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Is there an elegant tensor-free proof of the fact that over a reduced Noetherian ring $A$, every finitely-generated $A$-module which is locally free, is projective?

EDIT: I would be content with the case where $A$ is a local ring.

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Isn't "locally free" the same thing as "free" for a local ring? Free modules are always projective, by the axiom of choice. –  Zhen Lin May 4 '12 at 0:28
Yes, I think so, but it requires proof. Locally free means that for a f.g. $A-$module $M$, at all primes $\mathfrak{p}$ in the spectrum, $M_\mathfrak{p}$ is free. –  user21725 May 4 '12 at 0:34

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up vote 2 down vote accepted

The proof I have in mind.

Since $M$ is finitely generated over a Noetherian ring, it is also finitely presented. Therefore, we have a short exact sequence $$0\rightarrow K\rightarrow F\xrightarrow{\sigma} M\rightarrow 0$$ where $K$ is a finitely generated and $F$ is a finitely generated free $A$-module. Given a prime ideal $\mathfrak{p}\subseteq A$, since localisation is exact (which can be proved without tensor products), the sequence $$0\rightarrow K_{\mathfrak{p}}\rightarrow F_{\mathfrak{p}}\xrightarrow{\sigma_{\mathfrak{p}}} M_{\mathfrak{p}}\rightarrow 0$$ is exact, and in fact is split exact because $M_{\mathfrak{p}}$ is free. Now, for each prime $\mathfrak{p}\subseteq A$, $$\operatorname{Hom}_A(M,F)_{\mathfrak{p}}\cong\operatorname{Hom}_{A_{\mathfrak{p}}}(M_{\mathfrak{p}},F_{\mathfrak{p}})\xrightarrow{\operatorname{Hom}(M_{\mathfrak{p}},\sigma_{\mathfrak{p}})}\operatorname{Hom}_{A_{\mathfrak{p}}}(M_{\mathfrak{p}},M_{\mathfrak{p}})\cong\operatorname{Hom}_A(M,M)_{\mathfrak{p}}$$ is surjective, hence $\operatorname{Hom}(M,F)\xrightarrow{\operatorname{Hom}(M,\sigma)}\operatorname{Hom}(M,M)$ is surjective as well (need only be checked at maximal ideals), containing the morphism $\operatorname{id}_M$ in its image.

Then, the original exact sequence is split exact, and as a direct summand of a free module, $M$ is projective. Note that I have not used the reducedness criterion.

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