I am trying to do an exercise that goes like this
Let $X$ be an algebraic curve and $D$ a divisor on $X$ such that $deg(D)>0$, recall that $dim L(D) \leq 1+deg(D)$, show that equality holds if and only if $X$ has genus zero.
Now my doubt is is it true that $dim L(D)\leq 1+deg(D)$? I know this is true if the disivor is non-negative, but I dont see why this is true for every divisor. Maybe an application of the Riemann-Roch theorem that I a missing or something ? All I see is that $dim L(D) \leq dim L(K-D)+deg(D)+1$, where $K$ is a canonical divisor, and I dont see how I can have control over the term $dim L(K-D)$ if at priori I dont know that the curve as genus $0$.
Thanks in advance.
Best Answer
Assume that $\mathrm{dim}(L(D))=\mathrm{deg}(D)+1$ for some divisor with positive degree. Then there exists a non-constant function $f$ in $\mathrm{dim}(L(D))$, hence $D+(f)\geq 0$ and we can replace $D$ by the non-negeative divisor $D+(f)$ of the same degree and dimension.
Let $D=P_1+\ldots +P_d$, where $P_k$ are prime divisors and $d=\mathrm{deg}(D)$. Then for every $k\in\{1,\ldots ,d-1\}$:
$\mathrm{dim}(L(P_1+\ldots +P_{k+1}))\leq\mathrm{dim}(L(P_1+\ldots +P_{k}))+1$.
Since $\mathrm{dim}(L(D))=\mathrm{deg}(D)+1$ there must be equality for all $k$; in particular we get $\mathrm{dim}(L(P_1))=2$. Hence there exists a meromorphic function $g$ on $X$ having pole divisor $P_1$. The induced map $g:X\rightarrow S^1$ is an isomorphism.