Irreducible component of a Cohen-Macaulay variety

Question

Is it true that an irreducible component of a Cohen-Macaulay variety is also Cohen-Macaulay? If not, then in what cases does this fact hold?

Answer 1

Start with your favorite example of an affine irreducible variety $X$ that is not Cohen-Macaulay. Embed $X$ in $\mathbb A^N$, and call $c$ its codimension. Now take $c$ general polynomials that contain $X$: their intersection $Y$ has codimension $c$, and contains $X$. Then $Y$ is a complete intersection (hence Cohen-Macaulay) subvariety of $\mathbb A^N$, and contains $X$ as an irreducible component.

Answer 2

For the second question, the answer is "sometimes". It depends on how $X$ intersects the other components. Let me sketch out one approach to this question.

Suppose that $Y = X \cup Z$ where $X$ is one irreducible component and $Z$ is the union of the other irreducible components. We have the following short exact sequence: $$0 \to O_Y \to O_X \oplus O_Z \to O_{X \cap Z} \to 0.$$

Fix a point $x \in X \cap Z$, we will explore whether $X$ is Cohen-Macaulay at $x$. Taking local cohomology, we have $$\dots \to H^{i-1}_x (O_W) \to H^i_x(O_Y) \to H^i_x(O_X) \oplus H^i_x(O_Z) \to H^i_x(O_{W}) \to H^{i+1}_x(O_Y) \to \dots$$ where $W = X \cap Z$ (sorry, LaTeX was acting up when I was putting $X \cap Z$ in subscripts above).

Anyway, since $Y$ is Cohen-Macaulay, the $H^i_x(O_Y) = 0$ for $i < \dim Y = d$, we need to prove the same for $X$. For example, if $Y$ and $W$ is Cohen-Macaulay EDIT: and $W$ is of dimension one smaller than $X$, and $H^{d-1}_x(O_W) \to H^{d}_x(Y)$ injects then this follows immediately. Alternately, $Y$ is CM and if $H^i_x(O_Z) \to H^i_x(O_W)$ is surjective for all $i$, then you also get the desired result.