0

Let $A$ a Noetherian local ring and $M$ a finite $A$-module. How can I prove that if $x$ is $M$-regular then $$\dim M=\dim M/xM+1?$$

I had a proof with the hypothesis that $\operatorname{Ann}(M/xM)=(\operatorname{Ann}(M),x)$ but I can't prove it's true in this case (and it probably isn't).

Edit: the hard part is proving $$\dim M\leq \dim M/xM+1$$

max_zorn
  • 4,875
karmalu
  • 1,320
  • Have you looked up some texts (e.g. Matsumura)? These are standard results proved in most books dealing with dimension. – Mohan Sep 12 '15 at 14:13
  • This is actually true over rings that are only semi-local when $x\in\mathcal{J}(R)$, as a reference see Theorem 6 on pages 250-251 of Lessons on Rings, Modules and Multiplicities by D.G. Northcott (1968) – Dave Jan 19 '21 at 17:26

1 Answers1

0

We have $\sqrt{\operatorname{Ann}(M)+(x)}=\sqrt{\operatorname{Ann}(M/xM)}$ so $\dim M/xM=\dim A/(\operatorname{Ann}(M),x)$.

Now $x$ isn't a minimal prime in $A/(\operatorname{Ann}(M))$ because it's $M$-regular. So $\operatorname{ht} x=1$ and $$\dim A/(\operatorname{Ann}(M),x)+1\leq \dim M$$

Now consider $x_2,\ldots, x_s$ a system of parameter in $A/(\operatorname{Ann}(M),x)$, then $(x, x_2,\ldots, x_s)$ is a system of parameter of $A/\operatorname{Ann}(M)$ which then has dimension $\leq s=A/(\operatorname{Ann}(M),x)+1$

karmalu
  • 1,320