9
$\begingroup$

Let $X$ be a Cohen-Macaulay scheme (let's say of finite type over a field). Let $X_{red}$ be the corresponding reduced scheme. Is it true that $X_{red}$ is also Cohen-Macaulay?

$\endgroup$
3
  • 3
    $\begingroup$ Not necessarily, see here: mathoverflow.net/questions/133657/… $\endgroup$ Aug 31, 2013 at 16:35
  • $\begingroup$ @Vesselin: Would you consider posting this as an answer? $\endgroup$ Aug 31, 2013 at 18:17
  • 1
    $\begingroup$ This was a question raised by Hartshorne in `Ample subvarieties' and answered in the negative in general by Cowsik and Nori. $\endgroup$
    – Mohan
    Sep 1, 2013 at 15:28

1 Answer 1

8
$\begingroup$

The simplest counter-example I know is the following: Hartshorne showed that if $k$ has positive characteristic, $k[s^4, s^3t, st^3,t^4]$ (which will be $X_{red}$) is a set-theoretic complete intersection (said complete intersection will be $X$). The former is well-known to be not CM (cheapest proof: $s^4,t^4$ form a s.o.p but not a regular sequence).

There are more examples of projective curves which are set-theoretic c.i. (you can find quite a few papers). Among them the ones which are not arithmetically CM give counter examples via taking the affine cone.

EDIT (to address the OP's new question below): this new situation is discussed in my answer quoted above by Vesselin, so you may want to take a look. To use that answer's notation, you need at least two height one primes, say $P,Q$, which are Cohen-Macaulay, and $a[P]+b[Q]=0$ in the class group of $Y$ (this takes care of the assumption that $X$ is set theoretically principal), but $[P]+[Q]$ is not CM. Such examples probably still exist (for example there are torsion classes which are not CM), but we may need a lot of luck (or hard work) to write one down.

$\endgroup$
2
  • $\begingroup$ Thanks a lot. Let me ask about a specific situation: assume that $Y$ is a Cohen-Macaulay scheme and $X$ is a divisor in it. Assume that we know the following: 1) Every component of $X$ is CM 2) Set-theoretically $X$ is the zero set of some regular function on $Y$. Can it be enough to conclude that $X$ is CM? If not, is there anything else one might require to gaurantee that $X$ is CM? $\endgroup$ Sep 1, 2013 at 6:16
  • 1
    $\begingroup$ Alexander, please see the edit. My guess is that one needs some serious effort to answer your question now. $\endgroup$ Sep 1, 2013 at 9:41

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.