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.

Author: hilbertthm90

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

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.