Applying the Hilbert Polynomial

October 29, 2009 by hilbertthm90 | 2 Comments

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 am a mathematics graduate student fascinated in how all my interests fit together.

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Applying the Hilbert Polynomial

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