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.

• I didn't follow every detail of your argument, but if you're convinced the original claim is true then I'm happy! Aug 21 '20 at 22:19
• @NickAddington Happiness is good, but not quite mathematics. What is that you didn't follow ? Aug 22 '20 at 0:51
• 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. Aug 22 '20 at 2:36
• 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. Aug 22 '20 at 2:36
• When you dualize the exact sequence $0\to I \to R \to R/I \to 0$, how do you get $Hom(I,R)\cong R/J$ ? 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.

• The question is whether $Y$ Cohen--Macaulay implies $Z$ Cohen--Macaulay.
– abx
Aug 21 '20 at 13:08
• Right, this (very nice) example shows that Y and Z can both be non-CM, but that's not what I'm asking. Aug 21 '20 at 14:28
• 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. Aug 21 '20 at 14:59