Bourbaki

Proposition 3

Tag 0D0B, permanent. Cite it as [Bourbaki, Tag 0D0B].

Commutative Algebra, chapter X, § 1. Profondeur

Let A be a ring, C a complex of A-modules bounded on the left and p an integer. Suppose that for every pair of integers (m, n) with m \geq n \geq p, the depth of the A-module C_m relative to the annihilator of H_n(C) is > m - n. Then one has H_n(C) = 0 for n \geq p.

Since C is bounded on the left, H_n(C) is zero for n sufficiently large. If the conclusion were false, there would exist an integer q \geq p such that H_n(C) = 0 for n > q and H_q(C) \neq 0. Denote by J the annihilator of H_q(C); one then has prof_A(J ; H_q(C)) = 0. Moreover, since Z_q(C) is a submodule of C_q, and since by hypothesis one has prof_A(J ; C_q) > q - q = 0, one has prof_A(J ; Z_q(C)) > 0. One then deduces from the exact sequence

0→Bq(C)→Zq(C)→Hq(C)→0 0 \to B_q(C) \to Z_q(C) \to H_q(C) \to 0

the equality prof_A(J ; B_q(C)) = 1 (No. 1, prop. 1). By the definition of q, B_n(C) is equal to Z_n(C) for every integer n > q. From the canonical exact sequences

0→Bn(C)→Cn→Bn−1(C)→0(n>q) 0 \to B_n(C) \to C_n \to B_{n-1}(C) \to 0 \quad (n > q)

and from the hypothesis prof_A(J ; C_n) > n - q, one obtains by induction the equality prof_A(J ; B_n(C)) = n - q + 1 for every n \geq q (loc. cit.). But this is absurd since B_n(C) is zero for n sufficiently large.

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

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