# Projective dimension of the residue field of a noetherian local ring. [duplicate]

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 identityFeb 24 '13 at 10:52

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

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

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.