TITLE: A grade $0$ module with finite projective dimension has annihilator $(0)$
QUESTION [1 upvotes]: Let $M$ be a finitely generated module of finite projective dimension
over a noetherian local ring $A$. Then if $M$ is of grade $0$, the
annihilator of $M$ is $(0)$.

A sketch proof of the above says that one takes a projective resolution of $M$. By reasoning on the ranks, one proves that $\operatorname{Supp}M=\operatorname{Spec}A$. If $I$ is the annihilator of $M$, one deduces that for every prime ideal $p\in \operatorname{Ass}A$, one has $IA_p=0$ thus $I=0$.
How to use reasoning on the ranks to show $\operatorname{Supp}M=\operatorname{Spec}A$?

REPLY [1 votes]: In fact, $M$ has a finite free resolution, so it has a rank.
(In the following I suppose that by grade of $M$ you actually mean the grade of its annihilator.) The hypothesis says that the annihilator does not contain a regular element, hence the rank of $M$ is strictly greater than zero. We have to show that in this case the annihilator is zero.
Notice that $M_{\mathfrak p}$ has also a finite free resolution, for every prime ideal $\mathfrak p$, and $\operatorname{rank}M_{\mathfrak p}=\operatorname{rank}M>0$. In particular, $M_{\mathfrak p}\ne0$. We get $\operatorname{Supp}M=\operatorname{Spec}A$. Now, let $I$ denote the annihilator of $M$, and let $\mathfrak p\in\operatorname{Ass}A$. We have $M_{\mathfrak p}\ne0$ and by Auslander-Buchsbaum we get that $M_{\mathfrak p}$ is free, so $I_{\mathfrak p}=0$. It follows that $I=0$.
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