Lemma 10.85.2. Let M be a countably generated R-module with the following property: if M = N \oplus N' with N' a finite free R-module, then any element of N is contained in a free direct summand of N. Then M is free.
Proof. Let x_1, x_2, \ldots be a countable set of generators for M. We inductively construct finite free direct summands F_1, F_2, \ldots of M such that for all n we have that F_1 \oplus \ldots \oplus F_ n is a direct summand of M which contains x_1, \ldots , x_ n. Namely, given F_1, \ldots , F_ n with the desired properties, write
M = F_1 \oplus \ldots \oplus F_ n \oplus N
and let x \in N be the image of x_{n + 1}. Then we can find a free direct summand F_{n + 1} \subset N containing x by the assumption in the statement of the lemma. Of course we can replace F_{n + 1} by a finite free direct summand of F_{n + 1} and the induction step is complete. Then M = \bigoplus _{i = 1}^{\infty } F_ i is free. \square
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