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?
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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 $p$. $\endgroup$– Jason StarrCommented Nov 29, 2018 at 14:16
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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.
answered Nov 29, 2018 at 22:14