# If a line bundle and its dual both have a section (on a projective variety) does this imply that the bundle is trivial?

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?

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 $s\in \Gamma(X,L)$ .
• 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