If $a_M$ is not locally nilpotent, why does there exist $x \in M$ and a prime ideal $\mathfrak{p}$ such that $(Ax)_\mathfrak{p} \neq 0$? Let $A$ be a commutative ring. I have a short question about the small result (Proposition 2.5 of Lang's Algebra, pg. 418, third edition) that if $M$ is an $A$ module, and $a\in A$, then $a_M$ defined by $x\mapsto ax$ for $x\in M$, (i.e. the left-multiplication map) is locally nilpotent if and only if $a$ lies in every prime ideal $\mathfrak{p}$ such that $M_\mathfrak{p}\neq 0$. Here $M_\mathfrak{p}$ is the localization of $M$ by $A\setminus \mathfrak{p}$.
In the converse statement, if $a_M$ is not locally nilpotent, then there is $x\in M$ such that $a^nx\neq 0$ for all $n\geq 0$. Let $S=\{1,a,a^2,\dots\}$ be a multiplicative set, so I know there exists a prime $\mathfrak{p}$ maximal among ideals not intersecting $S$. Why does it follow that $(Ax)_\mathfrak{p}\neq 0$, so that $M_\mathfrak{p}\neq 0$? Lang states this in one line, so I feel it must be obvious.

I think I understand this now. It follows from the definitions that if $x$ is an element of $M$ with $\frak{a}$ its annihilator, and $p$ a prime ideal of $A$, then $(Ax)_p\neq 0$ iff $p\supset\frak{a}$. 
Now $S$ is disjoint from the annihilator of $x$, so the annihilator of $x$ is contained in some ideal maximal among those not intersecting $S$, which happens to be prime. Take $p$ to be this prime ideal, and so $(Ax)_p\neq 0$.
