Bourbaki

Proposition 4

Tag 0DY6, permanent. Cite it as [Bourbaki, Tag 0DY6].

Algebra, chapter X, § 6. Utilisation de résolutions non canoniques

Let M be a left A-module, a:R→Ma : R \to M a left resolution of M such that the sequence

(7)

0→HomgrA(R,E′)→Homgr(s~,1)HomgrA(R,E)→Homgr(r~,1)HomgrA(R,E′′)→0 0 \rightarrow \mathrm{Homgr}_A(R, E') \xrightarrow{\mathrm{Homgr}(\tilde{s}, 1)} \mathrm{Homgr}_A(R, E) \xrightarrow{\mathrm{Homgr}(\tilde{r}, 1)} \mathrm{Homgr}_A(R, E'') \rightarrow 0

is exact. Then the following diagram is commutative.

H(HomgrA(R,E′′))→∂((7))H(HomgrA(R,E′))φ(R,E′′)↓↓φ(R,E′)ExtA(M,N′′)→δ(M,(5))ExtA(M,N′) \begin{array}{ccc} \mathrm{H}(\mathrm{Homgr}_A(R, E'')) & \xrightarrow{\partial((7))} & \mathrm{H}(\mathrm{Homgr}_A(R, E')) \\ \varphi(R, E'') \downarrow & & \downarrow \varphi(R, E') \\ \mathrm{Ext}_A(M, N'') & \xrightarrow{\delta(M, (5))} & \mathrm{Ext}_A(M, N') \end{array}

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.