Bourbaki

Proposition 5

Tag 042S, permanent. Cite it as [Bourbaki, Tag 042S].

Commutative Algebra, chapter I, § 2. Flat modules

Let 0→E′→vE→wE′′→00 \to E' \xrightarrow{v} E \xrightarrow{w} E'' \to 0 be an exact sequence of right A\mathbf{A}*-modules. Suppose* E′′E'' is flat. Then, for EE to be flat it is necessary and sufficient that E′E' be flat.

Let u:F′→Fu : F' \to F be an injective homomorphism of left A\mathbf{A}-modules. Consider the diagram

E′⊗F′→v⊗1F′E⊗F′→w⊗1F′E′′⊗F′↓lE′⊗u↓lE⊗u↓lE′′⊗uE′⊗F→v⊗1FE⊗F→w⊗1FE′′⊗F \begin{array}{ccccccccc} E' \otimes F' & \xrightarrow{v \otimes 1_{F'}} & E \otimes F' & \xrightarrow{w \otimes 1_{F'}} & E'' \otimes F' \\ \downarrow l_{E'} \otimes u & & \downarrow l_E \otimes u & & \downarrow l_{E''} \otimes u \\ E' \otimes F & \xrightarrow{v \otimes 1_F} & E \otimes F & \xrightarrow{w \otimes 1_F} & E'' \otimes F \end{array}

It is commutative and its rows are exact (no. 1, Lemma 1). Since E′′E'' is flat, l_{E''} \otimes u is injective; moreover, Proposition 4 proves that v \otimes 1_{F'}, and v \otimes 1, are injective. This being so, if E is flat, l_E \otimes u is injective, hence also

(lE⊗u)∘(v⊗1F′)=(v⊗1F)∘(lE′⊗u);(l_E \otimes u) \circ (v \otimes 1_{F'}) = (v \otimes 1_F) \circ (l_{E'} \otimes u);

it follows that l_{E'} \otimes u is injective and consequently E' is flat. Conversely, if E' is flat, l_{E'} \otimes u is injective; then it follows from § 1, no. 4, Corollary 1 to Proposition 2 that l_E \otimes u is injective and so E is flat.

Remarks (1) It is possible for E and E' to be flat without E'' being so, as the example of the \mathbf{Z}-modules E = \mathbf{Z}, E' = n\mathbf{Z}, E'' = \mathbf{Z}/n\mathbf{Z} (n \geq 2), shows. (2) A submodule of a flat module is not necessarily a flat module (Exercise 3).

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

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