Studying some topics in Algebraic Geometry I've bumped into the following question:
Let $A$ be a regular ring. Is $A$ integrally closed?
Someone said me that with the hypothesis $A$ local regular ring it is true (Can anyone give me a proof of this assert or give me some references?), but what about the general case?
There are some counterexamples?
Thank you
Best Answer
A regular ring is, by definition, a noetherian ring whose localizations with respect to maximal ideals are (local) regular.
Similarly, a ring is called normal if all localizations with respect to maximal ideals are integrally closed domains.
Clearly, a regular ring is normal (see Remark). If moreover, $R$ is a regular integral domain, then it is integrally closed as being $\bigcap_{\mathfrak m\in\operatorname{Max}R} R_{\mathfrak m}$, an intersection of integrally closed domains having the same field of fractions.
Remark. A simpler proof that local regular rings are integrally closed (that is, without proving they are UFDs) follows from the Proposition 2.2.3 combined with Proposition 2.2.6 and Theorem 2.2.22 in Bruns and Herzog. (For those who don't have the book: local regular rings are integral domains and Cohen-Macaulay, hence, in particular, they satisfy the conditions $(R_1)$ and $(S_2)$, and by Serre's normality criterion it follows that they are integrally closed.)