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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?

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    $\begingroup$ If $x\notin\mathfrak m$, then $(x)+\mathfrak m$ contains a non-zerodivisor. Same for $(y)+\mathfrak m$. The product of two non-zerodivisors is a non-zerodivisor. But the product belongs to... $\endgroup$ – user26857 Mar 12 '17 at 10:33
  • $\begingroup$ @user26857: Nice proof. But how can I go in my original direction? $\endgroup$ – spiritfire Mar 12 '17 at 12:26
  • $\begingroup$ An already existing version of this question with a good answer: math.stackexchange.com/q/627655/29335 $\endgroup$ – rschwieb Mar 12 '17 at 13:14