Theorem 1

Tag 0DX3, permanent. Cite it as [Bourbaki, Tag 0DX3].

Algebra, chapter X, § 5. Modules d’extensions

The right-unbounded sequence of homomorphisms of $k$ -modules

0 \longrightarrow \operatorname{Hom}_A(M, N') \xrightarrow{\operatorname{Hom}(1,u)} \operatorname{Hom}_A(M, N) \xrightarrow{\operatorname{Hom}(1,v)} \operatorname{Hom}_A(M, N'')

\phantom{0} \xrightarrow{\delta_{(M,\mathscr{E})}} \operatorname{Ext}_A^1(M, N') \to \cdots \xrightarrow{\delta^{n-1}(M,\mathscr{E})} \operatorname{Ext}_A^n(M, N') \xrightarrow{\operatorname{Ext}^n(1,u)} \operatorname{Ext}_A^n(M, N)

\phantom{0} \xrightarrow{\operatorname{Ext}^n(1,v)} \operatorname{Ext}_A^n(M, N'') \xrightarrow{\delta^n(M,\mathscr{E})} \operatorname{Ext}_A^{n+1}(M, N') \to \cdots

is exact.

Consider indeed the diagram on page X.91. It is commutative by the remark 3 of $X$ , p. 88 and def. 2 ; moreover the bottom row is exact ( $X$ , p. 30, th. 1), and the vertical arrows are bijective ( $X$ , p. 88, prop. 5).

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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