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For an integral domain $R$, the intersection of two non-zero ideals is also non-zero, because the product of any two non-zero elements is non-zero.

Is the converse true, i.e. if $R$ has the property that the intersection of any two non-zero ideals is also non-zero, is $R$ an integral domain? If not, what is a natural equivalent condition?

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    $\begingroup$ Any ring of the form $\mathbb{Z}/p^n\mathbb{Z}$ for $n>1$ would be a counterexample. $\endgroup$
    – Zev Chonoles
    Commented Apr 4, 2016 at 7:00

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$R$ is an integral domain if and only if $R$ is reduced and the intersection of two nonzero ideals is nonzero.

Sketch of Proof.

Necessity.check

Sufficiency. If $ab=0$, then $I=aR\cap bR =0$ since $x=at=bs\in I$ implies $(at)^2=atbs=0$ and so $x=0$.

PS Just joined the site so my apologies if not the correct way to answer. Any counter-example to your question (e.g. the comments) must possess non-zero nilpotents.

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