characterizing free modules by exterior power 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? 
 A: 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?
A: 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)
A: 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).
