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.


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