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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.

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up vote 5 down vote accepted

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 $F_0=0$.

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Can you expand/explain how you reduce to the case $X=\mathrm{Spec}(k)$? – Baptiste Calmès Jul 19 '11 at 9:52
@Baptiste: We may assume that $X$ is affine and I think that Neil means $k$ just to be a ring (not a field, in which case nothing has to be shown). @Neil: Thanks! I'll try to generalize this method of proof. – Martin Brandenburg Jul 19 '11 at 10:30
Neil, your proof generalizes to arbitrary $n$: Let $\wedge^n F$ be free of rank $1$. Writing a generator as sums of pure wedge products and localizing, we may arrange that the generator is pure, say $v_1 \wedge ... \wedge v_n$. Define $\phi : F \to (\wedge^n F)^n \cong A^n$ by $x \mapsto (v_1 \wedge ... \wedge v_{i-1} \wedge x \wedge v_{i+1} \wedge ... \wedge v_n)_{1 \leq i \leq n}$. Then $\phi$ is linear and satisfies $\phi(v_i)=e_i$. – Martin Brandenburg Jul 19 '11 at 15:31
Thus $v_1,...,v_n$ span a free submodule $U$ of $F$ and for $V:=\ker(\phi)$ we have $F = U \oplus V$. Thus $\wedge^n F$ is the direct sum of $\wedge^n U, \wedge^{n-1} U \otimes V, ... , \wedge^n V$. However, by construction $\wedge^n U \to \wedge^n F$ is an isomorphism. Thus all the other summands are zero, in particular $\wedge^{n-1} U \otimes V = 0$. Since $\wedge^{n-1} U$ is a free module of rank $\binom{n}{n-1}=n>0$, it follows $V=0$. Thus $F=U$ is free. – Martin Brandenburg Jul 19 '11 at 15:44

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 $\mathcal{G}=0$.

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Great! This equals Nick's proof (generalized to arbitrary $n$ as I've done in the comments there). – Martin Brandenburg Jul 19 '11 at 15:40
Very nice, Torsten: this explains a lot! – Georges Elencwajg Jul 19 '11 at 16:26

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