Bourbaki

Remark 2

Tag 0E04, permanent. Cite it as [Bourbaki, Tag 0E04].

Algebra, chapter X, § 7. Produit de composition

One can also use the resolutions l(P)l(P) and l(Q)l(Q).

Analogously, if NN is a second left AA-module, one defines a composition product (μ,γ)↦μ∘γ(\mu, \gamma) \mapsto \mu \circ \gamma denoted

cP:M,N:ExtAr(M,N)×TormA(P,M)→Torm−rA(P,N) c_{P:M,N}: \mathrm{Ext}_A^r(M, N) \times \mathrm{Tor}_m^A(P, M) \to \mathrm{Tor}_{m-r}^A(P, N)

by the commutative diagram

ExtAr(M,N)×TormA(P,M)→cP:M,NTorm−rA(P,N)1×σP,M,r↓↓σP,N,m−rExtAr(M∘,N∘)×TormA∘(M∘,P∘)→cM∘,N∘,P∘Torm−rA∘(N∘,P∘) \begin{array}{ccc} \mathrm{Ext}_A^r(M, N) \times \mathrm{Tor}_m^A(P, M) & \xrightarrow{c_{P:M,N}} & \mathrm{Tor}_{m-r}^A(P, N) \\ 1 \times \sigma_{P,M,r} \downarrow & & \downarrow \sigma_{P,N,m-r} \\ \mathrm{Ext}_A^r(M^\circ, N^\circ) \times \mathrm{Tor}_m^{A^\circ}(M^\circ, P^\circ) & \xrightarrow{c_{M^\circ,N^\circ,P^\circ}} & \mathrm{Tor}_{m-r}^{A^\circ}(N^\circ, P^\circ) \end{array}

where σ\sigma denotes the commutation isomorphisms (X, p. 71).

If μ∈ExtAr(M,N)\mu \in \mathrm{Ext}_A^r(M, N) is the class of the morphism g:L(M)→L(N)(−r)g : L(M) \to L(N) (-r), and if γ∈TormA(P,M)\gamma \in \mathrm{Tor}_m^A(P, M) is the class of the cycle z=∑zijz = \sum z_{ij}, where zij∈Li(P)⊗ALj(M)z_{ij} \in L_i(P) \otimes_A L_j(M), then μ∘γ\mu \circ \gamma is therefore the class of the cycle ∑(−1)ir(1⊗g)(zij)\sum (-1)^{ir} (1 \otimes g)(z_{ij}).

One can also represent γ\gamma by a cycle y∈P⊗Lm(M)y \in P \otimes L_m(M), and μ∘γ\mu \circ \gamma is the class of the cycle (1⊗g)(y)∈P⊗Lm−r(M)(1 \otimes g)(y) \in P \otimes L_{m-r}(M).

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

Cited by

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Cites

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Translated with machine help, not yet reviewed by a person.