# Bound on the dimension of $L(D)$ for a compact Riemann Surface

this is homework, but after banging my head against the wall for a day, I feel justified asking for a hint.

Let $X$ a compact Riemann Surface, $D,E \in Div(X)$ divisors on $X$ with $degE\geq 0$.

Show that $dimL(D+E) \leq dimL(D) + deg(E)$.

I think if I can show that for $E = 1x$, $dim(D+1x) \leq DimL(D) + 1$, then this follows by induction, but I don't even know how to show this. Ideas?

Please note that this class is rather elementary, so I am seeking a rather elementary answer.

Edit: Based off the hint, let $f_1$ and $f_2$ linearly independent in $L(D+x)$ and let the degree of $x$ in $D$ be given by $k$. Then in local coordinates on a neighborhood of $x$ we can write

$$f_1(z) = a/(z-x)^{k+1} + h(z)$$ and $$f_2(z) = b/(z-x)^{k+1} + g(z)$$

for $h,g\in L(D)$ and $a,b\in \mathbb{C}$. Observe that for any values of $a$ and $b$ there are coefficients $\alpha, \beta \in \mathbb{C}$ both not zero such that $\alpha f_1 +\beta f_2 \in L(D)$ (this requires enumerating a couple cases and showing we can get rid of the order $k+1$ pole) so at least one of the two is in $L(D)$. But then the dimension of $L(D+x)$ is at most one more than $L(D)$ since we have shown there is no two lin. indep. functions in $L(D+x)$ that are both not in $L(D)$.

To show $dimL(D+E) \leq dimL(D) + deg(E)$ for any positive $E$ we simply use induction.

I can also prove this using the same rough argument about the local laurent expansion using the other hint, but I will add that later. It should be exact because we can map $f(z) = a/(z-x)^{k+1} + h(z)$ to $a$ and the kernel will be $h\in L(D)$.

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The ultimate goal is to show that $L(D)$ is finite dimensional, but I have not even gotten past this step. – AnonymousCoward Mar 22 '11 at 2:54

Suppose that if you have two functions $f_1$ and $f_2$ in $L(D+x)$, show that for some scalars $\alpha$ and $\beta$, not both zero, that $\alpha f_1 + \beta f_2$ belongs in $L(D)$. (For this you have to go back to the very definitions of $L(D)$ and $L(D+x)$, and think about what it means for a meromorphic function to belong to these spaces.) From this, show that $\dim L(D + x) \leq \dim L(D) + 1$.
Hint: construct an exact sequence $0 \to L(D) \to L(D+x) \to \mathbb{C}$ where the last map is the residue at $x$ (at least if $x \notin supp(D)$; the argument can be modified if $x \in supp(D)$ accordingly).