# Canonical divisor of a curve base point free (if g>0)

Is there a way to prove that the canonical divisor $W$ of an algebraic function field in one variable $F$ over a field $K$ (that is the function field of an algebraic curve) of genus $g>0$ is base point free, without using CliffordÂ´s theorem?
Note that K is not necessarily algebraically closed! in that case I know how to solve the problem.

Equivalently, I have to show that for any place $P$ of $F/K$ there exists an holomorphic differential $\omega$ of $F/K$ such that $v_P(\omega)=0$, that is the support of the associated canonical divisor $(\omega)$ to $\omega$ does not contain the place $P$.

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Could you remind me (us) what Clifford's theorem says ? – Damian Rössler Nov 15 '11 at 15:47
if $A$ is a divisor of $F/K$ such that $0\leq\deg(A)\leq 2g-2$, then $$l(A)\leq 1+\deg(A)/2$$ where as usual $l(A)$ is the dimension over $K$ of the vector space $L(A)=\{x\in F|(x)+A\geq0\}$. – GiulioP Nov 15 '11 at 16:23
Surely the case where $K$ is not algebraically closed follows from the case where $K$ is algebraically closed? – Daniel Loughran Nov 15 '11 at 16:38
In the algebraically closed case (that is enough for your purpose, as remarked above) it follows directly by Riemann-Roch, without using Clifford's theorem. – rita Nov 15 '11 at 16:42
exactly, that is what I meant... I used clifford´s theorem (as in the exercises of Stichtenoth book) to solve the problem, but I would like a simple proof which avoids that. Essentially is the fact that if $l(A)=\deg(A)+1$ in genus greater than zero then $A$ is principal, which I can´t prove. Is it possible to prove it without extending the base field to the algebraic closure, as it seems that you suggest? It seems that the result is contained in the book of Deuring, "Lectures in the theory of alf. func.", Lemma Lectures 10, §20, but I don´t understand the proof... – GiulioP Nov 15 '11 at 17:21