I am currently reading Mumford's abelian varieties and Milne's notes on them and I have a problem understanding the proof that they are projective. Both of them use that a complete linear system of divisors on the abelian variety $V$ defines a morphism $V\rightarrow \mathbb{P}_k^n$ ($k$ can be assumed algebraically closed). The problem I have with this is that they seem to assume that the $k$-vector space $$ L(D)\cup \{0\}=\{f\in k(V)^{\times}|~div(f)+D\geq 0\}\cup \{0\} $$ is finite dimensional. Can anybody help me with this?
This is true for projective schemes $X$ by theorem 5.19 in Hartshorne that says that $\Gamma(X,F)$ is finite-dimensional for any coherent sheaf $F$ and our vector space is $\Gamma(X,\mathcal{O}(D))$. However we don't know that $V$ is projective yet. I think the crucial property of $V$ we need to use here is completeness.
I'll include definition and some basic properties: An abelian variety is per definition a complete connected group variety. They are non-singular, geometrically irreducible and commutative.
This is probably a little bit of an overkill, but anyway.
It seems that more generally, if $f: X \rightarrow S$ is proper, where $S=\operatorname{Spec}(A)$, with $A$ a noetherian ring, then for every coherent sheaf $\mathcal{F}$ the cohomology groups
$$ H^{i}(X,\mathcal{F}) $$
are $A$-modules of finite type, for every $i \in \Bbb{N}$. Here is an arxiv-reference; the author attributes the aforementioned result to Grothendieck.
What you are asking about is the particular case $i=0$, $X=V$ an abelian variety, $S=\operatorname{Spec}(k)$ and $\mathcal{F}=\mathcal{O}(D)$; then $$\Gamma(V,\mathcal{O}(D)) \cong H^0(V,\mathcal{O}(D))$$
is a finite-dimensional $k$-vector space.