Bourbaki

Exercise 2, § 6

Algebra, X, p. 189

Let A,BA, B be two kk-algebras (associative and with unit); let MM be a left AA-module, PP a right BB-module, NN an (A,B)(A, B)-bimodule. Consider the homomorphism

σ:Hom⁡B(N,P)⊗AM→Hom⁡B(Hom⁡A(M,N),P) \sigma : \operatorname{Hom}_B(N, P) \otimes_A M \to \operatorname{Hom}_B(\operatorname{Hom}_A(M, N), P)
such that σ(u⊗m)(v)=u∘v(m)\sigma(u \otimes m)(v) = u \circ v(m) for u∈Hom⁡B(N,P),v∈Hom⁡A(M,N),m∈Mu \in \operatorname{Hom}_B(N, P), v \in \operatorname{Hom}_A(M, N), m \in M (II, p. 190, exercise 6). (a) Replacing MM by a projective resolution, define with the aid of σ\sigma a kk-homomorphism
τ:Tor⁡A(Hom⁡B(N,P),M)→Homgr⁡B(Ext⁡A(M,N),P). \tau : \operatorname{Tor}^A(\operatorname{Hom}_B(N, P), M) \to \operatorname{Homgr}_B(\operatorname{Ext}_A(M, N), P) .
(b) Suppose henceforth that AA is left noetherian and that the AA-module MM is of finite type. Now prove that if the BB-module PP is injective, τ\tau is an isomorphism. (c) Now prove that there exist two spectral sequences EE and ′E'E (X, pp. 175-177, exercises 14 to 17), converging to the same graded kk-module, such that
E2pq=Tor⁡pA(Ext⁡Bq(N,P),M)′E2pq=Ext⁡Bp(Ext⁡A−q(M,N),P). E_2^{pq} = \operatorname{Tor}_p^A(\operatorname{Ext}_B^q(N, P), M) \qquad 'E_2^{pq} = \operatorname{Ext}_B^p(\operatorname{Ext}_A^{-q}(M, N), P) .

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.