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Let $R$ be a Noetherian ring and $M$ a non-zero finite $R$-module with finite projective dimension equal to $n$. For any $P$ inside the support of $M$ we have that $\operatorname{projdim} M_P \le \operatorname{projdim} M$.

Question 1: Is it true that there exists some $P \in \operatorname{Supp} M$ such that $\operatorname{projdim} M_P = \operatorname{projdim} M$? If yes, please provide either proof or reference.

Question 2: If the answer to Question 1 is no, then can we "nicely" characterize modules $M$ having the required property described in Question 1?

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2 Answers 2

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Hint. Try to prove that $\operatorname{pd} M=\sup\{\operatorname{pd}M_P:P\in\operatorname{Supp}M\}$. (Since the property of a module to be $0$ localizes, the equality holds without any finiteness hypothesis on $R$ and $M$, only $\operatorname{pd} M<\infty$ is necessary.)

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  • $\begingroup$ Here is my proof: Let $pd M=n$ and let $0\rightarrow P_n \rightarrow \cdots \rightarrow P_0 \rightarrow M \rightarrow 0$ be a projective resolution. Suppose $sup \left\{ pd M_Q | Q \in Spec R\right\}=m<n$. Then for any $Q \in Spec R$ we get an exact sequence $0\rightarrow (P_n)_Q \rightarrow \cdots \rightarrow (P_{s+1})_Q \rightarrow 0$, where $s=pd M_Q$ and $s+1 \le m<n$. But $(P_{s+1})_Q$ is either zero or projective and so by induction $(P_n)_Q=0$ for all $Q \in Spec R$. Thus $P_n=0$ contradiction. Correct? $\endgroup$
    – Manos
    Commented Jan 8, 2014 at 15:56
  • $\begingroup$ @Manos Almost. You asked the sup for primes in Supp M, so have to consider only these, but when a prime is not in Supp then $M_P=0$ and thus the whole resolution is such. $\endgroup$
    – user119598
    Commented Jan 8, 2014 at 19:50
  • $\begingroup$ @Benja Have you noticed that $pd(M)$ is assumed finite? $\endgroup$
    – user119598
    Commented Jan 10, 2014 at 6:05
  • $\begingroup$ @Manos It seems to me that there is an issue with your proof. For one thing, it does not assume that R is Noetherian or M is f.g. Second, I don't see how the induction goes through: to start with, why do you have this exact sequence? It is true that the kernel of $(P_{s - 1})_Q \to (P_{s - 2})_Q$ is projective, but it does not need to be $(P_{s})_Q$. A similar issue exists when trying to do the inductive step $\endgroup$ Commented Dec 28, 2022 at 16:46
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In this case (i.e. $R$ Noetherian, $M$ finite), $\text{projdim}_R M = \sup \{\text{projdim}_{R_m} M_m \mid m \in \text{mSpec}(R) \}$, so if $\text{projdim} M < \infty$, then this value is achieved locally.

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