While reading through Chapter X of Lang's Algebra, I came across following definitions and results. In my opinion, something seems wrong in the proof of Proposition 2.5. (page 418, third edition).
Some prior results:
Proposition 2.1. Let $S$ be a multiplicative subset of $A$, and assume that $S$ does not contain $0$. Then there exists an ideal of $A$ which is maximal in the set of ideals not intersecting $S$, and any such ideal is prime.
Lemma 2.4. Let $x$ be an element of an $A$-module $M$. Let $\mathfrak{p}$ be a prime ideal of $A$. Then $(Ax)_\mathfrak{p} \neq 0$ if and only if $\mathfrak{p}$ contains the annihilator of $x$.
Let $a \in A$ and $M$ be an $A$-module. We denote by $a_M$ the homomorphisms $x \mapsto ax$, $x \in M$. We shall now say that $a_M$ is locally nilpotent if for each $x \in M$ there exists an integer $n(x)\geq 1$ such that $a^{n(x)}x=0$.
Now, the statement and proof of Proposition 2.5.:
Proposition 2.5. Let $M$ be a module, $a \in A$. Then $a_M$ is locally nilpotent if and only if $a$ lies in every prime ideal $\mathfrak{p}$ such that $M_\mathfrak{p} \neq 0$.
Proof. The first part of the proof shows that when $a_M$ is locally nilpotent, $a$ lies in every prime ideal $\mathfrak{p}$ such that $M_\mathfrak{p} \neq 0$.
Conversely, suppose $a_M$ is not locally nilpotent, so there exists $x \in M$ such that $a^nx = 0$ (I'm guessing this has to be a "$\neq$" and therefore the rest of the proof isn't valid anymore) for all $n\geq 0$. Let $S=\{1,a,a^2,\dots\}$, and using proposition 2.1 let $\mathfrak{p}$ be a prime not intersecting $S$. Then $(Ax)_{\mathfrak{p}} \neq 0$, so $M_\mathfrak{p} \neq 0$ and $a \notin \mathfrak{p}$, as desired.
Am I right that it has to be a "$\neq$"? If not, can someone fix the proof?
You are right about $a^nx \neq0$. But then the proof is correct.
You might want to note additionally that $(Ax)_{\mathfrak{p}} \neq 0$ since $x/1\neq0$. And, $a \notin \mathfrak{p}$ since $a \in S$.