The Stacks project

Proof. Let $K/k$ be a field extension such that $X_ K$ has an ample invertible sheaf $\mathcal{L}$. The morphism $X_ K \to X$ is surjective. Hence $X$ is quasi-compact as the image of a quasi-compact scheme (Properties, Definition 28.26.1). Since $X_ K$ is quasi-separated (by Properties, Lemma 28.26.7) we see that $X$ is quasi-separated: If $U, V \subset X$ are affine open, then $(U \cap V)_ K = U_ K \cap V_ K$ is quasi-compact and $(U \cap V)_ K \to U \cap V$ is surjective. Thus Schemes, Lemma 26.21.6 applies.

Write $K = \mathop{\mathrm{colim}}\nolimits A_ i$ as the colimit of the subalgebras of $K$ which are of finite type over $k$. Denote $X_ i = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(A_ i)$. Since $X_ K = \mathop{\mathrm{lim}}\nolimits X_ i$ we find an $i$ and an invertible sheaf $\mathcal{L}_ i$ on $X_ i$ whose pullback to $X_ K$ is $\mathcal{L}$ (Limits, Lemma 32.10.3; here and below we use that $X$ is quasi-compact and quasi-separated as just shown). By Limits, Lemma 32.4.15 we may assume $\mathcal{L}_ i$ is ample after possibly increasing $i$. Fix such an $i$ and let $\mathfrak m \subset A_ i$ be a maximal ideal. By the Hilbert Nullstellensatz (Algebra, Theorem 10.34.1) the residue field $k' = A_ i/\mathfrak m$ is a finite extension of $k$. Hence $X_{k'} \subset X_ i$ is a closed subscheme hence has an ample invertible sheaf (Properties, Lemma 28.26.3). Since $X_{k'} \to X$ is finite locally free we conclude that $X$ has an ample invertible sheaf by Divisors, Proposition 31.17.9. $\square$