Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 12 down vote accepted

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 $c$ 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)$ .

share|cite|improve this answer
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 6 '12 at 8:54
@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 6 '12 at 9:17
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 6 '12 at 10:35

Your Answer


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.