The Next Inequality

Considering it has been at least a post removed, I’ll bring us back to our situation. We have a local Noetherian ring (R, \frak{m}). Our notation is that \delta(R) is the least number of generators of an \frak{m}-primary ideal (which was shown to be independent of choice of ideal here). The goal for the day is to show that d(R)\geq \dim R.

Suppose \frak{q} is \frak{m}-primary. We’ll prove something more general. Let M be a finitely generated R-module, x\in R a non-zero divisor in M and M'=M/xM. Then the claim is that \deg\chi_q^{M'}\leq \deg\chi_q^M -1.

Since x is not a zero-divisor, we have an iso as R-modules: xM\cong M. Define N=xM. Now take N_n=N\cap \frak{q}^nM. Since \frak{q}^nM is a stable \frak{q}-filtration of M, by Artin-Rees we get that (N_n) is a stable \frak{q}-filtration of N.

For each n we have 0\to N/N_n \to M/\frak{q}^nM\to M'/\frak{q}^nM'\to 0 exact.

Thus we get l(N/N_n)-l(M/\frak{q}^nM)+l(M'/\frak{q}^nM')=0. So if we let g(n)=l(N/N_n), we get for large n: g(n)-\chi_q^M(n)+\chi_q^{M'}(n)=0.

But (N_n) is also a stable \frak{q}-filtration of M, since N\cong M. We already showed that the degree and leading coefficient of g(n) depends only on M and \frak{q} and not on the filtration. Thus g(n) and \chi_q^M(n) have the same degree and leading coefficient, so the highest powers kill eachother which gives \deg\chi_q^{M'}\leq \deg \chi_q^M-1.

In particular, we will need that R as an R-module gives us d(R/(x))\leq d(R)-1.

Now we prove the goal for today. For simplicity, let d=d(R). We will induct on d. The base case gives that l(R/\frak{m}^n) is constant for large n. In particular, there is some N such that \frak{m}^n=\frak{m}^{n+1} for all n>N. But we are local, so \frak{m}=J(R) and hence by Nakayama, \frak{m}^n=0. Thus for any prime ideal \frak{p}, we have \frak{m}^k\subset \frak{p} for some k, so take radicals to get \frak{m}=\frak{p}. Thus there is only one prime ideal and we actually have an Artinian ring and hence have \dim R=0.

Now suppose d>0 and the result holds for \leq d-1. Let p_0\subset p_1\subset \cdots \subset p_r be a chain of primes. Choose x\in p_1\setminus p_0. Define R'=R/p_0 and \overline{x} be the image of x in R'.

Note that since R' is an integral domain, and \overline{x} is not 0, it is not a zero-divisor. So we use our first proof from today to get that d(R'/(\overline{x}))\leq d(R')-1.

Let \frak{m}' be the maximal ideal of R'. Then R'/\frak{m}' is the image of R/\frak{m}, so l(R/\frak{m}^n)\geq l(R'/\frak{m}'^n) which is precisely d(R)\geq d(R'). Plugging this into the above inequality gives d(R'/(\overline{x}))\leq d(A)-1=d-1.

So by the inductive hypothesis, \dim(R'/\overline{x})\leq d-1. Take our original prime chain. The images form a chain \overline{p}_1, \ldots , \overline{p}_r in R'/(\overline{x}). Thus r-1\leq d-1\Rightarrow r\leq d. Since the chain was arbitrary, \dim R\leq d(R).

A nice corollary here is that the dimension of any Noetherian local ring is finite. Another similar corollary is that in any Noetherian ring (drop the local) the height of a prime ideal is finite (and hence primes satisfy the DCC), since ht(p)=\dim A_p which is local Noetherian.


Applying the Hilbert Polynomial

Let’s start applying to some specific situations now. Suppose R is a Noetherian local ring with maximal ideal \frak{m}. Let \frak{q} be an \frak{m}-primary ideal. Let M be a finitely-generated R-module, and (M_n) a stable \frak{q}-filtration of M.

Don’t panic from the set-up. I think I haven’t talked about filtrations. All the stable \frak{q}-filtration means is that we have a chain of submodules M=M_0\supset M_1\supset \cdots \supset M_n\supset \cdots such that \frak{q}M_n=M_{n+1} for large n.

The goal for the day is to prove three things.

1) M/M_n has finite length for all n\geq 0.

Define G(M)=\bigoplus \frak{q}^n/\frak{q}^{n+1} and G(M)=\bigoplus M_n/M_{n+1}. We have a natural way to make G(M) into a finitely-generated graded G(R)-module. The multiplication in the ring comes from the following. If x_n\in\frak{q}^n, then let the image in \frak{q}^n/\frak{q}^{n+1} be denoted \overline{x_n}. We take \overline{x_n}\overline{x_m}=\overline{x_nx_m}. This does not depend on representative.

We’ll say G_n(M) is the n-th grade: M_n/M_{n+1}. Now G_0(R)=R/q is an Artinian local ring and each G_n(M) is a Noetherian R-module annihilated by \frak{q}. Thus they are all Noetherian R/\frak{q}=G_0(R)-modules. So by the Artinian condition we get that each G_n(M) is of finite length. Thus l_n=l(M/M_n)=\sum_{r=0}^{n-1} l(G_r(M))<\infty.

2) For large n, l(M/M_n) is a polynomial g(n) of degree \leq s where s is the least number of generators of \frak{q}.

Suppose x_1, \ldots, x_s generate \frak{q}. Then \{\overline{x_i}\} in \frak{q}/\frak{q}^2 generate G(R) as an R/\frak{q}-algebra. But l is an additive function on the filtration, so by last time we saw thatfor large n there is some polynomial such that f(n)=l(G_n(M))=l(M_n/M_{n+1}), and each \overline{x_i} has degree 1, so the polynomial is of degree \leq s-1.

Thus we get that l_{n+1}-l_n=l(G_n(M))=f(n). So from two posts ago, we get for large n that l_n is some polynomial g(n) of degree \leq s.

3) Probably the most important part is that the degree and leading coefficient of g(n) depends only on M and \frak{q} and not on the filtration.

Let (\overline{M_n}) be some other stable \frak{q}-filtration with polynomial \overline{g}(n)=l(M/\overline{M_n}). Since any two stable \frak{q}-filtrations have bounded difference, there is an integer N such that M_{n+N}\subset \overline{M_n} and \overline{M_{n+N}}\subset M_n for all n\geq 0. But this condition on the polynomials says that g(n+N)\geq \overline{g}(n) and \overline{g}(n+N)\geq g(n), which means that \lim_{n\to\infty}\frac{g(n)}{\overline{g}(n)}=1. Thus they have the same degree and leading coefficient.

That seems to be enough for one day. Unfortunately, I haven't quite got to the right setting that I want yet.