Showing that $\Gamma(X,\mathcal{O}_X)=k$

Question

Let $X$ be an integral scheme, proper over an algebraically closed field $k$. It is known that $\Gamma(X,\mathcal{O}_X)=k$. In the book that I am reading, the author claims that this fact follows from the following fact:

If an integral affine scheme $\operatorname{Spec}A$ is proper over a field $k$, then $A$ is also a field, i.e. $\operatorname{Spec}A$ is a single point.

My question is, how do we use this fact to deduce that $\Gamma(X,\mathcal{O}_X)=k$? I have been pondering over this for awhile and am unable to get why. Any hints given would be greatly appreciated!

Answer 1

Any proper morphism is of finite type by definition, so $A$ is a field which is finitely generated as an algebra over $k$. By Zariski's lemma, any such field must be finitely generated as a module over $k$. Since $k$ is algebraically closed by assumption, we conclude $A = k$.

Answer 2

I’m unsure what the author meant, but there’s a standard proof as follows: a global section $f \in \mathcal{O}_X(X)$ induces a natural morphism $\hat{f}: X \rightarrow \mathbb{A}^1$. Composing with the natural open immersion $i$ from the affine space to the projective space, we get a morphism of proper $k$-schemes $\hat{f}’: X \rightarrow \mathbb{P}^1$, so it is closed. Thus the (topological) image $I$ of this morphism is a proper closed irreducible subset of $\mathbb{P}^1$, because $X$ is irreducible.

Therefore $I$ is a $k$-point of $\mathbb{P}^1$ (because $k$ is algebraically closed) thus $\hat{f}$ is a morphism to a $k$-point, thus $f$ is an element of $k$ plus a nilpotent – as $X$ reduced, $f \in k$ and we are done.