15
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ I am especially interested in the meaning of being normal in dimension no less than 2.(on curves being normal is pretty clear now.) $\endgroup$
    – 7-adic
    Apr 1, 2010 at 8:41
  • 1
    $\begingroup$ See this question: mathoverflow.net/questions/12688/nonsingular-normal-schemes $\endgroup$ Apr 1, 2010 at 13:32
  • $\begingroup$ This could be an answer, but a comment suffices. See Theorem 36 of Matsumura's "Commutative Algebra" (p.121): A regular local ring is an integrally closed integral domain. $\endgroup$
    – Jose Capco
    Feb 27, 2019 at 14:28

2 Answers 2

26
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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

http://www.math.iitb.ac.in/atm/caag1/jayanthan.pdf

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.

$\endgroup$
1
  • 1
    $\begingroup$ Eisenbud's book "Commutative Algebra (with a view to Algebraic Geometry)" has a comprehensible and well-presented treatment too, building it up from scratch. $\endgroup$
    – Ravi Vakil
    Apr 4, 2010 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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