Let $\mathscr{L}$ be a line bundle on a curve $C$ such that $h^0(C, \mathscr{L}) = 1$ and deg $\mathscr{L} = 0$. Why does it imply that $\mathscr{L}$ is the trivial line bundle?
I found some explanation in Vakil's notes here, which I don't quite understand. We know that degree of $\mathscr{L}$ can be computed by counting zeros and poles of any rational section. Say $s$ is a section of $\mathscr{L}$, then $s$ must have no pole, and since $\mathscr{L}$ has degree 0, $s$ must also have no zero. That means $s$ is invertible. But why does this then imply that $\mathscr{L}$ is the trivial line bundle?
Best Answer
In the language of sheaves, you can define an isomorphism of sheaves $\mathscr{O}_C \rightarrow \mathscr{L}$ where on each open $U$ sends $a \mapsto as|_U$.