Lemma 7

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

Commutative Algebra, chapter I, § 2. Flat modules

Let E be a right A-module, F a left A-module and F', F'' two submodules of F such that F = F' + F''. Then the intersection of the canonical images of E \otimes F' and E \otimes F'' in E \otimes F is equal to the canonical image of E \otimes (F' \cap F'').

Consider the diagram

\begin{array}{ccccccc} 0 & \longrightarrow & F' \cap F'' & \longrightarrow & F' & \longrightarrow & F'/(F' \cap F'') \longrightarrow 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \longrightarrow & F'' & \longrightarrow & F' + F'' & \longrightarrow & (F' + F'')/F'' \longrightarrow 0 \end{array}

where the unspecified arrows are the canonical injections and surjections and j the canonical isomorphism defined in Algebra, Chapter I, § 6, no. 13, Theorem 6. This diagram is commutative and its rows are exact. We derive (since F = F' + F'') a commutative diagram

\begin{array}{ccc} E \otimes (F' \cap F'') & \longrightarrow & E \otimes F' \longrightarrow E \otimes (F'/(F' \cap F'')) \\ \downarrow & & \downarrow \\ E \otimes F'' & \longrightarrow & E \otimes F \longrightarrow E \otimes (F/F'') \end{array}

The rows of this diagram are exact (no. 1, Lemma 1) and 1 \otimes j is an isomorphism. Our assertion is then a special case of § 1, no. 4, Proposition 1, (i) (cf. Exercise 5).

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

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Proposition 6, § 2

Lemma 7

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