Bourbaki

Example 3

Tag 0DCP, permanent. Cite it as [Bourbaki, Tag 0DCP].

Commutative Algebra, chapter X, § 10. Cohomologie locale, dualité de Grothendieck

Let Ω\Omega be a dualizing A-module. For i≠dim⁡(A)i \neq \dim(A), one has ExtAi(A/α,Ω)=0\mathrm{Ext}_A^i(A/\alpha, \Omega) = 0 for every ideal of definition α\alpha of A (§ 8, No. 5, example 3), whence HAi(Ω)=0H_A^i(\Omega) = 0; for i=dim⁡(A)i = \dim(A), the A-module HAi(Ω)H_A^i(\Omega), which is isomorphic to lim←⁡ExtAi(A/mAn,Ω)\varprojlim \mathrm{Ext}_A^i(A/\mathfrak{m}_A^n, \Omega), is a Matlis module (loc. cit., example 6).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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