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Assume $M$ is a (finitely generated) $A$-module such that $\wedge^n M$ is free of rank $1$ for some $n \geq 1$. Does it follow that $M$ is free of rank $n$? Or at least locally free of rank $n$?

In general, is there a way of characterizing local free modules via exterior powers and tensor products?

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The conditions "locally free", "finitely generated projective" and "dualizable" are equivalent, and the latter one can be formulated in terms of tensor products (namely of the "unit" $A \to M \otimes M^*$ and the "counit" $M^* \otimes M \to A$ satisfying the two triangular identities).

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This isn't a complete answer, but just an idea about your first question. Let $P$ be a rank one projective module over the commutative ring $A$, and $P^*=\mathrm{Hom}_A(P,A)$ be its dual. Then for $M=P\oplus P^*$, $\bigwedge^2 M\cong P\otimes_A P^*\cong A$ is free. There must be $P$ for which $M$ isn't free, but I can't think of any off the top of my head.

If $A$ is a Dedekind domain then $M$ is free. Taking $A=C^\infty(N)$ where $N$ is a smooth manifold, then $P$ would correspond to a line bundle on $N$. If $M$ is free then the direct sum of this line bundle and its dual would be trivial. Surely there are manifolds and line bundles for which this isn't true?

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    Real line bundle $\oplus$ dual line bundle is always trivial. Because of following two facts. Fact A: Line bundle is isomorphic to dual line bundle by choosing metric on it .Fact B: line bundles have 2-torsion because they are classified by stiefel-whitney class which have 2-torsion, because they belong to $H^1(M, \mathbb Z /(2) )$2010-11-20
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    Thanks Robin. I'm also sure that there is such an example. However, your $M$ is locally free of rank $2$. So this does not answer yet my 2nd question, which is probably more difficult ...2010-11-20
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Following variation of Robin Chapman idea works in algebraic geometry. Take $X$ elliptic curve minus point : it is affine variety. For general point $P \in X$ take $L=\mathcal O(P)$. Then $M=L\oplus L^*=\mathcal O(P) \oplus \mathcal O(-P)$ has properties: A) $\Lambda ^2 (M)=\mathcal O$ is free of rank one. B) $M$ is not free because it has no sections except zero (since $\mathcal O(P)$ and $\mathcal O(-P)$ have no sections except zero)