Bourbaki

Corollary 1

Tag 043E, permanent. Cite it as [Bourbaki, Tag 043E].

Commutative Algebra, chapter I, § 2. Flat modules

For a right AA-module EE to be flat, it is necessary and sufficient that the following condition hold:

(RP) If (ei)i∈I(e_i)_{i \in I} and (bi)i∈I(b_i)_{i \in I} are two finite families of elements of EE and AA respectively such that ∑eibi=0\sum e_i b_i = 0, there exists a finite set JJ, a family (xj)j∈J(x_j)_{j \in J} of elements of EE and a family (aji)(a_{ji}) (j∈J,i∈Ij \in J, i \in I) of elements of AA such that ∑ajibi=0\sum a_{ji} b_i = 0 for all j∈Jj \in J and ei=∑j∈Jxjajie_i = \sum_{j \in J} x_j a_{ji} for all i∈Ii \in I.

Condition (RP) is just condition (R) of Proposition 13 applied to the module F=AsF = A_s.

Loosely speaking, (RP) states: every "relation" between the bib_i, with coefficients in EE, is a linear combination (with coefficients in EE) of "relations" between the bib_i with coefficients in AA.

Let us consider in particular a homomorphism of AA to a ring BB, which makes BB into a right AA-module. We know (no. 3, Proposition 1) that this is equivalent to saying that this AA-module is flat, or that it is flat for every left AA-module AsmA_s^m (m⩾1m \geqslant 1). Applying condition (R) of Proposition 13 with E=BE = B and F=AsmF = A_s^m we obtain the following condition:

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

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Transcribed from the printed volume. Not translated.