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A commutative ring $R$. The set $\Sigma$ consists of all ideals in which every element is a zero-divisor. Clearly $\Sigma$ has a maximal element $M$. I want to prove that $M$ is a prime ideal of $R$. (Atiyah & MacDonald Ex.1.14)

Here is my idea: suppose that $M$ is not prime. Choose $a, b\in R-M$ such that $ab=m\in M$, and argue that $(M\cup {a})$ is an ideal consisting of zero-divisors. But how to prove this? I have no idea.

user150248
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    Found the duplicate in the related questions list. The top two hits when I searched for "zero divisor maximal prime" were dupes. Please search first. – rschwieb Sep 20 '17 at 13:28
  • @Arthur I think you must need to update the page or something. The duplicate I found was the one I closed it as shortly before posting the comment. Whether or not the related questions list is the same isn't very important, I think, I just meant to explain the method I used looking for duplicates. – rschwieb Sep 20 '17 at 13:32

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