Bourbaki

Exercise 23, § 2

Commutative Algebra, I, 47

(a) Let 0→R→L→E→00 \to R \to L \to E \to 0 be an exact sequence of left A-modules, where L is a free A-module; let (ei)(e_i) be a basis of L. Show that the following conditions are equivalent:

(a) E is flat.

(b) For all x∈Rx \in R, if aa is the right ideal generated by the components of xx with respect to the basis (eα)(e_\alpha), then x∈Rax \in Ra.

(y) For all x∈Rx \in R, there exists a homomorphism ux:L→Ru_x : L \to R such that ux(x)=xu_x(x) = x.

(8) 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 the Corollary to Proposition 7 of no. 6.)

(b) Let a be a left ideal of A such that A/aA/a is a flat A-module. Show that, for every finitely generated left ideal b⊂ab \subset a, there exists x∈Ax \in A such that

b⊂Ax⊂a b \subset Ax \subset a

(use condition (8) of (a)).

(c) Derive from (a) a new proof of the result of Exercise 15.

(d) Let rr be the Jacobson radical of AA and 0→R→L→E→00 \to R \to L \to E \to 0 an exact sequence of left AA-modules such that LL is free. Suppose that EE is flat and RR is contained in rLrL. Show that R=0R = 0 (in the notation of (a) observe that ara_r is a finitely generated ideal and that ar=ax,ra_r = a_{x,r}).

(e) Let EE be a finitely generated flat AA-module; suppose that there exists a two-sided ideal bb of AA contained in the Jacobson radical of AA such that E/bEE/bE is a free (A/b)(A/b)-module. Show that EE is then a free AA-module (observethat there exists a finitely generated free AA-module LL such that L/bLL/bL is isomorphic to E/bEE/bE and use Algebra, Chapter VIII, § 6, no. 3, Corollary 4 to Proposition 6; then apply (d)).

Solution

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