Can somebody help proofing the following lemma.
Let $x$ be an element of a module $M$, and let $\mathfrak{a}$ be its annihilator. Let $\mathfrak{p}$ be a prime ideal of $A$. Then $(Ax)_{\mathfrak{p}} \neq 0$ if and only if $\mathfrak{p}$ contains $\mathfrak{a}$.
Thus far I know that the kernel of the following map $$\phi: Ax \to (Ax)_{\mathfrak{p}};m \mapsto \frac{m}{1}$$ contains those elements of $Ax$ that are annihilated by an element of $A-\mathfrak{p}$.
I guess I found a proof. Can someone verify my steps?
Let $S=A - \mathfrak{p}$.
At first, suppose $(Ax)_{\mathfrak{p}}=0$. Particularly $\frac{x}{1}=0$. Hence there exists a $s \in S$ and therefore $s \notin \mathfrak{p}$ such that $sx=0$ which means $s \in \mathfrak{a}$. In conclusion $\mathfrak{p}$ does not contain $\mathfrak{a}$ since $s \in \mathfrak{a}$ and $s \notin \mathfrak{p}$.
Conversly, let $a$ be an element in $\mathfrak{a}$ but not in $\mathfrak{p}$, especially $s \in S$. Let $\frac{mx}{s}$ be a random element of $(Ax)_{\mathfrak{p}}$, because there exists an element $s' \in S$ such that $s'mx=0$ (indeed $amx=0$), $\frac{mx}{s}=0$. Hence $(Ax)_{\mathfrak{p}}=0$.