Proposition 2

Let $M$ be a module over a ring $A$. Every maximal element of the set of ideals $\operatorname{Ann}(x)$ of $A$, where $x$ runs through the set of elements $\neq 0$ of $M$, belongs to $\operatorname{Ass}(M)$.

Let $a = \operatorname{Ann}(x)$ ($x \in M, x \neq 0$) be such a maximal element; it is sufficient to show that $a$ is prime. As $x \neq 0$, $a \neq A$. Let $b, c$ be elements of $A$ such that $bc \in a$ and $c \notin a$. Then $cx \neq 0$, $b \in \operatorname{Ann}(cx)$ and $a \subset \operatorname{Ann}(cx)$. As $a$ is maximal, $\operatorname{Ann}(cx) = a$, whence $b \in a$, so that $a$ is prime.