# Why are canonical modules supported everywhere?

Let $A$ be a local CM ring, and $\omega$ a canonical module of $A$. Here are two properties of $\omega$ from Bruns & Herzog:

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

These properties seem to imply that $\operatorname{Supp}\omega=\operatorname{Spec}A$. As Graham Leuschke pointed out, this is not a property of maximal CM modules. Why, then, are canonical modules supported everywhere?

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See (1.7) on page 87 of Some basic results on canonical modules. For a local CM ring condition (b) there holds.

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I just realized that $\operatorname{Supp}\omega=\operatorname{Spec}A$ can be deduced from the isomorphism $\operatorname{Hom}(\omega,\omega)\simeq A$! –  ashpool Jan 1 '12 at 23:09
In general, $\text{Hom}(\omega, \omega)$ is the S2-ification of $A$ if I recall correctly (see the linked paper of Aoyama, or a different paper by the same author). –  Karl Schwede Jan 1 '12 at 23:27
@ Mahdi Majidi-Zolbanin $\operatorname{Ass}\omega=\operatorname{Ass}A$ from (1.7) also follows from the isomorphism $\operatorname{Hom}(\omega,\omega)=A$. Thanks for the reference! –  ashpool Jan 1 '12 at 23:38
$$R = k[[x,y,z]]/\langle x \rangle \cap \langle y, z \rangle$$
is only be supported at one of the minimal primes of $R$. The dualizing complex behaves better though.