Atiyah's Introduction to Commutative Algebra, chapter 1 exercise 14: In a commutative ring $A$ with identity, let ∑ be the set of all ideals in which every element is a zero-divisor. Show that the set ∑ has maximal elements and that every maximal element of ∑ is a prime ideal. Hence the set of zero-divisors in A is a union of prime ideals.
The existence of maximal elements is guaranteed by Zorn's lemma. To prove that every maximal element of ∑ is prime: Given $\mathfrak m$ a maximal element of ∑ and $xy \in \mathfrak m$. There exists a nonzero $z$ such that $xyz=0$. If $xz=0$, then $(x)\in$ ∑. If $xz\neq0$, then $(y)\in$ ∑. So we suppose $(x)\in$ ∑. HERE IS THE PROBLEM：Intuitively I want to prove $(x,\mathfrak m)\in$ ∑ and by maximality of $\mathfrak m$, $x\in\mathfrak m$. But how to prove that every element in $(x,\mathfrak m)$ is a zero-divisor? For example, if given $ax+bz$ where $a,b\in A, z\in \mathfrak m$ and $xs=zt=st=xz=0, xt\neq0,zs\neq0 $, how to prove that $ax+bz$ is a zero-divisor?