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Let $C$ be a hyperelliptic curve of genus $g$ and let $D$ be a divisor on $C$ of degree $g+1$. Assume that the linear system $|D|$ is base-point-free. Now add a $2$-torsion point $[E]$ to $D$. I would like to know if the linear system $|D+E|$ is again base-point-free.

I think of the hyperelliptic curve as the smooth intersection of a $2$-dimensional rational normal scroll and a quadric. If it helps, I would be happy if the above statement is true for a general hyperelliptic curve in this sense, meaning that I get to choose a generic quadric.

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This will never happen. The linear systems of degree $g+1$ on $C$ with a base point are of the form $g^1_2+F$, with $F$ effective of degree $g-1$; in $J^{g+1}C$, they form a divisor $\Delta $ which is a copy of the theta divisor. For any $E\neq 0$ in $JC$ (2-torsion or not), the translate $\Delta -E$ is different from $\Delta $, therefore there exists $D\in (\Delta -E)\smallsetminus \Delta $; then $|D|$ is base-point free, but $|D+E|$ belongs to $\Delta $, hence has a base point.

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  • $\begingroup$ Thank you, that sounds cool. Why is it true that every linear system of degree $g+1$ with a base point is of the form $g^1_2+F$ as you wrote? Can you give a short argument or a reference? It sounds very helpful to me. $\endgroup$
    – Rainer Sinn
    Commented May 18, 2016 at 19:00
  • $\begingroup$ Let $|D|$ be your linear system and $p$ a base point; then $|D-p|$ has dimension $\geq 1$, so it contains a divisor of the form $\sigma p+E$, where $\sigma $ is the hyperelliptic involution and $E$ is effective. Thus $D\sim g^1_2+E$. $\endgroup$
    – abx
    Commented May 18, 2016 at 19:52

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