪ pr.

(2) If R is standard and R1 ⊆ p1 ∪ ... ∪ pr, then m ⊆ p1 ∪ ... ∪ pr.

n!/i Proof. (1) Let a ∈ m. Write a = a1 + ··· + an with ai ∈ Ri. Put bi = ai . Then each bi is homogeneous of degree n!. Let

n Vi = {λ ∈ k | λ1b1 + ··· + λnbn ∈ pi}.

n n These are k-subspaces of k and we have k = V1 ∪ ... ∪ Vr, since Rn! ⊆ p1 ∪ ... ∪ pr. n Therefore k equals one of them say V1. Then bi ∈ p1 for every i. Therefore ai ∈ p1 for every i and so a ∈ p1. (2) Similar.

22 Proof of Theorem 3.3.10: We first prove s∗(R) ≤ dim (R) by induction on dim (R). If dim (R) = 0, then m is a minimal prime ideal. Since all the minimal prime ideals of R are homogeneous, m is the only prime ideal of R. Therefore some positive power of m is ∗ zero. This implies that Rn = 0 for n >> 0, whence rank k(R) < ∞ and s (R) = 0. Now, let dim (R) > 0. Let p1,..., pr be all the minimal prime ideals of R for which dim (R) = dim (R/pi), 1 ≤ i ≤ r. Then p1,..., pr are homogeneous and m 6⊆ p1 ∪ ... ∪ pr. Therefore by the lemma there exists a homogeneous element a of positive degree such that a 6∈ p1 ∪ ∗ ... ∪ pr. Hence, by 3.1.2, dim (R/(a)) ≤ dim (R) − 1 and so s (R/(a)) ≤ dim (R) − 1, by induction. From this the inequality s∗(R) ≤ dim (R) follows. ∗ To prove the other inequality, put s = s (R) and let a1, . . . , as be homogeneous ele- 0 0 ments of positive degree such that the k-rank of R = R/(a1, . . . , as) is finite. Put m = 0 0 0 m/(a1, . . . , as). Then some positive power of m is 0. Hence R = Rm0 = Rm/(a1, . . . , as)Rm is of finite length and so s ≥ dim (Rm) = dim (R), by 3.1.1 and 3.3.9. 2

Let d = dim (R). A set a1, . . . , ad of homogeneous elements of positive degree such that R/(a1, . . . , as) is of ﬁnite rank as a k-vector space is called a homogeneous system of parameters of R.

Corollary 3.3.12. Suppose R is standard. Then there exists a homogeneous system of parameters of R consisting of elements of degree 1.

Proof. This follows from the proof of 3.3.10 in view of part (2) of 3.3.11.

Assume that R is standard. Recall that the Hilbert function H(R) of R is deﬁned by H(R)(n) = rank k(Rn). By 2.1.7 H(R) is a polynomial function of degree ≤ rank k(R1) − 1. Pn In analogy with the case of a local ring, let us deﬁne P (R)(n) = i=0 rank k(Ri). Then H(R) = DP (R) in the notation of Section 2.1. Therefore by 2.1.5, P (R) is a polynomial function of degree ≤ rank k(R1). Put d(R) = deg(P (R)).

Theorem 3.3.13. If R is standard, then dim (R) = d(R).

Proof. Put A = Rm. Then R is naturally isomorphic to gr mA(A) (check this). Therefore d(R) = d(A). Also, dim (R) = dim (A) by 3.3.9. So the assertion follows from the Dimen- sion Theorem for a local ring.

Theorem 3.3.14. If R is standard then grdim (R) = dim (R) = s∗(R) = d(R).

Proof. Collect together the results proved above.

Corollary 3.3.15. Let B be a Noetherian local ring with maximal ideal n. Then dim (B) = dim (gr n(B)).

Proof. dim (B) = d(B) = d(gr n(B)) = dim (gr n(B)).

Exercise 5. Let k be a ﬁeld and let Xij, i = 1, 2; j = 1, 2, 3 be indeterminates over k. Let I be the ideal of k[Xij] generated by the 2 × 2 minors of the matrix (Xij). Compute the dimension of k[Xij]/I.

23 References

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