A regular local ring is a domain. Is a regular ring (a ring whose localization at every prime ideal is regular) also a domain? I am unable to find/construct a proof or a counterexample. Any help would be appreciated.


No. E.g. choose two regular domains and take their product; this will be regular, but not a domain.

This is more or less the general case, as I will explain: In general, a Noetherian ring, all of whose localizations at its prime ideals are domains, is a finite product of domains (and of course a finite product of domains has this localization property). So a regular Noetherian ring will be a finite product of regular domains (and conversely any such product will be regular).

Geometrically, one can think of this as follows: regularity of $A$ is a local property on Spec $A$, and Spec $A\times B$ is equal to Spec $A \coprod$ Spec $B$. So locally Spec $A\times B$ looks like either Spec $A$ or Spec $B$. In particular, local properties, such as regularity (or the condition that the localization at prime ideals be a domain) can't detect global properties (like $A$ itself being a domain).

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    $\begingroup$ Thanks a lot for the elaborate explanation. $\endgroup$ – Dev Bappa Jan 23 '11 at 23:50
  • $\begingroup$ The geometric point of view also illuminates the claim that if every localization is a domain the ring is a product of domains: if $\operatorname{Spec} A$ has more than one irreducible component, then if the components have any point of contact this point will yield a local ring that is not a domain. Thus the irreducible components of \operatorname{Spec}A $ must be completely disjoint. The decomposition into connected components will then correspond to a product decomposition of the ring. $\endgroup$ – Ben Blum-Smith Feb 26 '17 at 19:52

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