A Mind for Madness

Musings on art, philosophy, mathematics, and physics

Applying the Hilbert Polynomial

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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.

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Author: hilbertthm90

I write about math, philosophy, literature, music, science, computer science, gaming or whatever strikes my fancy that day.

2 thoughts on “Applying the Hilbert Polynomial

  1. Some typos:

    1. The definition of stable q-filtration. I think you have qM_n \subset M_{n+1} for all $n$ and equality holds only for big n. Otherwise there is only one stable filtration as opposed to your third proprosition.

    2. I guess in the setting part you want to have A to be your noetherian ring. (Since you are using G(A) thereafter)

    3. In the proof of 1, you want \frak{q} instead of \frak{a}.

  2. Thanks. I’m really trying to hold to this R notation (since that is how I started forever ago) even though the literature tends to favor A, so slipping is bound to occur.

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