Regular Ring is Integrally Closed?

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

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.)

• The way you wrote your answer is slightly weird. Your third sentence states that learly a regular ring is normal, but that that is true is actually almost the question you are answering! – Mariano Suárez-Álvarez Jan 16 '15 at 22:09
• @MarianoSuárez-Alvarez Maybe after reading the question carefully could find my answer less "weird". (The OP mentioned that he knows the result for local rings, but is still interested in a proof.) – user26857 Jan 16 '15 at 22:18

If you have a regular local ring, it is normal. Auslander and Buchsbaum proved that a regular local ring is a unique factorization domain, and a unique factorization domain (not necessarily local) is normal.

(JP) Serre then reproved this by proving something mor general : that regular is equivalent to having finite global dimension, that is, that any module over the ring has a finite length projective resolution.

I would say that normality (being normal) and regularity coincide in dimension 1, but not in higher dimension. (Think of the cone $T^2 = XY$ in $\mathbf{A}_k^3$ (dimension 2 for the ring) where you would have normality but not the regularity by construction.)

If you do know that local regular rings are integrally closed, you know that a regular ring has all its localizations at prime ideals integrally closed. Therefore, the only piece of knowledge you need to reach the conclusion you want is that

an integral domain all of whose localizations at prime ideals is integrally closed is integrally closed

which is part of the content of Proposition 5.13 in the book by Atiyah and MacDonald.

You can replace prime by maximal throughout.

• The reason for why people like Bruns and Herzog define a regular ring as being a noetherian ring whose localizations with respect to maximal ideals are (local) regular is the old question posed by Serre (I think) whether the localizations (at primes) of a local regular ring are (local) regular. – user26857 Jan 16 '15 at 23:04