If $P \in \operatorname{Supp}(M)$ prove that $P$ contains a prime ideal $Q$ with $Q \in \operatorname{Ass}_R(M)$. My problem is below,
Let $M$ be an $R$-module.
The set of prime ideals $P$ of $R$ for which the localization $M_P$ is nonzero is called the support of $M$, denoted $\operatorname{Supp}(M)$.
The set of prime ideals $Q$ of $R$ for which $Q$ is an annihilator for some element $m \in M$, denoted $\operatorname{Ass}_R(M)$.

Suppose that $R$ is Noetherian. If $P \in \operatorname{Supp}(M)$ prove that $P$ contains a prime ideal $Q$ with $Q \in \operatorname{Ass}_R(M)$.

My attempt,
Suppose there exists a prime ideal $P$ such that $M_P \neq 0$.
Then there exists $x \in M$ such that $\operatorname{Ann}(x) \subset P$.
Let $\mathcal{S} = \{ \operatorname{Ann}(x) : x \in M, \operatorname{Ann}(x) \subseteq P \}$.
Take a maximal in $\mathcal{S}$ and I tried to prove that it is a prime ideal.
But it fails until now.
Can anybody help me? Thank you.
 A: Consider the set 
$\Phi=\left\{\operatorname{ann}(\xi): 0\neq \xi \in M_P\right\}$.
This set of ideals is nonempty, since $M_P \neq 0$. Notice also that
for every $\operatorname{ann}(\xi) \in \Phi$ we have $\operatorname{ann}(\xi) \subset P$. Let 
$I = \operatorname{ann}(\xi)$ be a maximal element of $\Phi$. Suppose $a,b$ are elements of $R$
such that $ab \in I$ and suppose that $a \not\in I$. There are two possibilities:
i) $\operatorname{ann}(a\xi) \not\subset P$ and ii) $\operatorname{ann}(a
\xi) \subset P$. If i) is true, then there exists an element $s \not\in P$ such that
$sa\xi=0$. Now notice that $s \xi \neq 0$ and that $\operatorname{ann}(s\xi) \subset P$. Since $\operatorname{ann}(s \xi) \in \Phi$ and $\operatorname{ann}(\xi) \subset \operatorname{ann}(s \xi)$, the maximality of $\operatorname{ann}(\xi)$ in $\Phi$ gives $\operatorname{ann}(\xi) = \operatorname{ann}(s \xi)$. But then this implies that $a \in \operatorname{ann}(\xi)$, which is a contradiction on our hypothesis that $a \not\in \operatorname{ann}(\xi)$. Hence it must be the case that ii) is true, i.e. $\operatorname{ann}(a \xi) \subset P$. But then again the maximality of $\operatorname{ann}(\xi)$ gives $\operatorname{ann}(\xi) = \operatorname{ann}(a \xi)$, which implies that $b \in \operatorname{ann}(\xi)$ and so we have proved that $\operatorname{ann}(\xi)$ is prime. 
