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Let $A$ be a local CM ring, and $M$ a maximal CM $A$-module. Is it true that $\operatorname{Supp}M=\operatorname{Spec}A$ ? This suspicion stems from such statements as:

  • If $\omega$ is a canonical module of $A$, then $\omega_{\mathfrak{p}}$ is a canonical module of $A_{\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$.
  • If $\omega$ is a canonical module of $A$, then $\mu_i(\mathfrak{p},\omega)=\delta_{i}^{\operatorname{ht}\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$, where $\mu$ denotes the Bass number.

And am I correct in understanding that maximal CM module is by definition nonzero?

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    $\begingroup$ As Graham's example shows, it's not true in general, but sometimes it's true. For example if $A$ is a domain and $M$ is finitely generated, then it's true, because if $M$ is maximal CM, then $\mathrm{dim}\ M = \mathrm{depth}\ M = \mathrm{dim}\ A$. If $I$ is the annihilator of $M$, then it follows $\mathrm{dim}\ A/I=\mathrm{dim}\ A$. Then since $A$ is a domain, $I=(0)$, and since $M$ is finitely generated, $\mathrm{Supp}(M)=V(I)=\mathrm{Spec}\:A$. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Jan 1, 2012 at 17:46

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No. $R = k[x,y]/(xy)$, $M = R/(x)$.

The zero module has infinite depth and support of dimension $-\infty$, so should not be considered MCM.

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  • $\begingroup$ Thanks! So if the statements about canonical module $\omega$ are true, then the reason why $\operatorname{Supp}\omega=\operatorname{Spec}A$ does not come from MCM property... Then I guess it comes from type=1 condition? $\endgroup$
    – ashpool
    Commented Jan 1, 2012 at 17:57
  • $\begingroup$ I think Graham's example is a good counter-example to the common phrase "MCM modules localize." $\endgroup$
    – ashpool
    Commented Jan 1, 2012 at 18:20
  • $\begingroup$ I guess the question is, then, Why are canonical modules supported everywhere? $\endgroup$
    – ashpool
    Commented Jan 1, 2012 at 19:14
  • $\begingroup$ Because they are faithful: $\operatorname{End}(\omega)=R$ implies that they have no annihilator. $\endgroup$
    – Graham Leuschke
    Commented Jan 1, 2012 at 21:05

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