Tell me more ×  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 $M_{\frak p}\ne0$.  share|improve this
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