Bourbaki

Definition 2

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

Commutative Algebra, chapter I, § 2. Flat modules

A right A-module E is called flat if it has the equivalent properties of Proposition 1.

Flat left A-modules are defined similarly. To say that a right A-module E is flat equivalent to saying that E, considered as a left A0A^0-module, is flat.

Remarks

(1) By Lemma 3 of no. 2, for a right A-module E to be flat, it is necessary and sufficient that, for every finitely generated left ideal aa of A, the canonical mapping E⊗Aa→EE \otimes_A a \to E (Proposition 1) with image Ea be injective.

(2) Let E be a flat right A-module. If M′M' is a submodule of a left A-module M, the canonical injection E⊗AM′→E⊗AME \otimes_A M' \to E \otimes_A M allows us to identify E⊗AM′E \otimes_A M' with a subgroup of E⊗AME \otimes_A M. This being so, let N be a left A-module, u:M→Nu : M \to N a homomorphism, K=Ker uK = \mathrm{Ker}\, u, and I=Im uI = \mathrm{Im}\, u. By considering the exact sequence

0⟶K⟶M→uN 0 \longrightarrow K \longrightarrow M \xrightarrow{u} N

it is easily seen (Proposition 1) that E⊗A(Ker u)E \otimes_A (\mathrm{Ker}\, u) is identified with Ker(1E⊗u)\mathrm{Ker}(1_E \otimes u). On the other hand, writing u′u' for the surjective homomorphism M→IM \to I with the same graph as uu, and ii for the canonical injection I→NI \to N, 1E⊗u′1_E \otimes u' is surjective (no. 1, Lemma 1) and 1E⊗i1_E \otimes i is injective since E is flat. As

1E⊗u=(1E⊗i)∘(1E⊗u′), 1_E \otimes u = (1_E \otimes i) \circ (1_E \otimes u'),

E⊗A(Im u)E \otimes_A (\mathrm{Im}\, u) is identified with Im(1E⊗u)\mathrm{Im}(1_E \otimes u).

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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