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Is there any reason that, on a projective variety X, if a line bundle L has a (non-zero) section and also its dual has a section then this implies that L is the trivial line bundle?

Jacob Bell
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  • Your question is not a duplicate (and I like this question), but perhaps the answer should be accompanied by this link which finishes the proof: https://math.stackexchange.com/q/1397283/346324 – Tanner Strunk Nov 23 '17 at 06:09

1 Answers1

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Yes, there is a reason for $L$ to be trivial and here it is:

Let $0\neq s\in \Gamma(X,L)$ and $0\neq \sigma\in \Gamma(X,L^*)$ be two non zero sections.
Then $s\otimes \sigma\in \Gamma(X,L\otimes L^*)=\Gamma(X,\mathcal O)$ is a constant since $X$ is complete: $s\otimes \sigma =c\in k$ (the base field).

Now, since $s$ and $\sigma$ are non-zero there is a non-empty open subset $U\subset X$ on which both do not vanish and on which $s\otimes \sigma=c $ does not vanish either: in other words $c\neq0\in k$ .
Since $s\otimes \sigma =c\neq 0$, a non-zero constant, vanishes nowhere we conclude that a fortiori $s$ vanishes nowhere, so that $L$ is trivial, as announced, since $ s\in \Gamma(X,L)$ .

Tanner Strunk
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Georges Elencwajg
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    Hence, we have to suppose that $X$ is an integral scheme, proper over $k$. In this case $\Gamma(X,\mathcal{O})$ is a finite field extension of $k$. – Andrea Jun 06 '12 at 08:54
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    @Andrea: yes. When a user mentions projective varieties I try to interpret his question in the most elementary way possible. It is then generally possible for more advanced readers to adapt the solution to a more sophisticated context, if they so wish. – Georges Elencwajg Jun 06 '12 at 09:17
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    thanks to both of you. I tried to play with tensoring them together but didn't think to restrict to some open subset! – Jacob Bell Jun 06 '12 at 10:35