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

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 $$M_{\frak p}\ne0$$.

$$\begin{array}{rl}p\notin V(\operatorname{Ann}(M))&\Leftrightarrow \operatorname{Ann}(M)\not\subseteq p\\&\Leftrightarrow \operatorname{Ann}(M_p)=\operatorname{Ann}(M)_p=A_p\\&\Leftrightarrow M_p=0\\&\Leftrightarrow p\notin \operatorname{Supp}(M).\end{array}$$
• Can someone say how $V(F_i(M))=\{p\in\text{Spec}(R)|\mu_{R_p}(M_p)>i\}$ is related to this problem? Oct 31 '18 at 3:37