# Supp$(M)=V($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 Supp$(M)$ for the support of the $A$-module $M$. We always have Supp$(M)\subseteq V($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 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 ann$M$. Since $M$ is finitely generated we have the equality (ann$M)_{\frak p}=$ 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 $M_{\frak p}\ne0$.

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