# Existence of divisor which linear system is base point free

Let $X$ be non-hyperelliptic of genus $g \geq 3$. Show that there exist $D=p_1+\cdots+p_{g+1}$ in $X$ such that the linear system $|D|$ is base point free and $\dim |D| = 1$.

My attempt:

Because $X$ is non-hyperelliptic, by Geometric Riemann Roch we have $\dim\text{span}(p_1,\cdots,p_{g+1})=\deg(D)-\dim|D|-1=g-1$. This shows that The indued map by canonical line bundle $\phi_K:X\to\mathbb{P}^{n-1}$ is embedding and surjective. Then it's remain to find a base point free $D$, or $h^0 (D\times\mathcal{O}(-p))=h^0(D)-1$ for any $p$. Then how do I proceed?

This is true for any general points $p_i$. Show that for general points $p_i$, $1\leq i\leq g$, $h^0(K-p_1-p_2-\cdots-p_i)=g-i$.
• @ryanriess Can you use this to show that for general points $p_i, 1\leq i\leq g+1$, $h^0(K-D)=0$ for $D$ sum of $g$ points among these $g+1$ points? Then show that $\sum p_i$ has the right property you desire? – Mohan May 4 '18 at 1:44
• @ryanriess The term is used somewhat loosely and understood in contexts. In this particular case, if you think of an effective divisor of degree $d$ as a point of $X\times\cdots \times X=X^d$, then general means they lie in an unspecified non-empty open set of $X^d$. – Mohan May 4 '18 at 13:18