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.
 A: 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 $(Ax)_p\neq 0$.
A: For any $A$-module $M$ let $S(M)$ be its support and $\mathfrak a(M)$ its annihilator. Let $LN$ be the set of those $a$ in $A$ such that $a_M$ is locally nilpotent. We want to prove 
$$
LN=\bigcap_{\mathfrak p\in S(M)}\ \mathfrak p.\qquad\qquad(1)
$$
The main lemma will be:
$(2)$ If $M$ is finitely generated, then $S(M)=V(\mathfrak a)$, with $\mathfrak a:=\mathfrak a(M)$.
Recall: $V(\mathfrak a):=\{\mathfrak p\in\text{Spec}(A)\ |\ \mathfrak p\supset\mathfrak a\}$.
Proof of $(2)$. Assume $\mathfrak p\supset\mathfrak a$. We must show $M_{\mathfrak p}\neq0$. Suppose by contradiction $M_{\mathfrak p}=0$. Let $x_1,\dots,x_n$ be generators of $M$. For each $i=1,\dots,n$ there is an $s_i$ which is $A$ but not in $\mathfrak p$ such that $s_ix_i=0$. Then $s_1\cdots s_n$ is in $\mathfrak a$ but not in $\mathfrak p$, contradiction.
Conversely, assume $M_{\mathfrak p}\neq0$. Then there is an $x$ in $M$ such that $sx\neq0$ for all those $s$ in $A$ which are not in $\mathfrak p$. Let $\alpha\in\mathfrak a$. As $\alpha x=0$, the element $\alpha$ is in $\mathfrak p$.
Proof of $(1)$. Let $(M_i)$ be a family of finitely generated submodules of $M$ whose sum is $M$. Put $\mathfrak a_i:=\mathfrak a(M_i)$.
By definition we have 
$$
LN=\bigcap_i\ r(\mathfrak a_i).
$$
Recall the equality 
$$
r(\mathfrak a_i)=\bigcap_{\mathfrak p\in V(\mathfrak a_i)}\ \mathfrak p.
$$
where $r(\mathfrak a_i)$ is the radical of $\mathfrak a_i$. By $(2)$, we have
$$
\bigcap_{\mathfrak p\in V(\mathfrak a_i)}\ \mathfrak p=\bigcap_{\mathfrak p\in S(M_i)}\ \mathfrak p.
$$
For formal reasons we have
$$
\bigcap_i\ \bigcap_{\mathfrak p\in S(M_i)}\ \mathfrak p=\bigcap_{\mathfrak p\in\bigcup S(M_i)}\ \mathfrak p.
$$
Now $(1)$ follows from the fact that $S(M)$ is the union of the $S(M_i)$.
A: An useful observation, which may help you conclude.
If the map $\alpha:m\in M\mapsto am\in M$ is locally nilpotent, then the map $1-\alpha:M\to M$ is invertible: you can use the geometric series to define its inverse. 
