# 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.