Why does $L(0)=k$?

Question

Definition 1: For every divisor $D=\sum_{P\in C}n_PP$ over a curve $C$, we define the vectorial space:

$L(D)\doteqdot\{f\in k(C);\text{ord}_P(f)\ge -n_P,\forall P\in C\}$

Furthermore, $L(D)$ is a vectorial space over $k$ and we denote $l(D)\doteqdot\dim L(D)$.

Following these definitions why $L(0)=k$?

Remark1: Fulton proves this using a corollary and a proposition, I would like to know if there would be a more direct (maybe easier?) proof of this fact, i.e., using only these definitions.

Thanks in advance

Usually, one takes $f\in k(C)^*$ and add zero to $L(D)$. But if $n_P\ge 0$ for each $P$ it is the same. — Jérémy Blanc, Jan 13 '15 at 23:30

score 4 · Answer 1 · edited Apr 13 '17 at 12:20

Using your definition, $L(0)$ is the set of elements $f\in k(C)$ such that $ord_P(f)\ge 0$ for each point $P$. Of course every constant function is like this. Then, you need to see that only constant functions are possible. Your element $f\in k(C)$ corresponds to a regular morphism $f\colon C\to \mathbb{A}^1$, so you need to show that only constant morphisms are possible.

This is true for any irreducible projective algebraic variety $C\subset \mathbb{P}^n$. Indeed, the image by a morphism of a projective variety is closed (see Do there exist regular functions on the projective space over a field? ), and so is here a constant as the image is in $\mathbb{A}^1$.

The tools I can use is only these definitions I wrote, thanks anyway. — user42912, Jan 14 '15 at 12:33

Why does $L(0)=k$?

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