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.



Characterization of locally free modules via exterior powers
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 $F\_0=0$. 


