Proposition. Let $R$ be a commutative ring and $I\subseteq R$ an ideal. Then $I$ is primary if and only if every zero divisor in $R/I$ is nilpotent.

Proof. ,,$\Rightarrow$ '' Assume, that we have $x\in R$ such that $x+I$ is a zero divisor in $R/I$ . In particular $x+I\neq 0+I$ and there is $y\in R$ , $y+I\neq 0+I$ such that $$0+I=(x+I)(y+I)=xy+I.$$ This is if and only if $xy\in I$ . Thus either $y\in I$ or $x^n\in I$ for some $n\in\mathbb{N}$ . Of course $y\not\in I$ , because $y+I\neq 0+I$ and thus $x^n\in I$ . Therefore $x^n+I=0+I$ , which means that $x+I$ is nilpotent in $R/I$ .

,,$\Leftarrow$ '' Assume that for some $x,y\in R$ we have $xy\in I$ and $x,y\not\in I$ . Then $$(x+I)(y+I)=xy+I=0+I,$$ so both $x+I$ and $y+I$ are zero divisors in $R/I$ . By our assumption both are nilpotent, and therefore there is $n,m\in\mathbb{N}$ such that $x^n+I=y^m+I=0+I$ . This shows, that $x^n\in I$ and $y^m\in I$ , which completes the proof. $\square$