# On the dimension of a module over a Noetherian ring

For a noetherian ring $R$, let $x\in R$ be a non-zerodivisor of an $R$-module $M$. Then this is a well-known fact:

$$\dim M/xM \leq \dim M-1.$$

I saw a proof for the case of finitely generated $M$. I wonder if there is a proof of this fact in the general case.

And, perhaps, is it possible to prove this fact using Krull's Principal Ideal Theorem?

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Hi! Remember that not everyone might recognize your abbreviations right away, so it's always good policy to type out things you normally abbreviate. –  rschwieb Dec 27 '12 at 14:50

Take $\mathfrak p\in\operatorname{Supp}(M/xM)$. I claim that $\mathfrak p$ is not minimal in $\operatorname{Supp}(M)$ and this is enough to prove $\dim M/xM \leq \dim M-1$.
It is obvious that $\mathfrak p\in\operatorname{Supp}(M)$. Assume that $\mathfrak p$ is minimal in $\operatorname{Supp}(M)$. Then $\mathfrak p$ is minimal in $\operatorname{Ass}(M)$ (why?), so there exists $z\in M$, $z\neq 0$, such that $\mathfrak p=\operatorname{Ann}(z)$. On the other side, $\mathfrak p\in\operatorname{Supp}(M/xM)\Rightarrow x\in\mathfrak p\Rightarrow xz=0\Rightarrow z=0$, a contradiction.