Bourbaki

Corollary 1

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

Commutative Algebra, chapter X, § 9. Modules dualisants

Let A be a noetherian ring, Ω\Omega a dualizing A-module, and B a finite A-algebra; suppose that the A-module B is Macaulay. The B-module ExtA(B,Ω)\mathrm{Ext}_A(B, \Omega) is dualizing.

Let Ω′\Omega' denote the BB-module ExtA(B,Ω)\mathrm{Ext}_A(B, \Omega). Let nn be a maximal ideal of BB; its inverse image in AA is a maximal ideal mm (V, §2§ 2, No. 1, prop. 1). The AmA_m-algebra Bm=Am⊗ABB_m = A_m \otimes_A B is finite, and is a Macaulay AmA_m-module; by the proposition, the BmB_m-module Ωm′\Omega'_m, which identifies with ExtAm(Bm,Ωm)\mathrm{Ext}_{A_m}(B_m, \Omega_m) (§3§ 3, No. 2, prop. 2), is dualizing. Since BnB_n is a ring of fractions of BmB_m, the BnB_n-module Ωn′\Omega'_n is dualizing, whence the corollary.

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

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