Lemma 71.10.2. Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be a quasi-separated algebraic space over $k$.

1. If there exists a field extension $k \subset K$ such that $X_ K$ is a scheme, then $X_{\overline{k}}$ is a scheme.

2. If $X$ is quasi-compact and there exists a field extension $k \subset K$ such that $X_ K$ is a scheme, then $X_{k'}$ is a scheme for some finite separable extension $k'$ of $k$.

Proof. Since every algebraic space is the union of its quasi-compact open subspaces, we see that the first part of the lemma follows from the second part (some details omitted). Thus we assume $X$ is quasi-compact and we assume given an extension $k \subset K$ with $K_ K$ representable. Write $K = \bigcup A$ as the colimit of finitely generated $k$-subalgebras $A$. By Limits of Spaces, Lemma 69.5.11 we see that $X_ A$ is a scheme for some $A$. Choose a maximal ideal $\mathfrak m \subset A$. By the Hilbert Nullstellensatz (Algebra, Theorem 10.34.1) the residue field $k' = A/\mathfrak m$ is a finite extension of $k$. Thus we see that $X_{k'}$ is a scheme. If $k' \supset k$ is not separable, let $k' \supset k'' \supset k$ be the subextension found in Fields, Lemma 9.14.6. Since $k'/k''$ is purely inseparable, by Lemma 71.10.1 the algebraic space $X_{k''}$ is a scheme. Since $k''|k$ is separable the proof is complete. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).