Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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 $F_0=0$.

share|improve this answer
    
Can you expand/explain how you reduce to the case $X=\mathrm{Spec}(k)$? –  Baptiste Calmès Jul 19 '11 at 9:52
1  
@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
1  
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
1  
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
add comment

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 $\mathcal{G}=0$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.