Lemma 33.15.4. Let k be a field. Let X be a scheme over k. If X_ K is proper over K for some field extension K/k, then X is proper over k.
Proof. Let K/k be a field extension such that X_ K is proper over K. Recall that this implies X_ K is separated and quasi-compact (Morphisms, Definition 29.41.1). 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 separated 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). By Limits, Lemma 32.13.1 there exists an i such that X_ i \to \mathop{\mathrm{Spec}}(A_ i) is proper. Here we use that X is quasi-compact and quasi-separated as just shown. Choose a maximal ideal \mathfrak m \subset A_ i. By the Hilbert Nullstellensatz (Algebra, Theorem 10.34.1) the residue field k' = A_ i/\mathfrak m is a finite extension of k. The base change X_{k'} \to \mathop{\mathrm{Spec}}(k') is proper (Morphisms, Lemma 29.41.5). Since k'/k is finite both X_{k'} \to X and the composition X_{k'} \to \mathop{\mathrm{Spec}}(k) are proper as well (Morphisms, Lemmas 29.44.11, 29.41.5, and 29.41.4). The first implies that X is separated over k as X_{k'} is separated (Morphisms, Lemma 29.41.11). The second implies that X \to \mathop{\mathrm{Spec}}(k) is proper by Morphisms, Lemma 29.41.9. \square
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