Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I try to keep a list of standard ring examples in my head to test commutative algebra conjectures against. I would therefore like to have an example of a ring which is normal but not Cohen-Macaulay. I've found a few in the past, but they were too messy to easily remember and use as test cases. Suggestions?

share|improve this question
    
Isn't there a theorem that says that a normal semigroup ring is always CM? So that puts a lower bound on the complexity of an example. –  Charles Siegel Oct 21 '09 at 14:06
    
Yes, there is. We definitely need to get beyond semigroup rings here. One idea I had was that one might be able to take a product of curves, in some particular projective embedding, and cone on that. But all suggestions are welcome! –  David Speyer Oct 21 '09 at 14:11
    
I guess you are requiring normal rings to be Noetherian, so that that isn't an easy example? –  Greg Muller Oct 21 '09 at 14:17
add comment

3 Answers

up vote 19 down vote accepted

Another family of examples is given by the homogeneous coordinate rings of irregular surfaces (ie 2-dimensional $X$ such that $H^1({\mathcal O}_X) \neq 0$); these surfaces cannot be embedded in any way so that their homogeneous coordinate rings become Cohen-Macaulay. Elliptic scrolls (such as the one in the previous answer) and Abelian surfaces in P4, made from the sections of the Horrocks-Mumford bundle, are such examples.

The point is that sufficiently positive, complete embeddings of any smooth variety (or somewhat more generally) will have normal homogeneous coordinate rings, and they will be Cohen-Macaulay iff the intermediate cohomology of the variety vanishes. All the examples above fall into this category. It's an interesting general question to ask how positive is "sufficiently positive".

share|improve this answer
    
Just to clarify this answer: the (normalization of the) cone over a smooth projective variety X is CM iff the middle (i.e. of degrees $0<d<\dim X$) cohomology of the sheaves $O_X(k)$ is zero, for all $k\in\mathbb Z$. Vanishing for $O_X$ only is insufficient. (Reference: D. Eisenbud, Commutative algebra with a view towards algebraic geometry, the section on local cohomology). –  VA. Dec 8 '09 at 19:55
add comment

Segre products of normal CM N-graded K-algebras A, B are always normal (since they're direct summands of the normal ring A\otimes_K B), but rarely CM.

A particular example is here:

http://www.math.purdue.edu/~walther/research/segre.ps

or also here:

http://www.mathstat.dal.ca/~faridi/research/enescu-faridi.pdf (Corollary 35)

Take A=k[x,y,z]/x^3+y^3+z^3 to be the Fermat cubic, and B=k[a,b]. Then the Segre product A#B is not CM, since the a-invariant of A is non-negative.

share|improve this answer
add comment

MJ Bertin en Anneaux d'invariantes d'Anneaux de polynômes, en caractéritique p . C.R.Acad.Sci.Paris Sér. A-B 264 1967 A653-A656

"... MJ Bertin has made use of Galois descent in order to construct an example of a factorial noetherian ring wich is not Cohen Macaulay..." Robert M. Fossum

The example is done completly in The Divisor Class Groups of a Krull Domain, Robert Fossum, Example 16.7 pages 87-88

share|improve this answer
add comment

Your Answer

 
discard

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.