TITLE: If the localization of a ring $R$ at every prime ideal is an integral domain, must $R$ be an integral domain?
QUESTION [29 upvotes]: Let $R$ be a commutative ring. Suppose that for every prime ideal $\mathfrak p$ of $R$, the localization $R_{\mathfrak p}$ is an integral domain. Must $R$ be a integral domain?

I was trying to think of counter-examples, but kept getting $R_{(0)}$ = the zero ring, which is not a domain.
Any guidance would be much appreciated. Thanks.

REPLY [3 votes]: A broad class of answers that encompasses some of the specific examples already mentioned:
It is known that a commutative ring is von Neumann regular iff its localizations at prime ideals are all fields.
But a von Neumann regular ring is only a domain if it is a field. Von Neumann regular rings are a huge class with lots of examples which aren't bound by chain conditions. The main reason they make good candidates is because their prime ideals are maximal (so the result of the localization is rigged.)