Proposition 66.16.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-compact, and $X$ is decent. Then $f$ is universally closed if and only if the existence part of the valuative criterion holds.

## 66.16 Valuative criterion

For a quasi-compact morphism from a decent space the valuative criterion is necessary in order for the morphism to be universally closed.

**Proof.**
In Morphisms of Spaces, Lemma 65.42.1 we have seen one of the implications. To prove the other, assume that $f$ is universally closed. Let

be a diagram as in Morphisms of Spaces, Definition 65.41.1. Let $X_ A = \mathop{\mathrm{Spec}}(A) \times _ Y X$, so that we have

By Morphisms of Spaces, Lemma 65.8.4 we see that $X_ A \to \mathop{\mathrm{Spec}}(A)$ is quasi-compact. Since $X_ A \to X$ is representable, we see that $X_ A$ is decent also, see Lemma 66.5.3. Moreover, as $f$ is universally closed, we see that $X_ A \to \mathop{\mathrm{Spec}}(A)$ is universally closed. Hence we may and do replace $X$ by $X_ A$ and $Y$ by $\mathop{\mathrm{Spec}}(A)$.

Let $x' \in |X|$ be the equivalence class of $\mathop{\mathrm{Spec}}(K) \to X$. Let $y \in |Y| = |\mathop{\mathrm{Spec}}(A)|$ be the closed point. Set $y' = f(x')$; it is the generic point of $\mathop{\mathrm{Spec}}(A)$. Since $f$ is universally closed we see that $f(\overline{\{ x'\} })$ contains $\overline{\{ y'\} }$, and hence contains $y$. Let $x \in \overline{\{ x'\} }$ be a point such that $f(x) = y$. Let $U$ be a scheme, and $\varphi : U \to X$ an étale morphism such that there exists a $u \in U$ with $\varphi (u) = x$. By Lemma 66.7.2 and our assumption that $X$ is decent there exists a specialization $u' \leadsto u$ on $U$ with $\varphi (u') = x'$. This means that there exists a common field extension $K \subset K' \supset \kappa (u')$ such that

is commutative. This gives the following commutative diagram of rings

By Algebra, Lemma 10.49.2 we can find a valuation ring $A' \subset K'$ dominating the image of $\mathcal{O}_{U, u}$ in $K'$. Since by construction $\mathcal{O}_{U, u}$ dominates $A$ we see that $A'$ dominates $A$ also. Hence we obtain a diagram resembling the second diagram of Morphisms of Spaces, Definition 65.41.1 and the proposition is proved. $\square$

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