Irreducible variety over an algebraically closed field which admits only constant regular functions

Question

Let $(X,\mathcal O_X)$ be an irreducible variety over an algebraically closed field $k$ such that $\mathcal O_X(X)\cong k$. Then is it true that $X$ is a complete variety ? If this is not true in general, what of we also assume $X$ is quasi-projective ?

Answer 1

Just to get this out of the unanswered queue.

We have the following nice proposition:

Theorem (Algebraic Hartog's Lemma): Let $X$ be an integral normal (Noetherian so I don't scare myself) scheme and let $Z\subseteq X$ be a closed subscheme of dimension at least $2$. Let $U:=X-Z$. Then, the restriction map $\mathcal{O}(X)\to \mathcal{O}(U)$ is an isomorphism.

You can find a proof of this, for example, in Vakil's book. The point is that if $A$ is a normal integral domain then $A$ is actually equal to the intersection of all the $A_\mathfrak{p}$ as $\mathfrak{p}$ ranges over all height $1$ primes. In particular, since our $U$ contains all codimension $1$ subsets, we know that any function will be in ever $A_\mathfrak{p}$ and thus in $A$.

Anyways, assume now that $X$ is geometrically integral and projective. Then, $\mathcal{O}(X)=k$. Let $Z\subseteq X$ be codimension at least $2$ and set $U:=X-Z$. Note then that $U$ is not projective. In fact, it's not even proper else the map $U\to X$ would have closed image, which it doesn't. That said, by Algebraic Hartog's Lemma we have that $\mathcal{O}(U)=k$.

So, for example, if $n\geqslant 2$ then for any closed point $x\in\mathbb{P}^n_k$ we have that $\mathbb{P}^n_k-\{x\}$ is not projective but has global sections $k$.