Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

On a scheme, being normal means that each stalk of the structure sheaf is a integrally closed domain. Being regular means that each stalk of the structure sheaf is a regular local ring.

As for a local ring, being regular or being integrally closed does not imply another.

What is their connection with each other and classical/usual intuition of being smooth(being regular on stalk of each closed points)?

Moreover, is there a smooth/regular variety which is not normal?

share|cite|improve this question
I am especially interested in the meaning of being normal in dimension no less than 2.(on curves being normal is pretty clear now.) – 7-adic Apr 1 '10 at 8:41

2 Answers 2

Dear 7-adic, yes there is an implication between the two notions.

For a local ring, regular implies normal. Actually Auslander and Buchsbaum proved in 1959 that a regular local ring is a UFD and it is an easy result that a UFD (local or not) is integrally closed. Serre then gave a completely different proof. He proved that regular is equivalent to having finite global (=homological) dimension . This finiteness means that any module over the ring has a finite projective resolution. I have heard it claimed that this was the beginning of the acknowledgment of the importance of homological algebra in commutative algebra.

An example.The cone $z^2=xy$ in affine 3-space (over a field, say) is normal but not regular: its very equation suggests that we don't have the UFD property and this intuition can be converted into a rigorous proof. Normality is a weak form of regularity. The two concepts coincide in dimension one but not in higher dimensions: the quadratic cone above shows this in dimension two.

Finally, smoothness is even stronger: it is a relative concept meaning regular and remaining regular after base change.

share|cite|improve this answer

You seem a bit confused. A regular* local ring is a UFD hence integrally closed. In other words, regular implies normal. See for instance

for a relatively elementary algebraic treatment.

*: I had previously included Noetherian here, but after checking on this I see I was being overly careful: it is part of the definition of a regular local ring that it be Noetherian.

share|cite|improve this answer
Eisenbud's book "Commutative Algebra (with a view to Algebraic Geometry)" has a comprehensible and well-presented treatment too, building it up from scratch. – Ravi Vakil Apr 4 '10 at 21:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.