Bourbaki

Lemma 3

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

Commutative Algebra, chapter I, § 2. Flat modules

For a tight AA-module EE to be MM-flat, it is necessary and sufficient that for every finitely generated submodule M′M' of MM the canonical homomorphism

lE⊗j:E⊗AM′→E⊗AM l_E \otimes j : E \otimes_A M' \to E \otimes_A M

(*j being the canonical injection M′→MM' \to M) be injective.

Suppose that this condition holds and let NN be any submodule of MM. Suppose that the canonical image in E⊗AME \otimes_A M of an element

z=∑xi⊗yi∈E⊗AN(xi∈E,yi∈N) z = \sum x_i \otimes y_i \in E \otimes_A N \quad (x_i \in E, y_i \in N)

is zero and let M′M' be the finitely generated submodule of NN generated by they.; as by hypothesis the composite mapping E⊗AM′→E⊗AN→E⊗AME \otimes_A M' \to E \otimes_A N \to E \otimes_A M is injective, the sum z′=∑xi⊗yiz' = \sum x_i \otimes y_i, considered as an element of E⊗AM′E \otimes_A M', is zero. As zz is the image of z′z', we have also z=0z = 0, whence the lemma.

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

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Cites

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