$\operatorname{Supp}(M)=V(\operatorname{Ann}M)$ if $M$ is finitely generated

Question

Let $A$ be a commutative ring with 1, set $V(\frak a)=$ the set of prime ideals of $A$ that contains $\mathfrak a$, and write $\operatorname{Supp}(M)$ for the support of the $A$-module $M$. We always have $\operatorname{Supp}(M)\subseteq V(\operatorname{Ann} M)$; if $M$ is finitely generated then the other inclusion also holds. Assume now $M$ is finitely generated.

It is ok to show that if $M_{\frak p}=0$ then $\operatorname{Ann} M\cap A\setminus\frak p\ne\varnothing$, I was just wondering if the following procedure for showing this inclusion is flawed:

Suppose $\frak p$ contains $\operatorname{Ann} M$. Since $M$ is finitely generated we have the equality $(\operatorname{Ann} M)_{\frak p}=\operatorname{ Ann} (M_p)$, hence the annihilator of $M_{\frak p}$ is contained in maximal ideal of $A_{\frak p}$ and is therefore different from (1), so $M_{\frak p}\ne0$.

score 4 · Answer 1 · edited Aug 15 '20 at 19:36

4

Your argument is valid. You can be more direct:

$$\begin{array}{rl}p\notin V(\operatorname{Ann}(M))&\Leftrightarrow \operatorname{Ann}(M)\not\subseteq p\\&\Leftrightarrow \operatorname{Ann}(M_p)=\operatorname{Ann}(M)_p=A_p\\&\Leftrightarrow M_p=0\\&\Leftrightarrow p\notin \operatorname{Supp}(M).\end{array}$$

edited Aug 15 '20 at 19:36

KReiser

65,137

answered Aug 03 '18 at 01:21

Rafael

3,789

Can someone say how $V(F_i(M))={p\in\text{Spec}(R)|\mu_{R_p}(M_p)>i}$ is related to this problem? – Username Unknown Oct 31 '18 at 03:37
Where in this proof do you use that $M$ is finitely generated? – David Lui May 22 '23 at 17:14
Localization commutes with taking annihilator only if $M$ is finitely generated. See for instance https://math.stackexchange.com/a/1942551/303887 – Rafael May 22 '23 at 22:11

$\operatorname{Supp}(M)=V(\operatorname{Ann}M)$ if $M$ is finitely generated

1 Answers1

Linked

Related