Bourbaki

Lemma 1

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

Algebra, chapter X, § 4. Produits de torsion

Let C be a complex of right A-modules and E a complex of left A-modules. We assume that H(C)=0H(C)=0 and that E is flat and bounded on the right. Then H(C⊗AE)=0H(C\otimes_A E)=0.

For k∈Zk\in\mathbf Z, let T(k)T^{(k)} be the subcomplex of C⊗AEC\otimes_A E such that

Tn(k)=∑p+q=nq≤kCp⊗AEq ; T_n^{(k)}=\sum_{\substack{p+q=n\\q\leq k}} C_p\otimes_A E_q\ ;

then T(k−1)⊂T(k)T^{(k-1)}\subset T^{(k)} and we have an exact sequence of complexes

0⟶T(k−1)→ikT(k)→πC⊗AEk(−k)⟶0 0\longrightarrow T^{(k-1)}\xrightarrow{i_k}T^{(k)}\xrightarrow{\pi}C\otimes_A E_k(-k)\longrightarrow 0

where iki_k is the canonical injection and where π\pi projects the preceding direct sum onto its factor Cn−k⊗AEk=(C⊗AEk(−k))nC_{n-k}\otimes_A E_k=(C\otimes_A E_k(-k))_n. By Cor. 2 above, we have H(C⊗AEk(−k))=0H(C\otimes_A E_k(-k))=0, hence iki_k is a homomorphism. We have T(k)=0T^{(k)}=0 for kk sufficiently small, since EE is bounded on the right, hence H(T(k))=0H(T^{(k)})=0 for all kk by induction on kk. Finally, the canonical morphism lim⁡⟶ T(k)→C⊗AE\underset{\longrightarrow}{\lim}\,T^{(k)}\to C\otimes_A E is obviously an isomorphism, hence H(C⊗AE)=0H(C\otimes_A E)=0 (X, p. 28, Prop. 1).

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

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