TITLE: Global sections of a proper variety over an arbitrary field
QUESTION [5 upvotes]: While doing some research, I stumbled upon the following fact that I'd taken for granted.
Theorem: Let $ X $ be a proper variety over a field $ k $ (variety = geometrically integral, separated, finite type). Then $ \Gamma(X, \mathcal{O}_X) $ is a finite field extension of $ k $.
The line of proof I have in mind is as follows: Let $ s \in \Gamma(X, \mathcal{O}_X) $ be a global section. It corresponds to a morphism $ s : X \rightarrow \mathbb{A}^1_k $. This morphism factors as $$ X \xrightarrow{(id_X, s)} X \times_k \mathbb{A}^1_k \xrightarrow{p_2} \mathbb{A}^1_k $$ where $ (id_X, s) $ is a section of the first projection $ p_1 : X \times_k \mathbb{A}^1_k \rightarrow X $. In the composition, the first morphism is a closed immersion (being a section of the separated morphism $ p_1 $) and the second is a closed map by properness. So $ s $ is a closed map.
The image of $ s $ is proper, closed and irreducible, hence is a single closed point in $ \mathbb{A}^1_k $, say given by an irreducible polynomial $ h(T) \in k[T] $. This shows that $ h(s) = 0 $ in $ \Gamma(X, \mathcal{O}_X) $ and hence the global sections form a field.
If $ k $ was assumed algebraically closed, then $ h $ must be of the form $ T - a $ for some $ a \in k $. In this case, we recover the result that $ \Gamma(X, \mathcal{O}_X) = k $. However if $ k $ is arbitrary, I don't see how to get finite dimensionality of $ \Gamma(X, \mathcal{O}_X) $ although $ s $ itself lies in a finite extension.
Question: How to complete the proof? Does one rely on Grothendieck's (hard) result that for a coherent sheaf on a proper variety, all cohomology groups are finite dimensional? Or is there a proof not using this result? Most references either just cite this fact or deal with the algebraically closed case only.

REPLY [2 votes]: This is a community wiki post recording the discussion from the comments so that this question might be marked as answered, once this post is accepted or upvoted.

If you assume $X$ is geometrically integral, then $\Gamma(X,\mathcal{O}_X)=k$. For an arbitrary connected and proper scheme $X$ of finite type over $k$, you can extend scalars to an algebraic closure, say $k\subset\overline{k}$. In this case the scheme $X_\overline{k}$ is proper and of finite type over $\overline{k}$ so $\Gamma(X_\overline{k},\mathcal{O}_{X_\overline{k}})$ is a finite dimensional $\overline{k}$-vector space of dimension the number of connected components. But $\Gamma(X,\mathcal{O}_X)\otimes_k\overline{k}=\Gamma(X_\overline{k},\mathcal{O}_{X_\overline{k}})$. – Eoin