Bourbaki

Exercise 6, § 9

Commutative Algebra, X, p. 176

Bourbaki marks this one as harder than the rest.

Let AA be a noetherian local ring, admitting a dualizing module Ω\Omega; let MM be a finitely generated AA-module, of finite projective dimension.

a) Proving that one has ToriA(Ω,M)=0\mathrm{Tor}_i^A(\Omega, M) = 0 for i>0i > 0. (Reason by induction on dim⁡(A)\dim(A), by considering an exact sequence 0→N→L→M→00 \to N \to L \to M \to 0 where LL is a finitely generated free module. If xx is a cancellable element of mAm_A, deduce from the induction hypothesis that ToriA(Ω,N)\mathrm{Tor}_i^A(\Omega, N) is zero for i>0i > 0, which entails ToriA(Ω,M)=0\mathrm{Tor}_i^A(\Omega, M) = 0 for i>1i > 1, and that the homothety of ratio xx is injective in Ω⊗AN\Omega \otimes_A N, which entails that every prime ideal pp associated with Tor1A(Ω,M)\mathrm{Tor}_1^A(\Omega, M) is a minimal ideal of Spec(A)\mathrm{Spec}(A); observe that at such a pp the ApA_p-module MpM_p is free).

b) Proving that the AA-module Ω⊗AM\Omega \otimes_A M has finite injective dimension.

c) Proving that the mapping θ:M→HomA(Ω,Ω⊗AM)\theta : M \to \mathrm{Hom}_A(\Omega, \Omega \otimes_A M) defined by θ(m)(w)=w⊗m\theta(m)(w) = w \otimes m for m∈M,w∈Ωm \in M, w \in \Omega is an isomorphism (reduce, with the aid of a projective resolution, to the case where MM is free).

d) The mappings [T]↦[HomA(Ω,T)][T] \mapsto [\mathrm{Hom}_A(\Omega, T)] and [M]↦[Ω⊗AM][M] \mapsto [\Omega \otimes_A M] define reciprocal bijections of one another between the set of isomorphism classes of finitely generated AA-modules and of finite injective dimension and the set of isomorphism classes of finitely generated AA-module and of finite projective dimension, \* and quasi-inverse equivalences of categories between the corresponding categories *.

EXERCISES

AC X.177

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.