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Let $R$ be a commutative Noetherian local ring with maximal ideal $\mathfrak m$.

Is it true that the projective dimension of $R/\mathfrak m$ is finite knowing that its injective dimension is finite? If yes, why?!?

I would need this to prove something else but I'm not sure I can use it. Can you help me please?

Thanks.

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    Maybe the "global dimension theorem" will help. cf. Weibel... haven't thought about this though, but that's just where I'd look first.2012-03-29

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If $\operatorname {injdim} (k)\lt \infty$, then $R$ is regular : this is stated as exercise 3.1.26 of Bruns-Herzog's Cohen-Macaulay rings.
From this follows from Serre's theorem that $R$ has finite global dimension and then of course this implies that $k$, like all $R$-modules, has finite projective dimension.

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    No need to use Serre's Theorem: use exercise 3.1.25 from the same book.2012-05-23
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This answer answers a different question than what was asked...

The global dimension of a local ring $R$ coincides with the projective dimension of its residue field $R/\mathfrak m$. If the latter is finite, then the former is of course finite and it follows that the injective dimension of all modules is finite.

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    just to make sure I understand correctly: this answers the converse of the question asked, right?2013-02-23