# Every line bundle on a complex algebraic curve has a meromorphic section

Every line bundle $$L$$ on a complex algebraic curve $$X$$ is of the form $$\mathcal{O}(D)$$, where $$D$$ is some divisor on $$X$$. This means $$L$$ has at least one nonzero meromorphic global section, i.e. $$H^0(L \otimes_{\mathcal{O}_X} \mathcal{M}_X) \neq 0$$ where $$\mathcal{M}_X$$ is the sheaf of meromorphic functions on $$X$$.

Is there any way to see this by directly calculating global sections of the sheaf $$L \otimes_{\mathcal{O}_X} \mathcal{M}_X$$ and without assuming $$L \cong \mathcal{O}(D)$$ for some divisor $$D$$? I've seen other arguments so I'm mostly just interested in a calculation of the above cohomology group.

More generally, how does one compute $$H^0(V \otimes_{\mathcal{O}_X} \mathcal{M}_X)$$, where $$V$$ is an arbitrary vector bundle on $$X$$?

• @user347489 I don't think it does. I'd like to prove directly that $L \otimes_{\mathcal{O}_X} \mathcal{M}_X$ has a global section without assuming $L \cong \mathcal{O}(D)$. I'll edit to clarify. – leibnewtz Feb 22 at 9:18
• Let $U$ an open where $X$ is trivialised, then $1 : U \to \mathcal O_{U}$ is a meromorphic section. – Nicolas Hemelsoet Feb 22 at 12:19