Chap 11: Dimension Theory
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MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 67 11. Dimension theory The dimension of a Noetherian local ring can be defined in several ways. The main theorem is that all of these descriptions give the same number. So far we have the Krull dimension dim A which is the length of the maximal chain of prime ideals. 11.1. Poincar´eseries and Hilbert functions. Suppose that m is a maximal ideal of A and Gr(A)= A = A/m m/m2 m2/m3 n ⊕ ⊕ ⊕··· is the associated graded ring. If A is Noetherian then each An = n n+1 m /m is a finite dimensional vector space over the field k = A0 = A/m and we have the Poincar´eseries ∞ n PGr(A)(t)= t dimk An n=0 11.1.1. length function. In greater generality we consider a graded ring ∞ A = A = A A A n 0 ⊕ 1 ⊕ 2 ⊕··· n=0 so that A is Noetherian and A0 is Artinian. Suppose that ∞ M = M = M M M n 0 ⊕ 1 ⊕ 2 ⊕··· n=0 is a finitely generated graded module over A. Then each Mn is a finitely generated A0-module. Since A0 is Artinian, each Mn has finite length (Mn) and we can define the Poincar´eseries of M to be ∞ n PM (t)= (Mn)t n 0 − Recall that the length of an A0-module N is defined to be the largest integer m = (N) so that there is a filtration N = N0 N1 N2 Nm =0 ··· by A0-submodules Nn. It can be shown that any two maximal fil- trations have the same length. The subquotients Nn/Nn+1 are called composition factors of N. They are simple modules which are unique up to isomorphism and permutation of indices. Given this property it is easy to see that the length function satisfies the property that, for any short exact sequence of f.g. A0-modules 0 M M M 0 → → → → 68 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA we have (M)=(M )+(M ). Any function on f.g. A0-modules with this property is called an additive function. Some trivial observations about the Poincar´eseries: (1) 0 PM (1) in general. (2) P ≤(1) = ≤∞(M ) < iff M = 0 for all but finitely many n. M n ∞ n (We are given that (Mn) < for all n 0.) ∞ ≥ (3) PM (1) = 0 iff M =0. 11.1.2. rational functions. Theorem 11.1. The Poincar´eseries of M is a rational function of the form f(t) PM (t)= s (1 tki ) i=1 − where f(t) is a polynomial in t with coefficients in .) Z Proof. The proof is by induction on s, the number of homogeneous generators xi of the ideal A+. If s = 0 then A = A0 and each Mn is an A-submodule of M. Since M is f.g. this implies that Mn = 0 for all but finitely many n. So n PM (t)=f(t)= (Mn)t is a polynomial in t with coefficients in Z. Now suppose that s 1 and consider the last generator xs Ak . ≥ ∈ s Multiplication by xs gives a graded map M M of degree ks.LetK be the kernel and L the cokernel of this map.→ Then, for each n 0we get an exact sequence: ≥ xs 0 Kn ks Mn ks Mn Ln 0 → − → − −→ → → where Kn ks = Mn ks =0forn<ks. Then − − (Kn ks ) (Mn ks )+(Mn) (Ln)=0 − − − − Multiply by tn n n n n (Kn ks )t (Mn ks )t + (Mn)t (Ln)t =0 − − − − Sum over all n tks P (t) tks P (t)+P (t) P (t)=0 K − M M − L So: ks PL(t) t PK (t) PM (t)= − 1 tks − Since xs annihilates K and L, these are graded modules over the graded ring A = A/(x ) whose positive grade ideal has s 1 generators s − x1, x2, , xs 1, the images of xi A. So, PL(t) and PK (t) are rational ··· − s 1∈ k functions with denominator − (1 t i ). The theorem follows. i=1 − MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 69 There is one special case when we can easily calculate f(t) by induc- tion. This is the case in which K = 0, i.e., x z = 0 for all z = 0 in M. s In this case we say that xs is M-regular. Then PM/xsM (t) PM (t)= 1 tks − Next, suppose that xs 1 is M/xsM-regular. Then − PM/xs 1xsM (t) − PM/xsM (t)= 1 tks 1 − − Definition 11.2. We say that (any sequence of elements of A) xk, ,xs is an M-regular sequence if this continues in the analogous way··· up to xk. When M = A, we call this a regular sequence for A. By induction we would get: PM/xkxk+1 xsM (t) PM (t)= s ··· (1 tki ) i=k − Example 11.3. Take A = M = K[x , ,x ] graded in the usual way 1 ··· d so that x1, ,xd span A1 = M1. Then x1, ,xd is an M-regular sequence and···M/x x M = K is 1-dimensional··· and 1 ··· d 1 n + d 1 P (t)= = − tn A (1 t)d d 1 − − Proof. (of the second equality) Start with 1 n 2 3 (1 t)− = t =1+t + t + t + − ··· and differentiate d 1 times to get − d n d+1 (d 1)!(1 t)− = n(n 1) (n d +2)t − − − − ··· − d n n d+1 n + d 1 n (1 t)− = t − = − t − d 1 d 1 − − ∞ nd 1 nd 2 = − − + tn (d 1)! − (d 3)! ··· n=0 − − 11.1.3. dimension using PM (t). Definition 11.4. Let d = d(M) be the smallest positive integer so that d lim(1 t) PM (t) < t 1 → − ∞ Proposition 11.5. d(M) s for all f.g. graded A-modules M. ≤ 70 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA Proof. 1 t 1 1 − k = k 1 1 t 1+t + + t − → k as t 1. So, (1 t)s−P (t)convergesas··· t 1. → − M → Lemma 11.6. If x A is M-regular then d(M)=d(M/xM)+1. ∈ k Proposition 11.7. If A has an M-regular sequence of length r then d(M) r. ≥ In Example 11.3, d = s = r. So, d(K[x1, ,xd]) = d. The idea is that d(M) is the number of generators of A ···which act as independent variables on M. 11.1.4. Hilbert function. The Hilbert function gives a formula for (Mn) for large n. Theorem 11.8. Suppose that ki =1for i =1, 2, ,s. Then (Mn) is a polynomial of degree d(M) 1 for large n. ··· − Proof. By Theorem 11.1 we have: f(t) (M )tn = n (1 t)s − where f(t) is a polynomial. By definition of d = d(M) we have f(t)= s d (1 t) − g(t) where g(1) = 0. So, − g(t) n + d 1 (M )tn = = − tng(t) n (1 t)d d 1 − − m Suppose that g(t)= a tk = a + a t + + a tm. Then k=0 k 0 1 ··· m m n n + d 1 n+k (M )t = a − t n k d 1 k=0 − When n m we get ≥ m n k + d 1 (M )= a − − n k d 1 k=0 − (When n<mwe take the sum over 0 k n.) Since the binomial coefficient is a polynomial in n of degree≤d ≤1 with leading coefficient − 1/(d 1)! regardless of the value of k, this formula for (Mn) is a polynomial− in n of degree d 1 with leading coefficient − 1 m g(1) a = =0 (d 1)! k (d 1)! − k=0 − MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 71 Definition 11.9. The Hilbert function of M is the polynomial func- tion hM (n)=(Mn) for sufficiently large n. 11.1.5. Samuel function. For each n consider the finite sum: g(n)=(M0)+(M1)+ + (Mn 1). · − Lemma 11.10. g(n) is a polynomial of degree d(M) for sufficiently large n. Proof. For m>m0 we have d 1 (Mm)=hM (m)=c0 + c1m + + cd 1m − ··· − Therefore n 1 − g(n)= hM (m)+const m=0 n 1 k for n>m0. But, anyone who has taught Calculus knows that m−=0 m is a polynomial in n of degree k + 1. The lemma follows. Suppose that A is a Noetherian local ring with unique maximal ideal m and quotient field K = A/m.LetM be any f.g.A-module. For any n, M/mnM is a f.g. module over the Artin local ring A/mn. Therefore it has a finite length: n 2 n 1 n (M/m M) = dim (M/mM mM/m M m − M/m M) K ⊕ ⊕···⊕ Proposition 11.11. For sufficiently large n, (M/mnM) is a polyno- mial in n of degree s = dim (m/m2). ≤ K Proof. Take the associated graded ring Gr(A)=A/m m/m2 m2/m3 ⊕ ⊕ ⊕··· 2 2 3 has positive ideal Gr(A)+ = m/m m /m generated by s ele- ments of degree 1. The associated Gr⊕ (A)-module⊕··· Gr(M)=M/mM mM/m2M ⊕ ⊕··· is a f.g.Gr(A)-module with d(Gr(M)) s by Proposition 11.5. So, (M/mnM) is a polynomial of degree d≤(Gr(M)) s for sufficiently ≤ large n by the lemma above. Definition 11.12. The Samuel function for M is defined to be the polynomial function given for large n by: M n χm (n)=(M/m M) deg χM (n)=d(Gr(M)) dim (m/m2) m ≤ K 72 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 11.2.