I am trying to prove the following question from here without relying on the Riemann-Roch theorem.
Let $X$ be a compact Riemann surface, and let $D$ be a divisor on $X$.
(i) If $\mathrm{deg}(D) = 0$, show that $\mathrm{dim}(L(D))$ is equal to $0$ or $1$, with the latter occurring if and only if $D$ is principal. Furthermore, any non-zero element of $L(D)$ has divisor $-D$.
(ii) If $\mathrm{deg}(D) \geq 0$, establish the bound $\mathrm{dim}(L(D)) \leq \mathrm{deg}(D) + 1$.
Here is my work so far:
If $D = 0$ then we have that $L(0) = 0$ as $X$ is compact, with the opening mapping theorem, implies that $L(0)$ is the space of constant functions. However, I am not sure what to do even for the simple case where $D = (P) – (Q)$. any $f \in L(D)$ must have at least a simple zero at $Q$ and at most a simple pole at $P$. How can I get further information about $f$? Similarly, the only one that I could prove so far for (ii) is the case when $D$ is trivial.
Thank you so much for your help.
Best Answer
i) If $\deg D=0$ and $\dim L(D)>0$, then $D$ is linearly equivalent to some effective divisor $D'$, having the same degree: $\deg D'=0$. But there is only one such divisor: $D'=0$ and so $D$ is principal.
Moreover $0=D'=D+(f)$, so $(f)=-D$.
ii) can be easily done by induction. i) takes care of the starting step.
Now assume that for any divisor $D$ of degree $n\geq 0$ it holds $\dim L(D) \leq n+1$, and let $D'$ be a divisor of degree $n+1$. Write it as $D+p$ ($p\in X$ a point) and apply Lemma 19: $$\dim L(D+p)\stackrel{Lemma \, 19}\leq 1+\dim L(D) \leq 1+1+\deg D= 1+ (n+1)$$