Example 2

Let $q$ be an ideal of $A$ such that there exists a single prime ideal $m$ (necessarily maximal) containing $q$; then, if $M$ is an $A$-module such that $qM \neq M$, $qM$ is m-primary with respect to $M$. For every element of $\operatorname{Ass}(M/qM)$ contains $q$, hence is equal to $m$ and $\operatorname{Ass}(M/qM) \neq 0$ ($§ 1$, no. 1, Corollary 1 to Proposition 2). In particular $q$ is an m-primary ideal in $A$.