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?
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asked Aug 31, 2013 at 15:13
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4$\begingroup$ Not necessarily, see here: mathoverflow.net/questions/133657/… $\endgroup$– Vesselin DimitrovCommented Aug 31, 2013 at 16:35
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$\begingroup$ @Vesselin: Would you consider posting this as an answer? $\endgroup$– Jason StarrCommented Aug 31, 2013 at 18:17
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1$\begingroup$ This was a question raised by Hartshorne in `Ample subvarieties' and answered in the negative in general by Cowsik and Nori. $\endgroup$– MohanCommented Sep 1, 2013 at 15:28
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1 Answer
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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.
answered Sep 1, 2013 at 0:46
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$\begingroup$ 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? $\endgroup$ Commented Sep 1, 2013 at 6:16
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1$\begingroup$ Alexander, please see the edit. My guess is that one needs some serious effort to answer your question now. $\endgroup$ Commented Sep 1, 2013 at 9:41