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Can you tell me, why the following is true: Let $C$ be a smooth, complete curve over an algebraically closed field. Let $D=P_1+...+P_n$ be an effective divisor that is linear combination of (not necessarily different) points $P_1,..., P_n \in C$. Let $P_n$ be not a basepoint of the linear system of $D$. Can you tell me why the dimension of $H^0(O_C(D))$ is smaller by at least $1$ than the dimension of $H^0(O_C(D-P_n))$, i.e. the dimension of the global sections drops by $1$ when removing $P_n$.

flag	asked Jul 11 2011 at 16:35 phil 91●2

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Since $P_n$ is not a base point of $|D|$, there is a section of $H^0(D)$ that does not vanish in $P_n$. So the inclusion $H^0(D-P_n) \subset H^0(D)$ is strict. On the other hand, imposing the passage through a point is at most one linear condition, so $H^0(D-P_n)$ has codimension exactly $1$ in $H^0(D)$. – Francesco Polizzi Jul 11 2011 at 16:57

Mike Skirvin · Answer 1 · 2011-07-11 16:48:58Z UTC

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This is exactly Proposition 3.1(a) in Chapter IV of Hartshorne. The proof, in my opinion, provides a good explanation of why the result is true.

answered Jul 11 2011 at 16:48

Mike Skirvin
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Removing non-basepoint from linear sytem.

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