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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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marked as duplicate by YACP, Alexander Gruber, Sasha, rschwieb, no identity Feb 24 '13 at 10:52

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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. –  Dylan Wilson Mar 29 '12 at 17:35

2 Answers 2

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. –  user26857 May 23 '12 at 20:37

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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I am using the basic fact that the global dimension of a ring is simultaneously the supremum of the projective dimensions of its modules and the sumpremum of the injective dimension of its modules. This is proved in any textbook about homological algebra. –  Mariano Suárez-Alvarez Mar 29 '12 at 17:38
    
I think I don't get your hint. The residue field has finite injective dimension and I would like to prove that also its projective dimension is finite... –  user11428 Mar 29 '12 at 17:53
    
just to make sure I understand correctly: this answers the converse of the question asked, right? –  robjohn Feb 23 '13 at 9:06

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