$(R,m)$ is a local Noetherian ring. $M$ is a finite $R$-module.
Here, using dualizing complex, Karl Schwede says that if $R=S/I$ where $S$ is regular of dimension $d$, then we have: "$H^i_m(M)$ is finitely generated iff the support of $Ext^{d-i}_S(M, S)$ has dimension zero".

Question: can one prove this, Not using dualizing complex

thank you

user 1
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1 Answers1


If $S$ is Gorenstein, $R=S/I$, and $M$ a finitely generated $R$-module, then $$H^i_{\mathfrak m}(M)\simeq\operatorname{Hom}_R(\operatorname{Ext}^{d-i}_S(M, S),E_R(R/\mathfrak m)).$$ (This is the Local Duality Theorem; see Brodmann and Sharp, Local Cohomology, Theorem 11.2.6.)
Then use the Matlis duality; see Bruns and Herzog, Cohen-Macaulay Rings, Theorem 3.2.13.

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  • Bruns and Herzog, Theorem 3.2.13 assume that R is complete local ring. in this case R is homomorphic image of regular ring. +1 – user 1 Jan 19 '15 at 17:14
  • No need to worry about this. One can take the completions of all objects we are working with and nothing is lost. – user26857 Jan 19 '15 at 19:02