Bourbaki

Exercise 14, § 1

Algebra, X, p. 169

a) Let 0→R→L→E→00 \to R \to L \to E \to 0 be an exact sequence of A-modules, where L is a free A-module; let (ea)(e_a) be a basis of L. Now prove that the following conditions are equivalent: α) E is flat. β) For every x∈Rx \in R, if axa_x is the right ideal generated by the components of x on the basis (ea)(e_a), one has x∈axRx \in a_x R. γ) For every x∈Rx \in R, there exists a homomorphism ux:L→Ru_x : L \to R such that ux(x)=xu_x(x) = x. δ) For every finite sequence (xi)1≤i≤n(x_i)_{1 \leq i \leq n} of elements of R, there exists a homomorphism u:L→Ru : L \to R such that u(xi)=xiu(x_i) = x_i for 1≤i≤n1 \leq i \leq n. (Use th. 1, p. 14.) b) Let r be the radical of A, and let 0→R→L→E→00 \to R \to L \to E \to 0 be an exact sequence of A-modules, such that L is free. Suppose that E is flat and that R is contained in rL. Now prove that one has R=0R = 0 (with the notations of a), observe that axa_x is an ideal of finite type and that one has ax=axra_x = a_x r. c) Let E be a flat A-module of finite type; suppose that there exists a two-sided ideal b of A contained in the radical of A, such that E/bEE/bE is a (A/b)(A/b)-module free. Now prove that E is then a free A-module (observe that there exists a free A-module of finite type L such that L/bLL/bL is isomorphic to E/bEE/bE and apply b)). In particular, every flat module of finite type over a local ring is free.

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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.