Let $A$ be a Noetherian ring, $M$ be a finitely generated $A$-module, and $\mathfrak{p}$ be a prime ideal in $A$.
I want to ask how to prove that $M\otimes A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\neq0$ if and only if $\text{Ann}(M)\subseteq\mathfrak{p}$.
Thanks.
(We don't need to assume that $A$ is Noetherian.)
$M_{\mathfrak{p}} \cong M \otimes_A A_{\mathfrak{p}}$ is finitely generated over $A_{\mathfrak{p}}$, so by Nakayama, we get $0 \neq M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}} \cong M \otimes_A A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ if and only if $M_{\mathfrak{p}} \neq 0$ which is by definition the case if and only if $\mathfrak{p} \in \operatorname{Supp}(M)$, so your question reduces to the relation $V(\operatorname{Ann}(M))=\operatorname{Supp}(M)$ which is shown for example here.
A complementary of Lukas's excellent answer.
Proposition. If $M$ is a finitely generated $A$-module, then $\operatorname{Supp}(M)=V(\operatorname{Ann}(M))$.
Proof. If $\mathfrak{p}\notin\operatorname{Supp}(M)$, then $M_\mathfrak{p}=0$. Then for any $m\in M$ and $a\in A-\mathfrak{p}$, there exists $s\in A-\mathfrak{p}$ such that $sm=0$. Thus for generators $m_1,...,m_n$, we can find $s_1,...,s_n\in A-\mathfrak{p}$ such that $s_1\cdots s_nm=0$ for any $m\in M$, thus $\mathfrak{p}$ doesn't contain $\operatorname{Ann}(M)$.
In the other side, if $\mathfrak{p}$ doesn't contain $\operatorname{Ann}(M)$, then there exists an element $t\in A-\mathfrak{p}$ such that $tm=0$ for any $m$. Therefore $M_\mathfrak{p}=0$.
