Base point free for $g^1_2$ for hyperelliptic curve

Question

Let $C$ be a curve of genus larger than 1. $C$ is called hyperelliptic if it contains a $g_2^1$ linear system, meaning that $D$ is of degree $2$ with $\dim |D|=1$ if $D$ is such a divisor in this linear system. Then the book said that it has to be base point free. I did not get this result. I used Riemann-Roch and showed that $l(K-D)>0$. But I did not see why it is base point free.

Answer 1

Say $D= p_0 + p$, and suppose $p_0$ is a base-point. The linear system is of $g^1_2$ type means we can vary $p$ as we like and still stay in the linear system. But $p_0 + p \sim p_0 + p'$ implies $p \sim p'$ for any $p$ and $p'$. This occurs only for the Riemann sphere, not a hyperelliptic curve.

Answer 2

In general for every divisor $D$ and point $P$ we have that $\dim|D-P|=\dim|D|$ if and only if $P$ is a base point (from Hartshorne IV 3.1). As a result, if $P$ is a base point of the $g^1_2$ then $\dim|g^1_2-P|=\dim|g^1_2|=1$ by definition. But we have $\deg(g^1_2-P)=\deg(g^1_2)-1=2-1=1$, and degree can be equal to dimension only if $g=0$ or $D=0$ (Hartshorne exercice IV 1.5). As here $g\geq1$ we have $g^1_2=P$ which is impossible for degree reason.