Proposition 67.16.1. 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.

## 67.16 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 66.42.1 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 66.41.1. Let $X_ A = \mathop{\mathrm{Spec}}(A) \times _ Y X$, so that we have

By Morphisms of Spaces, Lemma 66.8.4 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 67.5.3. 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 67.7.3 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 10.50.2 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 66.41.1 and the proposition is proved. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)