Suppose I have a reduced l.c.i. scheme with two irreducible components: $X = Y \cup Z$. I want to say that if $Y$ is Cohen-Macaulay then $Z$ is as well.

I think this follows from Eisenbund Theorem 21.23 (which has a typo: the first $J = (0:_A I)$ should be deleted). Or from Peskine and Szpiro, "Liaison des variétés algébriques," Proposition 1.3, which is essentially the same.

Am I understanding correctly?


The question is local. So, let $R$ be a local ring which is Gorenstein. $I,J\subset R$ define $Y,Z$ as in your question. Then you have an exact sequence $0\to I\to R\to R/I\to 0$ and we are assuming that $R/I$ is Cohen-Macaulay. Notice that all $R,R/I,R/J$ have the same dimension $d$. Dualizing, one gets $0\to\omega_{R/I}\to R\to R/J\to 0$. This implies that the depth of $R/J\geq d-1$. By going modulo a general set of $d-1$ elements in the maximal ideal, one can reduce to the case where $d=1$. Now dualize again to get, $0\to \operatorname{Hom}_R(R/J,R)\to R\to R/I\to\operatorname{Ext}^1_R(R/J,R)\to 0$. It is clear by naturality, that the map $R\to R/I$ is onto and thus the ext is zero. This says that depth of $R/J>0$ which is what we wanted.

  • $\begingroup$ I didn't follow every detail of your argument, but if you're convinced the original claim is true then I'm happy! $\endgroup$ Aug 21 '20 at 22:19
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    $\begingroup$ @NickAddington Happiness is good, but not quite mathematics. What is that you didn't follow ? $\endgroup$
    – Mohan
    Aug 22 '20 at 0:51
  • $\begingroup$ It was laziness rather than confusion: your reduction to the case d=1 is the kind of thing I've done before, but I have to spend time with it to really convince myself. $\endgroup$ Aug 22 '20 at 2:36
  • $\begingroup$ Anyway I think I have three proofs now: yours and the two references in the original question, which are a bit different. So I'm satisfied. Thanks. $\endgroup$ Aug 22 '20 at 2:36
  • $\begingroup$ When you dualize the exact sequence $0\to I \to R \to R/I \to 0$, how do you get $Hom(I,R)\cong R/J$ ? $\endgroup$
    – user521337
    Jul 17 at 18:56

I have access to neither of your references but here, it seems to me, is a counterexample. Take a smooth quadric surface $Q$ in $\mathbb P^3$, a smooth curve $C$ in $Q$ of bidegree $(1,3)$ and another smooth curve $D$ in $Q$ of bidegree $(3,1)$. Each of $C,D$ is a twisted quartic in $\mathbb P^3$. Take $Y,\ Z$ and $X$ to be the affine cones over $C,\ D$ and $C\cup D$, respectively. $C\cup D$ is a $(2,4)$ complete intersection in $\mathbb P^3$, so $X$ is l.c.i. Moreover, $X=Y\cup Z$, while $Y,Z$ are cones over twisted quartics, so not Cohen--Macaulay.

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    $\begingroup$ The question is whether $Y$ Cohen--Macaulay implies $Z$ Cohen--Macaulay. $\endgroup$
    – abx
    Aug 21 '20 at 13:08
  • 1
    $\begingroup$ Right, this (very nice) example shows that Y and Z can both be non-CM, but that's not what I'm asking. $\endgroup$ Aug 21 '20 at 14:28
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    $\begingroup$ Ah! Indeed, I misread the question. Thank you to abx and the OP for pointing that out. I will leave up this answer, however, since deleting it would leave a meaningless thread. $\endgroup$
    – inkspot
    Aug 21 '20 at 14:59

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