Bourbaki

Example 3

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

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

Let Ω\Omega be an AA-module and dd an integer ⩾0\geqslant 0 such that Ext⁡Ai(κA,Ω)\operatorname{Ext}_A^i(\kappa_A, \Omega) is zero for i≠di \neq d and has length 1 for i=di = d. Set T(M)=Ext⁡Ad(M,Ω)T(M) = \operatorname{Ext}_A^d(M, \Omega) for every AA-module MM of finite length and T(f)=Ext⁡Ad(f,1Ω)T(f) = \operatorname{Ext}_A^d(f, 1_\Omega) for every AA-linear mapping ff between AA-modules of finite length. Then Ext⁡Ai(M,Ω)=0\operatorname{Ext}_A^i(M, \Omega) = 0 for every AA-module MM of finite length and every i≠di \neq d, and conditions FD 1) to FD 5) are satisfied: it is enough, in fact, to apply the preceding example to the case where CC is the canonical injective resolution of Ω\Omega.

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

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