Let $(R,\mathfrak m)$ be a local Cohen-Macaulay ring of positive dimension. Let $M$ be a finitely generated maximal Cohen-Macaulay module i.e. $\operatorname{depth} M=\dim R$. If $\operatorname{Supp}(M)=\operatorname{Spec}(R)$, then is it true that $M$ is faithful?

My thoughts: Since $\operatorname{Supp}(M)=\operatorname{Spec}(R)$, so the radical of the annihilator ann$_R(M)$ is exactly the nilradical of $R$, so every element in the annihilator of $M$ is nilpotent. We would be done if $R$ were reduced, but that's not necessarily the case.

Please help.


1 Answer 1


Let $R=K[X,Y]/(X^2Y)$ and $M=R/(xy)$. Then $M$ is MCM, $\operatorname{Ann}_R(M)=(xy)$, and $\operatorname{Supp}(M)=\operatorname{Spec}(R)$ since every prime ideal of $R$ contains $x$ or $y$.

  • $\begingroup$ I just noticed that you want $R$ local. Then localize at $(x,y)$ or replace $K[X,Y]$ by $K[[X,Y]]$. $\endgroup$
    – user26857
    Aug 29, 2021 at 20:22
  • 1
    $\begingroup$ As a note, one can also take any proper cyclic module over an Artinian local ring; examples abound. $\endgroup$ Aug 29, 2021 at 20:29

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