Line bundle from $\mathcal{O}(p)\cong \mathcal{O}(q)$

Question

Let $X$ be a compact Riemann surface. If $\mathcal{O}(p)\cong \mathcal{O}(q)$. How to see there exists a line bundle $L$ and $s_1,s_2$ two sections of $L$ such that $s_1$ vanishes only at $p$ and $s_2$ vanishes only at $q$ (and with multiplicity one).

Answer 1

This is somewhat tautological.

Denote $\mathcal{L}=\mathcal{O}(p)=\mathcal{O}(q)$.

Let's show that $\mathcal{L}$ does the job.

Start with the standard restriction sequence:

$$0 \to \mathcal{O}(-p) \to \mathcal{O} \to \mathcal{O}_p \to 0$$ and twist by $\mathcal{O}(p)$ to get $$0 \to \mathcal{O} \to \mathcal{O}(p) \to \mathcal{O}_p \to 0.$$

Taking the cohomology sequence we get $$0 \to H^0 (\mathcal{O}) \xrightarrow{\phi}H^0(\mathcal{O}(p)) \xrightarrow{\psi} H^0 (\mathcal{O}_p)$$ and using that $H^0 (\mathcal{O}) \neq 0$ we get a nonzero section $\phi(1)$ of $\mathcal{O}(p) = \mathcal{L}$ which first of all vanishes at $p$ (since it is in the image of $\phi$ and therefore in the kernel of $\psi$), second it has no poles (since it is a global section of a line bundle) and third must have 1 zero in total since $\deg \mathcal{O}(p) = 1$.

Repeating this construction with $q$ instead of $p$ and using $\mathcal{O}(q) = \mathcal{L}$ will finish the proof.