Bourbaki

Example 2

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

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

Let CC be an injective complex of AA-modules and dd an integer such that Hi(Homgr⁡A(κA,C))H^i(\operatorname{Homgr}_A(\kappa_A, C)) is zero for i≠di \neq d and is of length 1 for i=di = d. For every AA-module MM of finite length, one has Hi(Homgr⁡A(M,C))=0H^i(\operatorname{Homgr}_A(M, C)) = 0 for i≠di \neq d; let us argue in fact by induction on the length of MM, supposed >0> 0; there exists an exact sequence of AA-modules 0→κA→M→N→00 \to \kappa_A \to M \to N \to 0, which gives rise to an exact sequence of complexes

0→Homgr⁡A(N,C)⟶Homgr⁡A(M,C)⟶Homgr⁡A(κA,C)⟶0 0 \to \operatorname{Homgr}_A(N, C) \longrightarrow \operatorname{Homgr}_A(M, C) \longrightarrow \operatorname{Homgr}_A(\kappa_A, C) \longrightarrow 0
and the conclusion follows from the induction hypothesis applied to NN.

Let T(M)=Hd(Homgr⁡A(M,C))T(M) = H^d(\operatorname{Homgr}_A(M, C)) for every AA-module MM of finite length, and T(f)=Hd(Homgr⁡A(f,1C))T(f) = H^d(\operatorname{Homgr}_A(f, 1_C)) for every AA-linear mapping ff between AA-modules of finite length; conditions FD 1) to FD 5) are satisfied.

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

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