This exercise is from Fulton's Algebraic Curves, exercise 8.13, page 101. The exercise reads as follows
Suppose $l(D)>0$ and let $f \neq 0, f \in L(D)$. Show that $f \notin L(D-P)$ for all but a finite number of $P$. So $l(D-P)=l(D)-1$ for all but a finite number of $P$.
Here $L(D)$ represents the linear system (or series if you prefer) of some divisor $D$, while $l(D)$ denotes the dimension of this vector space.
My Attempt
As $l(D) > 0$, without loss of generality we may assume that $D$ is effective, so we can write $D = \sum_{Q \in I} n_QQ $ where $I$ is some finite set of points and all of the $n_Q > 0$. My guess is that we require $P \in I$ for the proposition to hold. Suppose instead that $P \notin I$. Now for $f \in L(D)$ we see that $val_P(f) \geq 0$, while for $g \in L(D-P)$, we need $val_P(g) \geq 1$. Additionally, if $P \in I$, we have $val_P(f) \geq -n_P$ and for any $g \in L(D-P)$ we have $val_P(g) \geq 1-n_P$.
In both of these cases it seems to me that there could be some $f$ which could be an element of $L(D-P)$ for any chosen $P$. Could someone show me what I'm missing with this problem, I feel like it should be straightforward but I can't figure out what to try next.
Best Answer
Write $D = \sum_{Q \in I} n_Q Q$ for some finite set of points $I$. Let $f \in L(D)$.
Assume that there is some infinite set $J$ such that for all $P$ in $J$, $f \in L(D - P)$. For each point $P \in J - I$, we have that $n_P = 0$. Since $f \in L(D - P)$, we have $v_p(f) + n_P - 1 \geq 0$, i.e; $v_P(f) \geq 1$ for all $P \in J - I$. This is impossible as $v_P(f)$ is nonzero only for finitely many points.
Thus $f \notin L(D - P)$ for all but a finite amount of $P$.