2
$\begingroup$

I am looking for an example of a commutative noetherian local ring $(A,m)$, and a maximal Cohen-Macaulay module $M$ over $A$ (in particular $M$ is finitely generated over $A$), such that for some $p \in Spec(A)$, we have that $M_p = 0$, and $A_p$ is also not a Cohen-Macaulay ring. Do there exist such examples?

$\endgroup$
1
  • 1
    $\begingroup$ Let $A$ be the local ring of a union of two $2$-planes intersecting in one point $x$ in a $4$-dimensional ambient space. Let $M$ be the local ring of one of the two irreducible components. Let $p$ be a prime ideal of a curve in the other component that contains $x$. If you really want $A$ to be not Cohen-Macaulay at $p$, then just add some embedded structure to $A$ along $p$. $\endgroup$ Nov 29 '18 at 14:16
3
$\begingroup$

Here's a concrete example, inspired by Jason Starr's comment but of smaller dimension.

Let $A=k[x,y,z]_{(x,y,z)} / (x^2z, xyz)$. Let $p=(x,y)A$. Let $M=A/zA$, thought of as a cyclic $A$-module. Then $M \cong k[x,y]_{(x,y)}$ is a maximal Cohen-Macaulay module over $A$, since $\dim A = \operatorname{depth} M = 2$. We have that $p$ is prime because $A/p \cong k[z]_{(z)}$. We have that $M_p = 0$ because $z \notin p$, so $z$ acts like a unit on $A_p$-modules but kills $M$. Finally, $A_p \cong k(z)[x,y]_{(x,y)} / (x^2, xy)$ has dimension 1 and depth 0.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.