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Let $R$ be a Noetherian regular integral domain of Krull dimension $n$. Let $M$ be a finite torsion-free $R$-module. Is this true that $M$ has projective dimension $<n$ ?

This would be a generalization of the fact that finite torsion-free modules over Dedekind rings are projective.

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Yes. If $\operatorname{pd}_A(M)=n $, there exists a maximal ideal $\mathfrak{m}$ of $A$ such that $\operatorname{pd}_{A_{\mathfrak{m}}}(M_{\mathfrak{m}})=n $. By the Auslander-Buchsbaum theorem, this implies $\operatorname{depth}(M_{\mathfrak{m}})=0 $, which means that $M_{\mathfrak{m}}$, hence also $M$, contain a nonzero element annihilated by $\mathfrak{m}$.

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    $\begingroup$ I think this can be done more elementarily if you use that the global dimension of R is n: we can embed M into a finite free R-module F (first embed into Frac(R)^n and then clear denominators) and resolve the quotient F/M. $\endgroup$
    – Anonymous
    Commented Sep 13, 2023 at 17:07

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