Bourbaki

Proposition 5

Tag 0E2Z, permanent. Cite it as [Bourbaki, Tag 0E2Z].

Algebra, chapter X, § 9. Complexes de Koszul

Let x=(x1,…,xn)x=(x_1,\ldots,x_n) be a sequence of elements of AA. If for i=1,…,ni=1,\ldots,n, the homothety of ratio xix_i in the AA-module M/(x1M+⋯+xi−1M)M/(x_1M+\cdots+x_{i-1}M) is injective, the sequence xx is completely secant for MM.

A sequence satisfying the conditions of the proposition is said to be regular for M, or M-regular. We shall note that this property depends in general on the order of the xix_i; for example the sequence (1,0)(1, 0) is always M-regular, whereas the sequence (0,1)(0, 1) is so only if M is zero. On the other hand, the fact that a sequence is completely secant does not depend on the order of the terms.

Let us prove the proposition by induction on nn, the case n=0n = 0 being immediate. Put x′=(x1,...,xn−1)x' = (x_1, ..., x_{n-1}); if the sequence xx is M-regular, the sequence x′x' is M-regular and multiplication by xnx_n in M/(x′)MM/(x')M is injective; by the induction hypothesis, we have Hi(x′,M)=0H_i(x', M) = 0 for i>0i > 0; it then follows from cor. 3 of X, p. 157, that Hi(x,M)=0H_i(x, M) = 0 for i>0i > 0.

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

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