# Remove a non-base point from a linear system of a curve, How the dim changes?

Suppose X is a smooth projective curve. And D is an effective divisor. If P is a non-base point of the linear system |D|, what can we say about dim|D-P| ?

I was reading Hartshorne Chapter IV, section 6, page 351, proof of Thm6.4.(Castelnuovo). Let X be a degree d and genus g in $\mathbb{P}^3$. D is a hyperplane section, and write $D=P_1+P_2+...+P_d$. Then we can show that $P_i$ is a base point of $|nD-P_1-...-P_{i-1}|$. He claim that: each time we remove a non-base point from a linear system, the dimension drops by 1. I don't quite get this claim

• I think it's easier to think about twisting a line bundle — I think (hope) it's all equivalent. If I have a line bundle $\mathcal{L}$ then $\mathcal{L}(-P)$ has either the same number of global sections or one less [from tensoring the exact sequence $0\to\mathcal{O}_X(-P) \to \mathcal{O}_X \to k(P) \to 0$ with $\mathcal{L}$ and looking at the associated LES]. There's an inclusion $H^0(\mathcal{L}(-P)) \subseteq H^0(\mathcal{L})$, and if it's proper then that means there's a section of $\mathcal{L}$ not vanishing at $P$. – Hoot Jan 19 '16 at 13:26
• Thank you, this sovles my question – Shuhang Feb 1 '16 at 18:36