Bourbaki

Exercise 6, § 1

Algebra, X, p. 168

now prove that an AA-module EE is flat if and only if it satisfies the following condition (cf. X, p. 15, remark) :

(ii′)(ii') For every finite family (ci)i∈I(c_i)_{i\in I} of elements of AA, every solution e=(ei)i∈Ie=(e_i)_{i\in I} of the equation ∑i∈Iciei=0\displaystyle\sum_{i\in I}c_i e_i=0 can be written b1z1+⋯+bnznb_1z_1+\cdots+b_nz_n, where b1,…,bn∈Eb_1,\ldots,b_n\in E and where, for r=1,…,nr=1,\ldots,n, zr=(zr,i)i∈Iz_r=(z_{r,i})_{i\in I} is a solution of the equation ∑i∈Izr,ici=0\displaystyle\sum_{i\in I}z_{r,i}c_i=0.

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.