# 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.