Let $C$ be a projective non-singular irreducible curve, let $D$ be a divisor on $C$. Suppose the Riemann-Roch Space is $L(D)=\langle f_1,...,f_n\rangle$. Define $\phi_D:C\to \mathbb{P}^{n-1}$ by $\phi_D(P)=(f_1(P):...,f_n(P))$. We say that $D$ is very ample if $\phi_D$ is an embedding.
Suppose $P\neq Q\in C$, $\phi_D(P)=(1:0:...:0),\phi_D(Q)=(0:1:...:0)$. I need to show that if $D$ is very ample, then $L(D-P)\neq L(D)$. I'm not sure how to go about that.
Some literatures incorporate a base point free property to the definition of very ample, but I don't have such a definition in this case.
Since $f_1$ achieves minimum order at $P$ across all $f\in L(D)$, is it true that $f_1\notin L(D-P)$? How to show this?
It is clear that $f_1\notin L(D-P)\iff L(D-P)\neq L(D)$, but I'm not sure how to prove either side of this equivalence.
Suppose that your complete linear system $|D|$ has base point, define the divisor $$ F=\min_{E\in |D|} E $$
Exercise: $|D-F|$ is base point free and $L(D)=L(D-F)$.
So we always can suppose that $|D|$ is base point free. In this case we can solve this with some propertys of the map $\phi_D$, for example $L(D-P)=L(D)$ implies that $P$ is base point of the complete linear system $|D|$.
Rmk: I think that for a good definition of very ample divisor it's necessary request that the complete linear system associated to this divisor is without base point.