Bourbaki

Proposition 3

Tag 0DP0, permanent. Cite it as [Bourbaki, Tag 0DP0].

Algebra, chapter X, § 1. Compléments d’algèbre linéaire

In order that the a-module E be flat, it is necessary and sufficient that, for every injective A-homomorphism u:M′→Mu : M' \to M of right A-modules, the mapping u⊗1:M′⊗AE→M⊗AEu \otimes 1 : M' \otimes_A E \to M \otimes_A E be injective.

If E is flat and u:M′→Mu : M' \to M is injective, the sequence 0→M′→uM0 \to M' \xrightarrow{u} M is exact, hence also the sequence 0⟶M′⊗AE→u⊗1M⊗AE0 \longrightarrow M' \otimes_A E \xrightarrow{u \otimes 1} M \otimes_A E, and u⊗1u \otimes 1 is injective. Conversely, consider the exact sequence (11); put M1′′=v(M)M''_1 = v(M), and let i:M1′′→M′′i : M''_1 \to M'' be the canonical injection and p:M→M1′′p : M \to M''_1 the mapping m↦v(m)m \mapsto v(m). The sequence M′→uM→pM1′′⟶0M' \xrightarrow{u} M \xrightarrow{p} M''_1 \longrightarrow 0 is exact; according to II, p. 58, prop. 5, the sequence M′⊗AE→u⊗1M⊗AE→p⊗1M1′′⊗AEM' \otimes_A E \xrightarrow{u \otimes 1} M \otimes_A E \xrightarrow{p \otimes 1} M''_1 \otimes_A E is therefore exact. Moreover, one has v=i∘pv = i \circ p, hence (v⊗1)=(i⊗1)∘(p⊗1)(v \otimes 1) = (i \otimes 1) \circ (p \otimes 1); if E satisfies the condition of the statement, then i⊗1i \otimes 1 is injective, hence

Ker⁡(v⊗1)=Ker⁡(p⊗1)=Im⁡(u⊗1) \operatorname{Ker}(v \otimes 1) = \operatorname{Ker}(p \otimes 1) = \operatorname{Im}(u \otimes 1)

and the sequence (12) is exact.

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

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