Bourbaki

Remark 2

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

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

For every ideal of definition α\alpha of AA, the AA-module ExtA(A/α,M)\mathrm{Ext}_A(A/\alpha, M) is annihilated by α\alpha, and A/αA/\alpha identifies with A^/αA^\widehat{A}/\alpha \widehat{A}; consequently, the graded AA-module ExtA(A/α,M)\mathrm{Ext}_A(A/\alpha, M) identifies with A^⊗AExtA(A/α,M)\widehat{A} \otimes_A \mathrm{Ext}_A(A/\alpha, M), hence also with ExtA^(A^/αA^,A^⊗AM)\mathrm{Ext}_{\widehat{A}}(\widehat{A}/\alpha \widehat{A}, \widehat{A} \otimes_A M) (A, X, p. 111, prop. 10). The set of ideals αA^\alpha \widehat{A}, for α∈D\alpha \in \mathscr{D}, contains the powers of mA^m_{\widehat{A}}, hence is cofinal in the set of ideals of definition of A^\widehat{A}; we therefore deduce from the foregoing a canonical isomorphism of graded AA-modules

HA(M)⟶HA^(A^⊗AM) H_A(M) \longrightarrow H_{\widehat{A}}(\widehat{A} \otimes_A M)

If the AA-module MM is of finite type, the AA-module A^⊗AM\widehat{A} \otimes_A M identifies with the completion M^\widehat{M} of MM (III, § 3, No. 4, th. 3), and we have an isomorphism HA(M)→HA^(M^)H_A(M) \to H_{\widehat{A}}(\widehat{M}), graded of degree 0.

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

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