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. $\square$

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. $\square$

Proof. By definition a morphism of schemes $g : Y \to T$ is quasi-projective if it is locally of finite type, quasi-compact, and there exists a $g$-ample invertible sheaf on $Y$. Let $K/k$ be a field extension such that $X_ K$ is quasi-projective over $K$. Let $\mathop{\mathrm{Spec}}(A) \subset X$ be an affine open. Then $U_ K$ is an affine open subscheme of $X_ K$, hence $A_ K$ is a $K$-algebra of finite type. Then $A$ is a $k$-algebra of finite type by Algebra, Lemma 10.126.1. Hence $X \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type. Since $X_ K \to \mathop{\mathrm{Spec}}(K)$ is quasi-compact, we see that $X_ K$ is quasi-compact, hence $X$ is quasi-compact, hence $X \to \mathop{\mathrm{Spec}}(k)$ is of finite type. By Morphisms, Lemma 29.39.4 we see that $X_ K$ has an ample invertible sheaf. Then $X$ has an ample invertible sheaf by Lemma 33.15.1. Hence $X \to \mathop{\mathrm{Spec}}(k)$ is quasi-projective by Morphisms, Lemma 29.39.4. $\square$

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. $\square$

Proof. A scheme over $k$ is projective over $k$ if and only if it is quasi-projective and proper over $k$. See Morphisms, Lemma 29.43.13. Thus the lemma follows from Lemmas 33.15.3 and 33.15.4. $\square$