Bourbaki

Proposition 7

Tag 0E33, permanent. Cite it as [Bourbaki, Tag 0E33].

Algebra, chapter X, § 9. Complexes de Koszul

a) Let N be a flat A-module. If the family x is completely secant for M, it is so for M⊗ANM \otimes_A N.

b) Let 0→M′→M→M′′→00 \to M' \to M \to M'' \to 0 be an exact sequence of AA-modules. If the family xx is completely transverse for M′M' and for M′′M'', it is so for MM.

The complex K.(x,M⊗AN)K.(x, M \otimes_A N) is isomorphic by definition to K.(x,M)⊗ANK.(x, M) \otimes_A N; since NN is flat, one deduces from this an isomorphism H.(x,M)⊗AN→H.(x,M⊗AN)H.(x, M) \otimes_A N \to H.(x, M \otimes_A N) (X, p. 66, cor. 2), whence a).

The complex K.(x)K.(x) being flat, one has an exact sequence of complexes

0→K.(x,M′)→K.(x,M)→K.(x,M′′)→0; 0 \to K.(x, M') \to K.(x, M) \to K.(x, M'') \to 0 ;

assertion b) follows from the associated exact homology sequence.

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

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