Lemma 33.15.2. Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_ K$ is quasi-affine for some field extension $K/k$, then $X$ is quasi-affine.

Proof. Let $K/k$ be a field extension such that $X_ K$ is quasi-affine. The morphism $X_ K \to X$ is surjective. Hence $X$ is quasi-compact as the image of a quasi-compact scheme (Properties, Definition 28.18.1). Since $X_ K$ is quasi-separated (as an open subscheme of an affine scheme) 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 $i$ such that $X_ i$ is quasi-affine (Limits, Lemma 32.4.12; here we use that $X$ is quasi-compact and quasi-separated as just shown). 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 is quasi-affine (Properties, Lemma 28.27.2). Since $X_{k'} \to X$ is finite locally free we conclude by Divisors, Lemma 31.17.10. $\square$

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