Lemma 69.21.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $Y$ is locally Noetherian and $f$ is of finite type. Then the following are equivalent

1. $f$ is universally closed,

2. $f$ satisfies the existence part of the valuative criterion,

3. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $|\mathbf{A}^ n \times X \times _ Y V| \to |\mathbf{A}^ n \times V|$ is closed for all $n \geq 0$,

4. for all diagrams (69.21.1.1) with $A$ a discrete valuation ring there there exists a finite separable extension $K'/K$ of fields, a discrete valuation ring $A' \subset K'$ dominating $A$, and a morphism $\mathop{\mathrm{Spec}}(A') \to X$ such that the following diagram commutes

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$
5. for all diagrams (69.21.1.1) with $A$ a discrete valuation ring there there exists a field extension $K'/K$, a valuation ring $A' \subset K'$ dominating $A$, and a morphism $\mathop{\mathrm{Spec}}(A') \to X$ such that the following diagram commutes

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$

Proof. Parts (1), (2), and (3) are equivalent by Lemma 69.20.2 and Morphisms of Spaces, Lemma 66.42.1. These equivalent conditions imply part (4) as Morphisms of Spaces, Lemma 66.41.3 tells us that we may always choose $K'/K$ finite separable in the existence part of the valuative criterion and this automatically forces $A'$ to be a discrete valuation ring by Krull-Akizuki (Algebra, Lemma 10.119.12). The implication (4) $\Rightarrow$ (5) is immediate. In the rest of the proof we show that (5) implies (1).

Assume (5). Chose an affine scheme $V$ and an étale morphism $V \to Y$. It suffices to show that the base change of $f$ to $V$ is universally closed, see Morphisms of Spaces, Lemma 66.9.5. Exactly as in the proof of Lemma 69.21.3 we see that assumption (5) is inherited by this base change; details omitted. This reduces us to the case discussed in the next paragraph.

Assume $Y$ is a Noetherian affine scheme and we have (5). To prove that $f$ is universally closed it suffices to show that $|X \times \mathbf{A}^ n| \to |Y \times \mathbf{A}^ n|$ is closed for all $n$ (by the discussion above). Since assumption (5) is inherited by the product morphism $X \times \mathbf{A}^ n \to Y \times \mathbf{A}^ n$ (details omitted) we reduce to proving that $|X| \to |Y|$ is closed.

Assume $Y$ is a Noetherian affine scheme and we have (5). Let $T \subset |X|$ be a closed subset. We have to show that the image of $T$ in $|Y|$ is closed. We may replace $X$ by the reduced induced closed subspace structure on $T$; we omit the verification that property (5) is preserved by this replacement. Thus we reduce to proving that the image of $|X| \to |Y|$ is closed.

Let $y \in |Y|$ be a point in the closure of the image of $|X| \to |Y|$. By Lemma 69.21.1 we may choose a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$

where $A$ is a discrete valuation ring and $K$ is its field of fractions mapping the closed point of $\mathop{\mathrm{Spec}}(A)$ to $y$. It follows immediately from property (5) that $y$ is in the image of $|X| \to |Y|$ and the proof is complete. $\square$

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).