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I have tried to formulate a question in which I was very curious, any hints suggestions are also welcomed. Thanks in advance.

Let $M$ be an $R$ module ($R$ commutative ring with unity). It is given that for every maximal ideal $m$ of $R$, $M_m$ has a finite projective resolution of fintely generated (f.g.) projective $R_m$ modules. Does this imply that $M$ also has a finite projective resolution of f.g. projective $R$ modules?

What I tried was to produce a projective module globally from localized rings. Let $P$ be a f.g. projective $S^{-1}R$ module there exists $S^{-1}R$ module $Q$ such that $P \oplus Q \cong S^{-1}(R^n)$. Now there always exists $P'$ and $Q'$ such that $S^{-1}P' \cong P$ and $S^{-1}Q' \cong Q$ (without taking either of them to be projective at this stage). Thus $S^{-1}(P'\oplus Q') \cong S^{-1}(R^n)$.

But now should this imply that $P' \oplus Q' \cong R^n$ ? Here I am stuck. I would be delighted if you could help me out in this problem.

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  • $\begingroup$ If it had a finite resolution, then all local resolutions will have length bounded by this. So a necessary (and sufficient) condition is that all local resolutions have length bounded by a number independent of the localization. It is possible to construct a counterexample if this is violated. $\endgroup$
    – Mohan
    Commented Nov 2, 2021 at 16:55

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