Is a degree zero divisor on a curve always basepoint-free?

Question

Let $X$ be a smooth projective curve and $D$ a divisor on $X$ of degree zero.

Is it always the case that $D$ is basepoint-free?

If not, then is there always some (positive) power $mD$ which is?

Answer 1

If there are no (nonzero) global sections of $O_X(D)$, then $D$ cannot be base point free. From theorem 19.2.4 in Vakil's Foundations of Algebraic Geometry, if $L$ is a degree 0 invertible sheaf and $L \neq O_C$ then $h^0(C, L) = 0$.

The proof of theorem 19.2.4 is : let $s$ be a global section of a degree 0 invertible sheaf $L$. Then, $s$ has no poles, and since $L$ is degree 0, it has no zeroes either. Therefore, the divisor corresponding to $s$ is $0$, and multiplication by $s$ is an isomorphism $L \simeq O_C$.

Therefore, choose any degree 0 invertible sheaf that is not the structure sheaf. That is a degree 0 invertible sheaf that is not base point free.