Characterization of locally free modules via exterior powers Let $X$ be a scheme and $\mathcal{F}$ be quasi-coherent module on $X$. It is clear that if $\mathcal{F}$ is locally free of rank $n$, then $\det(\mathcal{F}) := \wedge^n \mathcal{F}$ is invertible, i.e. locally free of rank $1$. But what about the converse?
Question. Assume $\wedge^n \mathcal{F}$ is invertible. Does it follow that $\mathcal{F}$ is locally free (necessarily of finite rank $n$)?
Of course we may assume that $X$ is affine. Then it is enough to prove that $\mathcal{F}$ is flat and of finite presentation, but I don't know how to prove either one. Also it seems to be hard to find counterexamples.
 A: I think that $\mathcal F$ is indeed locally free of rank $n$:
Pick a point $x\in X$. It will be enough to show that there is a neighbourhood
of $x$ on which $\mathcal F$ is free of rank $n$. Now, the exterior power
commutes with pullbacks (aka scalar extensions) so that in particular the fibre
(in the sense of pullback to $\mathrm{Spec}k(x)$) of $\Lambda^m\mathcal F$ at
$x$ equals $\Lambda^m\mathcal{F}_x$. This shows that $\mathcal{F}_x$ is an
$n$-dimensional vector space. After possibly shrinking $X$ we may assume that
there is a an $\mathcal{O}_X$-map $f\colon \mathcal{O}_X^n\to \mathcal F$ which induces
an isomorphism on fibres at $x$. Thus $\Lambda^nf$ is a map between locally free
modules (of rank $1$) that gives an isomorphism on fibres at $x$ and hence is an
isomorphism in a neighbourhood of $x$ so that we may assume that it is a global
isomorphism. The wedge product induces pairings
$\mathcal{F}\times\Lambda^{n-1}\mathcal{F}\to \Lambda^{n}\mathcal{F}$ and
$\mathcal{O}_X^n\times\Lambda^{n-1}\mathcal{O}_X^n\to
\Lambda^{n}\mathcal{O}_X^n$, the latter being a perfect pairing. Composing the
second with $\Lambda^nf$ gives a pairing
$\mathcal{O}_X^n\times\Lambda^{n-1}\mathcal{O}_X^n\to \Lambda^{n}\mathcal{F}$.
As $\Lambda^\ast f$ is multiplicative we get that the composite
$$\mathcal{O}_X^n\xrightarrow{f}\mathcal{F}\to
\mathrm{Hom}(\Lambda^{n-1}F,\Lambda^{n}\mathcal{F})\xrightarrow{\Lambda^{n-1}f^*}\mathrm{Hom}(\Lambda^{n-1}\mathcal{O}_X^n,\Lambda^{n}\mathcal{F})$$
equals the map induced by the pairing for $\mathcal{O}_X^n$. This is an
isomorphism (as $\mathcal{O}_X^n$ is free of rank $n$ and $\Lambda^nf$ is an
isomorphism) so we get that $f$ is split injective and we may write
$\mathcal{F}$ as `\mathcal{O}_X^n\bigoplus \mathcal G$ for some quasi-coherent sheaf
$\mathcal{G}$. Now, $\Lambda^n(\mathcal{O}_X^n\bigoplus \mathcal{G})$ splits up
as
$$
\bigoplus_{i+j=n}\Lambda^i\mathcal{O}_X^n\bigotimes \Lambda^j\mathcal{G}
$$
and $\Lambda^nf$ is the inclusion into the $j=0$ factor. As that inclusion is an
isomorphism, the other factors are zero but $\Lambda^{n-1}\mathcal{O}_X^n\bigotimes
\Lambda^1\mathcal{G}$ has $\mathcal{G}$ as a direct factor and hence $\mathcal{G}=0$.
A: Here's an argument for $n=2$.  After further localisation if necessary, we may assume that $X=\text{spec}(k)$ and that we have an isomorphism $\alpha:\Lambda^2(F)\simeq k$.  As this is surjective we see that $1$ can be written as a sum of terms $\alpha(u\wedge v)$; after yet more localisation we may assume that some such term is invertible, and then we can adjust the choice of $v$ to ensure that $\alpha(u\wedge v)=1$.  Now define $\beta:F\to k^2$ by $\beta(x)=(\alpha(x\wedge v),\alpha(u\wedge x))$, so $\beta(u)=(1,0)$ and $\beta(v)=(0,1)$.  It follows that $u$ and $v$ generate a free submodule of $F$ and that $F=ku\oplus kv\oplus F_0$ where $F_0=\ker(\beta)$.  This means that we can define a split monomorphism $\gamma:F_0\to\Lambda^2(F)$ by $\gamma(x)=u\wedge x$ but $\alpha\gamma=0$ by the definition of $F_0$ and $\alpha$ is an isomorphism so we must have $F_0=0$.  
