Bourbaki

Exercise 7, § 7

Algebra, X, p. 199

Bourbaki marks this one as harder than the rest.

Soient A′A' a kk-algebra (associative and unital) and B=A⊗kA′B = A \otimes_k A'. a) Let M be a left A-module, Q a right A-module, M′M' a left A′A'-module, Q’ a right A′A'-module. Let (P,p)(P, p) (resp. (P′,p′)(P', p'), resp. (R,r)(R, r)) be a projective resolution of the A-module M (resp. of the A′A'-module M′M', resp. of the B-module M⊗kM′M \otimes_k M'). Prove that there exists a morphism of B-complexes

u:P⊗kP′→R, u : P \otimes_k P' \to R ,

unique up to homotopy, such that r∘u=p⊗p′r\circ u=p\otimes p'. Define using uu a kk-homomorphism graded of degree zero

T:Tor⁡A(Q,M)⊗kTor⁡A′(Q′,M′)⟶Tor⁡B(Q⊗kQ′,M⊗kM′). \mathrm{T}:\operatorname{Tor}^{A}(Q,M)\otimes_k\operatorname{Tor}^{A'}(Q',M')\longrightarrow\operatorname{Tor}^{B}(Q\otimes_k Q',M\otimes_k M').

Prove that T\mathrm{T} does not depend on the choice of the resolutions P,P′,RP,P',R and of the morphism uu. If a∈Tor⁡A(Q,M)a\in\operatorname{Tor}^{A}(Q,M) and a′∈Tor⁡A′(Q′,M′)a'\in\operatorname{Tor}^{A'}(Q',M'), put T(a⊗a′)=aTa′\mathrm{T}(a\otimes a')=a\mathrm{T}a'.

b) Suppose in addition that the kk-modules A and A’ are projective and that one has Tor⁡nk(M,M′)=0\operatorname{Tor}^{k}_{n}(M,M')=0 for n>0n>0. Prove that P⊗kP′P\otimes_k P' is then a projective resolution of the B-module M⊗kM′M\otimes_k M'; deduce, for every A-module N and every A’-module N’, a kk-homomorphism graded of degree zero

∨:Ext⁡A(M,N)⊗kExt⁡A′(M′,N′)⟶Ext⁡B(M⊗kM′,N⊗kN′). \vee:\operatorname{Ext}_{A}(M,N)\otimes_k\operatorname{Ext}_{A'}(M',N')\longrightarrow\operatorname{Ext}_{B}(M\otimes_k M',N\otimes_k N').

Prove that ∨\vee is independent of the choice of the resolutions PP and P′P'. If b∈Ext⁡A(M,N)b\in\operatorname{Ext}_{A}(M,N) and b′∈Ext⁡A′(M′,N′)b'\in\operatorname{Ext}_{A'}(M',N'), we note ∨(b⊗b′)=b∨b′\vee(b\otimes b')=b\vee b'.

c) Let A’’ be a third kk-algebra (associative and unital), M’’ and N’’ left A’’-modules, Q’’ a right A’’-module. We identify the kk-algebras A⊗k(A′⊗kA′′)A\otimes_k(A'\otimes_k A'') and (A⊗kA′)⊗kA′′(A\otimes_k A')\otimes_k A'', as well as the kk-modules M⊗k(M′⊗kM′′)M\otimes_k(M'\otimes_k M'') and (M⊗kM′)⊗kM′′(M\otimes_k M')\otimes_k M'', N⊗k(N′⊗kN′′)N\otimes_k(N'\otimes_k N'') and (N⊗kN′)⊗kN′′(N\otimes_k N')\otimes_k N'', Q⊗k(Q′⊗kQ′′)Q\otimes_k(Q'\otimes_k Q'') and (Q⊗kQ′)⊗kQ′′(Q\otimes_k Q')\otimes_k Q''. Prove the formulas

aT(a′Ta′′)=(aTa′)Ta′′pour a∈Tor⁡A(Q,M),a′∈Tor⁡A′(Q′,M′), a\mathrm{T}(a'\mathrm{T}a'')=(a\mathrm{T}a')\mathrm{T}a'' \qquad\text{pour }a\in\operatorname{Tor}^{A}(Q,M),\quad a'\in\operatorname{Tor}^{A'}(Q',M'),
a′′∈Tor⁡A′′(Q′′,M′′) a''\in\operatorname{Tor}^{A''}(Q'',M'')
b∨(b′∨b′′)=(b∨b′)∨b′′pour b∈Ext⁡A(M,N),b′∈Ext⁡A′(M′,N′), b\vee(b'\vee b'')=(b\vee b')\vee b'' \qquad\text{pour }b\in\operatorname{Ext}_{A}(M,N),\quad b'\in\operatorname{Ext}_{A'}(M',N'),
b′′∈Ext⁡A′′(M′′,N′′). b''\in\operatorname{Ext}_{A''}(M'',N'').

d) Let

σ:Tor⁡A⊗kA′(Q⊗kQ′,M⊗kM′)⟶Tor⁡A′⊗kA(Q′⊗kQ,M′⊗kM) \sigma:\operatorname{Tor}^{A\otimes_k A'}(Q\otimes_k Q',M\otimes_k M') \longrightarrow \operatorname{Tor}^{A'\otimes_k A}(Q'\otimes_k Q,M'\otimes_k M)

and

τ:Ext⁡A⊗kA′(M⊗kM′,N⊗kN′)⟶Ext⁡A′⊗kA(M′⊗kM,N′⊗kN) \tau:\operatorname{Ext}_{A\otimes_k A'}(M\otimes_k M',N\otimes_k N') \longrightarrow \operatorname{Ext}_{A'\otimes_k A}(M'\otimes_k M,N'\otimes_k N)

the isomorphisms deduced from the commutation isomorphisms

A⊗kA′⟶A′⊗kA,M⊗kM′⟶M′⊗kM,N⊗kN′⟶N′⊗kN,Q⊗kQ′⟶Q′⊗kQ. A\otimes_k A'\longrightarrow A'\otimes_k A,\quad M\otimes_k M'\longrightarrow M'\otimes_k M,\quad N\otimes_k N'\longrightarrow N'\otimes_k N,\quad Q\otimes_k Q'\longrightarrow Q'\otimes_k Q.

Prove the formulas

σ(aTa′)=(−1)pqa′Tapour a∈Tor⁡pA(Q,M),a′∈Tor⁡qA′(Q′,M′) \sigma(a\mathrm{T}a')=(-1)^{pq}a'\mathrm{T}a \qquad\text{pour }a\in\operatorname{Tor}^{A}_{p}(Q,M),\quad a'\in\operatorname{Tor}^{A'}_{q}(Q',M')
τ(b∨b′)=(−1)pqb′∨bpour b∈Ext⁡Ap(M,N),b′∈Ext⁡A′q(M′,N′). \tau(b\vee b')=(-1)^{pq}b'\vee b \qquad\text{pour }b\in\operatorname{Ext}^{p}_{A}(M,N),\quad b'\in\operatorname{Ext}^{q}_{A'}(M',N').

e) Let (E):0→M2→M1→M0→0(\mathscr{E}):0\to M_2\to M_1\to M_0\to0 be an exact sequence of A-modules, such that the sequence

(E⊗M′):0→M2⊗kM′→M1⊗kM′→M0⊗kM′→0 (\mathscr{E}\otimes M'):0\to M_2\otimes_k M'\to M_1\otimes_k M'\to M_0\otimes_k M'\to0

be exact. We denote δ\delta (resp. Δ\Delta) the connecting homomorphism ∂(Q,E)\partial(Q,\mathscr{E}) (resp. ∂(Q⊗kQ′,E⊗M′)\partial(Q\otimes_k Q',\mathscr{E}\otimes M'). Now prove that one has (δa)Ta′=Δ(aTa′)(\delta a)\mathrm{T}a'=\Delta(a\mathrm{T}a') for a∈Tor⁡A(Q,M0)a\in\operatorname{Tor}^{A}(Q,M_0), a′∈Tor⁡A′(Q′,M′)a'\in\operatorname{Tor}^{A'}(Q',M'). Exercise prove the analogous formulas for the product ∨\vee.

f) If kk is a field, show that T\mathrm{T} is an isomorphism of graded kk-modules; the same holds for ∨\vee if, in addition, A and A’ are left noetherian and if the A-module M and the A’-module M’ are of finite type.

g) If the modules Tor⁡nA(Q,M)\operatorname{Tor}^{A}_{n}(Q,M) are identified with the sets Tn(Q,M)T_n(Q,M) defined in exercise 6, prove the formula

(P,p,q)T(P′,p′,q′)=(P⊗kP′,p⊗p′,q⊗q′). (P,p,q)\mathrm{T}(P',p',q')=(P\otimes_k P',p\otimes p',q\otimes q').
Solution

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Machine translation of the English, by gpt-5-6-mini, 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.