Bourbaki

Proposition 2

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

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

Let A be a local Macaulay ring, of dimension d, and let Ω\Omega be a dualizing A-module. The homomorphism ω:HAd(A)→D(Ω)\omega : H_A^d(A) \to D(\Omega) is bijective.

The homomorphism ω\omega is the limit of the inductive system of mappings (ωα)α∈Dcs(\omega_\alpha)_{\alpha \in \mathscr{D}_{cs}}, where

ωα:ExtAd(A/α,A)⟶HomA(Ω,ExtAd(A/α,Ω)) \omega_\alpha : \mathrm{Ext}_A^d(A/\alpha, A) \longrightarrow \mathrm{Hom}_A(\Omega, \mathrm{Ext}_A^d(A/\alpha, \Omega))

associates to an element uu of ExtAd(A/α,A)\mathrm{Ext}_A^d(A/\alpha, A) the mapping x↦fx∘ux \mapsto f_x \circ u (A, X, p. 114). It is therefore enough to prove that each of the mappings ωα\omega_\alpha is bijective.

Let α\alpha be an ideal of Dcs\mathscr{D}_{cs}, generated by a completely secant sequence x=(x1,…,xd)x = (x_1, \ldots, x_d) for AA. The Koszul complex K∙(x,A)K^\bullet(x, A) furnishes a projective resolution of A/αA/\alpha; for every AA-module MM, the AA-module Hd(HomgrA(K∙(x,A),M))H^d(\mathrm{Homgr}_A(K^\bullet(x, A), M)) identifies canonically with M/αMM/\alpha M (A, X, p. 155). From this one deduces an isomorphism (A, X, p. 100)

φM:M/αM⟶ExtAd(A/α,M). \varphi_M : M/\alpha M \longrightarrow \mathrm{Ext}_A^d(A/\alpha, M) .
Let x∈Ωx \in \Omega. Taking account of loc. cit., p. 103, prop. 2, one has a commutative diagram

A/α→φAExtAd(A/α,A)↓fx↓Ext(1A/α,fx)Ω/αΩ→φΩExtAd(A/α,Ω) \begin{array}{ccc} A/\alpha & \xrightarrow{\varphi_A} & \mathrm{Ext}_A^d(A/\alpha, A) \\ \downarrow f_x & & \downarrow \mathrm{Ext}(1_{A/\alpha}, f_x) \\ \Omega/\alpha \Omega & \xrightarrow{\varphi_\Omega} & \mathrm{Ext}_A^d(A/\alpha, \Omega) \end{array}

where fxf_x is the homomorphism deduced from fxf_x by passing to quotients. It follows that if for every AA-module MM one identifies ExtAd(A/α,M)\mathrm{Ext}_A^d(A/\alpha, M) with M/αMM/\alpha M by means of φM\varphi_M, the homomorphism ωα\omega_\alpha is identified with the AA-linear mapping of A/αA/\alpha into HomA(Ω,Ω/αΩ)\mathrm{Hom}_A(\Omega, \Omega/\alpha \Omega) which sends 1 to the canonical surjection, that is again with the canonical mapping A/α⟶EndA/α(Ω/αΩ)A/\alpha \longrightarrow \mathrm{End}_{A/\alpha}(\Omega/\alpha \Omega). But since the A/αA/\alpha-module Ω/αΩ\Omega/\alpha \Omega is dualizing (§ 9, No. 2, prop. 4), this mapping is bijective (§ 9, No. 4, prop. 6), which proves the proposition.

Let us identify the Matlis bidual D(D(Ω))D(D(\Omega)) with Ω^\widehat{\Omega} by the isomorphism α^Ω\widehat{\alpha}_\Omega (§ 8, No. 3, th. 2, b)).

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

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