Bourbaki

Example 6

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

Commutative Algebra, chapter X, § 8. Dualité des modules de longueur finie

Let us take up again the hypotheses and notations of Example 3. One obtains that I=lim→⁡ExtAd(A/mAn,Ω)I = \varinjlim \mathrm{Ext}_A^d(A/\mathfrak{m}_A^n, \Omega) is a Matlis AA-module. For every AA-module of finite length MM, there is a canonical AA-isomorphism

0M:ExtAd(M,Ω)⟶HomA(M,I); 0_M : \mathrm{Ext}_A^d(M, \Omega) \longrightarrow \mathrm{Hom}_A(M, I);
moreover this isomorphism is EndA(M)\mathrm{End}_A(M)-linear (Theorem 3, c)).

In particular, if the ring AA is Gorenstein of dimension dd, the AA-module lim→⁡ExtAd(A/mAn,A)\varinjlim \mathrm{Ext}_A^d(A/\mathfrak{m}_A^n, A) is a Matlis module.

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

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