Example of divisor $D$ such that $\deg D >0$ and $\ell(D)=0$ It is easy to see that if a divisor $D$ on a projective curve $C$ over a field $K$ has negative degree, then $\ell(D) = \dim_K \{f \in K(C) \mid div(f)+D\ge 0\}$ is zero. However, I suppose that the converse is not true. Can someone give me the simplest example of a divisor $D$ on some curve $C$ satisfying $\deg(D)>0$ but $\ell(D)=0$? 
 A: Let $X$ be a smooth curve of genus at least 2, and pick three distinct points $P_1,P_2,Q_1$. Let $D=P_1+P_2-Q_1$. Suppose that $D$ is linearly equivalent to an effective divisor. Since $D$ is of degree one, such an effective divisor must be a point. Call this point $Q_2$.
By the definition of linear equivalence, this means that there's a rational function on $X$ with zeroes at $P_1,P_2$ and poles at $Q_1,Q_2$. But this is the same as a degree 2 morphism $X\to\Bbb P^1$ sending $P_1,P_2$ to the same point. If $X$ is not hyperelliptic (ie does not possess a degree 2 map to $\Bbb P^1$), then this is enough.
In the case that $X$ is hyperelliptic and there exists a degree 2 map to $\Bbb P^1$ taking $P_1,P_2$ to the same point, we have a little more work to do. We'll adjust our choice of $P_2$ so that $P_1$ and $P_2$ no longer are mapped to the same point by any degree 2 morphism to $\Bbb P^1$. In this case, the morphism is given by $|K_X|$. Choose a new $P_2$, called $P_2'$, which is different from our original choice of $P_2$. If $P_1+P_2'\in|K_X|$, then we have $P_2\sim P_2'$, which would imply that $X\cong \Bbb P^1$, and else, we have $P_1+P_2'\notin |K_X|$. 
A: Let $X$ be a genus 2 smooth curve. Take three general points $P,Q,R$ and let $D=P+Q-R$. One can be a bit more precise about these points if one wants to, using the hyperelliptic involution on $X$. 
