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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 $\operatorname{Supp}M=\operatorname{Spec}A$?

user26857
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Flying pencil
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1 Answers1

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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 $I=0$.

user26857
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  • Why is the rank of $M$ constant? How is this implied by the annihilator of $M$? – Flying pencil Jan 17 '22 at 22:11
  • Do you know what rank is? – user26857 Jan 18 '22 at 06:06
  • I only know the rank of a module $M$ defined locally for a prime $p$ as the rank of $M_p/pM_p$. I did read your answer, I'm not sure how the nonexistence of a regular element implies the rank is strictly greater than $0$. – Flying pencil Jan 18 '22 at 14:32
  • This is not the rank I and the book or article you read are using. Read in Bruns and Herzog what rank means in this case – user26857 Jan 18 '22 at 14:37
  • The proposition in Bruns and Herzog gives that for any $p\in \mathrm{Ass} A$, $M_p$ is nonzero free. But I'm not sure how to relate $\mathrm{Ass} M$ and $\mathrm{Ass} A$. I see that they have the same minimal associated primes, but is it true that $\mathrm{Ass} M=\mathrm{Ass} A$? – Flying pencil Jan 31 '22 at 03:31
  • It was an obvious typo in my answer. Fixed it. – user26857 Jan 31 '22 at 03:53