Bourbaki

Proposition 3

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

Algebra, chapter X, § 7. Produit de composition

Let

(6)

0→P→Sm→Sm−1→…→S1→λN→0 0 \to P \to S_m \to S_{m-1} \to \ldots \to S_1 \xrightarrow{\lambda} N \to 0
(7)
0→N↠Rn→Rn−1→…→R1→M→0 0 \to N \twoheadrightarrow R_n \to R_{n-1} \to \ldots \to R_1 \to M \to 0
be two exact sequences of left A-modules with respective classes θ ∈ Ext^m_A(N, P) and θ' ∈ Ext^n_A(M, N). The class in Ext^{m+n}_A(M, P) associated with the exact sequence (8)
0→P→Sm→…→S1→μ∘λRn→…→R1→M→0 0 \to P \to S_m \to \ldots \to S_1 \xrightarrow{\mu \circ \lambda} R_n \to \ldots \to R_1 \to M \to 0
is the composition product θ ∘ θ'.

Choose commutative diagrams

0⟶P⟶I0(P)⟶…⟶Im−1(P)→δPm−1Im(P)lP↑w0↑wm−1↑wm↑0⟶P⟶Sm⟶…⟶S1→λN⟶0, \begin{array}{cccccccccccc} 0 & \longrightarrow & P & \longrightarrow & I^0(P) & \longrightarrow & \ldots \longrightarrow & I^{m-1}(P) & \xrightarrow{\delta_P^{m-1}} & I^m(P) \\ & & ^{l_P}\uparrow & & ^{w^0}\uparrow & & & ^{w^{m-1}}\uparrow & & ^{w^m}\uparrow \\ 0 & \longrightarrow & P & \longrightarrow & S_m & \longrightarrow & \ldots \longrightarrow & S_1 & \xrightarrow{\lambda} & N & \longrightarrow & 0 , \end{array}
0⟶N→eNI0(N)⟶…⟶In−1(N)⟶In(N)1N↑v0↑vn−1↑vn↑0⟶N→μRn⟶…⟶R1⟶M⟶0. \begin{array}{cccccccccccc} 0 & \longrightarrow & N & \xrightarrow{e_N} & I^0(N) & \longrightarrow & \ldots \longrightarrow & I^{n-1}(N) & \longrightarrow & I^n(N) \\ & & ^{1_N}\uparrow & & ^{v^0}\uparrow & & & ^{v^{n-1}}\uparrow & & ^{v^n}\uparrow \\ 0 & \longrightarrow & N & \xrightarrow{\mu} & R_n & \longrightarrow & \ldots \longrightarrow & R_1 & \longrightarrow & M & \longrightarrow & 0 . \end{array}

Since Im(P)I^m(P) is injective, there exists a homomorphism h0 ⁣:I0(N)→Im(P)h^0 \colon I^0(N) \rightarrow I^m(P) such that wm=h0∘eNw^m = h^0 \circ e_N; according to X, p. 49, Proposition 3 bis, h0h^0 extends to a morphism of complexes h ⁣:I(N)→I(P) (−m)h \colon I(N) \rightarrow I(P)\, (-m). Thus wm=h0∘eN=h0∘v0∘μw^m = h^0 \circ e_N = h^0 \circ v^0 \circ \mu, hence

δIm−1∘wm−1=wm∘λ=h0∘v0∘(μ∘λ), \delta_I^{m-1} \circ w^{m-1} = w^m \circ \lambda = h^0 \circ v^0 \circ (\mu \circ \lambda),

and the following diagram is commutative:

0⟶P⟶I0(P)⟶⋯⟶Im−1(P)⟶Im(P)⟶Im+1(P)⟶⋯⟶Im+n(P)↑iP↑w0↑wm−1↑t0↑t1↑tn0⟶P⟶Sm⟶⋯⟶S1→μ∘λRn⟶Rn−1⟶⋯⟶M⟶0 \begin{array}{ccccccccccccccccccc} 0&\longrightarrow&P&\longrightarrow&I^0(P)&\longrightarrow&\cdots&\longrightarrow&I^{m-1}(P)&\longrightarrow&I^m(P)&\longrightarrow&I^{m+1}(P)&\longrightarrow&\cdots&\longrightarrow&I^{m+n}(P)\\ &&\uparrow^{i_P}&&\uparrow^{w^0}&&&&\uparrow^{w^{m-1}}&&\uparrow^{t^0}&&\uparrow^{t^1}&&&&\uparrow^{t^n}\\ 0&\longrightarrow&P&\longrightarrow&S_m&\longrightarrow&\cdots&\longrightarrow&S_1&\xrightarrow[\mu\circ\lambda]{}&R_n&\longrightarrow&R_{n-1}&\longrightarrow&\cdots&\longrightarrow&M&\longrightarrow&0 \end{array}

where t0=h0∘v0t^0=h^0\circ v^0, t1=(−1)mh1∘v1t^1=(-1)^m h^1\circ v^1, …\ldots, ti=(−1)mihi∘vit^i=(-1)^{mi}h^i\circ v^i, …\ldots, tn=(−1)mnhn∘vnt^n=(-1)^{mn}h^n\circ v^n.

The class θ\theta associated to (6) is that of (−1)m(m+1)/2wm∈Homgr⁡Am(N,I(P))(-1)^{m(m+1)/2}w^m\in\operatorname{Homgr}_A^m(N,I(P)), corresponding by the isomorphism αN,P\alpha_{N,P} to the class of (−1)m(m+1)/2h∈Homgr⁡Am(I(N),I(P))(-1)^{m(m+1)/2}h\in\operatorname{Homgr}_A^m(I(N),I(P)); the class θ′\theta' associated to (7) is that of (−1)n(n+1)/2vn∈Homgr⁡n(M,I(N))(-1)^{n(n+1)/2}v^n\in\operatorname{Homgr}^n(M,I(N)), the class associated to (8) is that of (−1)(m+n)(m+n+1)/2tn∈Homgr⁡m+n(M,I(P))(-1)^{(m+n)(m+n+1)/2}t^n\in\operatorname{Homgr}^{m+n}(M,I(P)), hence the conclusion, according to the definition of the composition product (X, p. 114) and the formula

m(m+1)/2+n(n+1)/2=(m+n)(m+n+1)/2−mn. m(m+1)/2+n(n+1)/2=(m+n)(m+n+1)/2-mn.

4. — Considérons un diagramme commutatif de AA-modules à lignes exactes

0⟶N⟶Rn⟶Rn−1⟶⋯⟶R1⟶M⟶0↓g↓↓↓↓f0⟶N′⟶Rn′⟶Rn−1′⟶⋯⟶R1′⟶M′⟶0. \begin{array}{ccccccccccccccc} 0&\longrightarrow&N&\longrightarrow&R_n&\longrightarrow&R_{n-1}&\longrightarrow&\cdots&\longrightarrow&R_1&\longrightarrow&M&\longrightarrow&0\\ &&\downarrow^{g}&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow^{f}&&\\ 0&\longrightarrow&N'&\longrightarrow&R'_n&\longrightarrow&R'_{n-1}&\longrightarrow&\cdots&\longrightarrow&R'_1&\longrightarrow&M'&\longrightarrow&0. \end{array}

Let θ\theta (resp. θ′\theta') be the class of the first (resp. second) row in

Ext⁡An(M,N)(resp. Ext⁡An(M′,N′)). \operatorname{Ext}_A^n(M,N)\quad\text{(resp. }\operatorname{Ext}_A^n(M',N')\text{)}.

In Ext⁡An(M,N′)\operatorname{Ext}_A^n(M,N'), we have θ′∘f=g∘θ\theta'\circ f=g\circ\theta.

Indeed, consider a commutative diagram

Ln(M)→dnLn−1(M)⟶⋯⟶L0(M)→pMM⟶0↓un↓↓↓10⟶N⟶Rn⟶⋯⟶R1⟶M⟶0↓g↓↓↓f0⟶N′⟶Rn′⟶⋯⟶R1′⟶M′⟶0↓↓↓↓vn0⟶N′→ϵN′I0(N′)→δ0⋯⟶In−1(N′)→δn−1In(N′) \begin{array}{ccccccccccccc} L_n(M)&\xrightarrow{d_n}&L_{n-1}(M)&\longrightarrow&\cdots&\longrightarrow&L_0(M)&\xrightarrow{p_M}&M&\longrightarrow&0\\ \downarrow^{u_n}&&\downarrow&&&&\downarrow&&\downarrow^{1}\\ 0&\longrightarrow&N&\longrightarrow&R_n&\longrightarrow&\cdots&\longrightarrow&R_1&\longrightarrow&M&\longrightarrow&0\\ &&\downarrow^{g}&&\downarrow&&&&\downarrow&&\downarrow^{f}\\ 0&\longrightarrow&N'&\longrightarrow&R'_n&\longrightarrow&\cdots&\longrightarrow&R'_1&\longrightarrow&M'&\longrightarrow&0\\ &&\downarrow&&\downarrow&&&&\downarrow&&\downarrow^{v^n}\\ 0&\longrightarrow&N'&\xrightarrow{\epsilon_{N'}}&I^0(N')&\xrightarrow{\delta^0}&\cdots&\longrightarrow&I^{n-1}(N')&\xrightarrow{\delta^{n-1}}&I^n(N') \end{array}

By definition, θ′∘f\theta'\circ f is the class of (−1)n(n+1)/2vn∘f∘pM∈Homgr⁡n(L(M),I(N′))(-1)^{n(n+1)/2}v^n\circ f\circ p_M\in\operatorname{Homgr}^n(L(M),I(N')), while g∘θg\circ\theta is the class of (−1)n(n+1)/2ϵN′∘g∘un(-1)^{n(n+1)/2}\epsilon_{N'}\circ g\circ u_n. According to Lemma 1, applied to the two extreme lines of the diagram, these two classes are equal.

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.