Bourbaki

Proposition 7

Tag 0DU5, permanent. Cite it as [Bourbaki, Tag 0DU5].

Algebra, chapter X, § 3. Résolutions

Let CC be an AA-complex and let a∈Za \in \mathbf{Z} be such that Hn(C)=0H_n(C) = 0 for n<an < a. a) There exists a free AA-complex LL such that Ln=0L_n = 0 for n<an < a and a homologism f:L→Cf : L \to C.

b) Suppose AA left noetherian and the AA-modules Hn(C)H_n(C), n∈Zn\in \mathbf{Z}, finitely generated. There exists a free AA-complex LL such that Ln=0L_n=0 for n<an<a and that LnL_n is a finitely generated AA-module for every nn, and a homomorphism f:L→Cf:L\to C.

Let C′C' be the subcomplex of CC such that Cn′=CnC'_n=C_n for n>an>a, Ca′=Za(C)C'_a=Z_a(C), Cn′=0C'_n=0 for n<an<a; then the canonical injection of C′C' into CC is a homomorphism. Replacing CC by C′C', one may therefore suppose that Cn=0C_n=0 for n<an<a. The statement then results from repeated application of the following lemma, for r=a,a+1,…r=a,a+1,\ldots :

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

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