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I would like to show the following result: for a noetherian local ring $A$, we have $\mathrm{gl.dim}_A=\mathrm{hd}_A (k)$. Notice that the left side term of the equality is the global dimension of $A$, while the right side term is the homological dimension of the residue field $k=A/\mathfrak{m}$. I think that for any finitely generated $A$-module $M$ it should hold $\mathrm{hd}_A(M) \leq \mathrm{hd}_A(k)$. In this way the result follows. Is it true what i claimed? If yes why? Alternatively how can i show the result?

Thanks to everybody.

Mamadness
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    For a finitely generated module $M$ as in your case, homological dimension of $M$ is $n$ if and only if $Tor^m_A(M,k)=0$ for all $m>n$. This immediately implies your inequality. – Mohan Jul 03 '18 at 18:02
  • I can see the $\Rightarrow$ of your statement. What about the $\Leftarrow$? – Mamadness Jul 04 '18 at 08:39
  • I solved it in this way: let $\mathrm{hd}A(k)=m$, then $\mathbf{Tor}{m+1}^A(k,k)=0$ and so $\mathrm{gl.dim}_A \leq m=\mathrm{hd}_A(k)$. It follows by definition of global dimension the equality desired. I used the Propositions 4.17 and 4.18 of "Homological Methods in Commutative Algebra". – Mamadness Jul 04 '18 at 08:54

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