Here is a problem I need help getting started with. I am brand new to algebraic geometry.
Let $C$ be a projective non-singular irreducible curve of genus $g$. Let $P_0\in C$. Prove
that any divisor $D \in Div(C)$ is linearly equivalent to a divisor of the form $D_0 – nP_0$,
where $D_0\ge 0$ has degree $g$. Of course, $n=g-\deg D$. Would $D_0$ have to be canonical?
I am given the Riemann-Roch Theorem but I don't know how to connect this question to the dimensions of the Riemann-Roch spaces.
What theorems should I be thinking about? Thanks.
Best Answer
Write $D'=D+(g-deg(D))P_0$ so the problem is to show that $D'$ is equivalente to a effective divisor, note that $deg(D')=g$ by Riemann-Roch theorem we have $$ \dim L(D')\geq 1 $$
take $f\in L(D')$ so by definition $D_0=div(f)+D'\geq 0$ and $D_0$ is equivalent to $D'$.
rmk$_1$: $L(A)=\{f\in \mathcal{M}(X):\textrm{ }ord_p(f)\geq A(p)\}$, and $div(f)=\sum_p ord_p(f)p$ is the divisor associated to a meromorphic function.
rmk$_2$: Of course that $deg(D_0)=g$ because the $deg(div(f))=0$.
rmk$_3$: $X$ just need to be a compact Riemann surface of genus $g$.