Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Maybe this is well-known, but suppose you have an invertible sheaf $\mathcal{L}$ on a scheme $X$. If $X$ is a projective space (or even projective bundle over an integral scheme, by a similar argument...I think), then if $H^0(X, \mathcal{L})\neq 0$ and $H^0(X, \mathcal{L}^\vee)\neq 0$ we get that $\mathcal{L}\simeq \mathcal{O}_X$. This is really simple because $\mathcal{L}\simeq \mathcal{O}_X(n)$, so you get that the two hypothesis mean $n\geq 0$ and $n\leq 0$, so $n=0$.

My guess is this must be true more generally. Are there conditions on $X$ such that any invertible sheaf with the property that having non-trivial global sections of both it and its dual implies it is trivial?

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

In general, this fails. Take $X = \mathrm{Spec}A$ for $A$ a Dedekind domain. Then line bundles correspond to equivalence classes of Weil divisors. A line bundle corresponds to some formal sum $\sum n_i \mathfrak{p}_i$. But you can always find something in $A$ whose divisor is bigger than this (just take it highly divisible at those primes). So any line bundle will satisfy your condition, but if $A$ is not a UFD it will not necessarily be trivial: the Picard group is the class group of $A$. (See 8.2 of http://people.fas.harvard.edu/~amathew/CRing.pdf for an explanation of this isomorphism and a general discussion of line bundles on affine schemes.)

This is true for $X$ an integral proper scheme over an algebraically closed field $k$. (It's in Mumford's Abelian Varieties.). You don't need the isomorphism of the Picard group with $\mathbb{Z}$ which is only anyway true when $X$ is all of projective space.

Suppose $\mathcal{L}$ is such a line bundle. Then there is a nonzero morphism $\mathcal{O}_X \to \mathcal{L}$ and a nonzero morphism $\mathcal{L} \to \mathcal{O}_X$. This is what it means for the space of global sections of $\mathcal{L}$ and its dual to not vanish. So we get a composition $$\mathcal{L} \to \mathcal{L},$$ which is given by a global regular function that can't be zero (as it isn't zero at the generic point). This is necessarily an element of the field $k$, so an isomorphism of line bundles. Similarly the composition $\mathcal{O}_X \to \mathcal{O}_X$ is an isomorphism. (I suppose the key property here is that a nonzero endomorphism of a line bundle is an isomorphism, and this in turn is clear from $\Gamma(X, \mathcal{O}_X) = k$, which is a consequence of properness.) It follows that $\mathcal{L} \to \mathcal{O}_X$ is an isomorphism and $\mathcal{L}$ is trivial.

This doesn't really need $k $ to be algebraically closed (though then $X$ geometrically integral over $k$ would be necessary to make the argument work, unless I'm missing something).

This is false without integrality hypotheses. (Take a disconnected scheme, and a line bundle trivial on one piece but not the other.)

share|improve this answer
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.