Lemma 66.42.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

$f$ is quasi-compact, and

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

Then $f$ is universally closed.

Lemma 66.42.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

$f$ is quasi-compact, and

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

Then $f$ is universally closed.

**Proof.**
By Lemmas 66.8.4 and 66.41.7 properties (1) and (2) are preserved under any base change. By Lemma 66.9.5 we only have to show that $|T \times _ Y X| \to |T|$ is closed, whenever $T$ is an affine scheme over $S$ mapping into $Y$. Hence it suffices to prove: If $Y$ is an affine scheme, $f : X \to Y$ is quasi-compact and satisfies the existence part of the valuative criterion, then $f : |X| \to |Y|$ is closed. In this situation $X$ is a quasi-compact algebraic space. By Properties of Spaces, Lemma 65.6.3 there exists an affine scheme $U$ and a surjective étale morphism $\varphi : U \to X$. Let $T \subset |X|$ closed. The inverse image $\varphi ^{-1}(T) \subset U$ is closed, and hence is the set of points of an affine closed subscheme $Z \subset U$. Thus, by Algebra, Lemma 10.41.5 we see that $f(T) = f(\varphi (|Z|)) \subset |Y|$ is closed if it is closed under specialization.

Let $y' \leadsto y$ be a specialization in $Y$ with $y' \in f(T)$. Choose a point $x' \in T \subset |X|$ mapping to $y'$ under $f$. We may represent $x'$ by a morphism $\mathop{\mathrm{Spec}}(K) \to X$ for some field $K$. Thus we have the following diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-{x'} \ar[d] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \ar[r] & Y, } \]

see Schemes, Section 26.13 for the existence of the left vertical map. Choose a valuation ring $A \subset K$ dominating the image of the ring map $\mathcal{O}_{Y, y} \to K$ (this is possible since the image is a local ring and not a field as $y' \not= y$, see Algebra, Lemma 10.50.2). By assumption there exists a field extension $K'/K$ and a valuation ring $A' \subset K'$ dominating $A$, and a morphism $\mathop{\mathrm{Spec}}(A') \to X$ fitting into the commutative diagram. Since $A'$ dominates $A$, and $A$ dominates $\mathcal{O}_{Y, y}$ we see that the closed point of $\mathop{\mathrm{Spec}}(A')$ maps to a point $x \in X$ with $f(x) = y$ which is a specialization of $x'$. Hence $x \in T$ as $T$ is closed, and hence $y \in f(T)$ as desired. $\square$

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