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. $\square$

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).