Bourbaki

Lemma 5

Tag 0E1G, permanent. Cite it as [Bourbaki, Tag 0E1G].

Algebra, chapter X, § 8. Dimension homologique

Let MM be a graded AA-module bounded below, and let

s:A⊗A0M/J0M→M s : A \otimes_{A_0} M/J_0 M \to M
be a graded AA-homomorphism such that 1⊗As:A0⊗A(A⊗A0M/J0M)→A0⊗AM1 \otimes_A s : A_0 \otimes_A (A \otimes_{A_0} M/J_0 M) \to A_0 \otimes_A M is the canonical isomorphism. Then ss is surjective. If Tor1A(A0,M)=0\mathrm{Tor}_1^A(A_0, M) = 0, ss is bijective.

We have an exact sequence

0→Ker s→A⊗A0MJ0M→M→Coker s→0 0 \to \mathrm{Ker}\ s \to A \otimes_{A_0} M J_0 M \to M \to \mathrm{Coker}\ s \to 0
and the graded AA-modules Ker s\mathrm{Ker}\ s and Coker s\mathrm{Coker}\ s are bounded below. From the exact sequence A0⊗A(A⊗A0M/J0M)→1⊗sA0⊗AM→A0⊗ACoker s→0A_0 \otimes_A (A \otimes_{A_0} M/J_0 M) \xrightarrow{1 \otimes s} A_0 \otimes_A M \to A_0 \otimes_A \mathrm{Coker}\ s \to 0, we deduce that A0⊗ACoker s=0A_0 \otimes_A \mathrm{Coker}\ s = 0, hence ss is surjective (Lemma 4). We then have an exact sequence
Tor1A(A0,M)→A0⊗AKer s→A0⊗A(A⊗A0M/J0M)→1⊗sA0⊗AM. \mathrm{Tor}_1^A(A_0, M) \to A_0 \otimes_A \mathrm{Ker}\ s \to A_0 \otimes_A (A \otimes_{A_0} M/J_0 M) \xrightarrow{1 \otimes s} A_0 \otimes_A M .
If Tor1A(A0,M)=0\mathrm{Tor}_1^A(A_0, M) = 0, then A0⊗AKer s=0A_0 \otimes_A \mathrm{Ker}\ s = 0 and ss is injective (Lemma 4).

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

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