Let $R$ be a commutative ring with unity such that every prime ideal contains no non-zero zero divisor (i.e. if $P$ is a prime ideal and $x,y \in P$ with $xy=0$ then either $x=0$ , or $y=0$). Then is $R$ an integral domain ? (I want every prime ideal $P$ to contain no non-zero divisor as a ring in it's own right, not that $P$ contains no non-zero divisor of $R$.)

This question is related to Let R be a commutative ring, and let P be a prime ideal of R. Suppose that P has no nontrivial zero divisors in it. Show that R is an integral domain., but is not exactly same.

Please help. Thanks in advance.

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    $\begingroup$ Hint: Let $x$ a zero divisor. Then, it is not invertible. Therefore, it is contained in a maximal ideal... $\endgroup$
    – user228113
    Commented Feb 20, 2016 at 11:29
  • $\begingroup$ @G.Sassatelli : I think you are might be heading in the same wrong direction due ( perhaps ) to lack of clarification in the question body ? ( although I mentioned it clearly in parenthesis ..) $\endgroup$
    – user228168
    Commented Feb 20, 2016 at 11:32

1 Answer 1


What if $R=K\times K$ with $K$ a field? There are only two prime ideals $K\times\{0\}$ and $\{0\}\times K$, and for both we have $x,y\in P$ with $xy=0$ then $x=0$ or $y=0$.


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