## 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.

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.

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

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

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

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[rd] & X_ A \ar[d] \\ & \mathop{\mathrm{Spec}}(A) }$

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

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

is commutative. This gives the following commutative diagram of rings

$\xymatrix{ K' & \mathcal{O}_{U, u} \ar[l] \\ K \ar[u] & \\ & A \ar[lu] \ar[uu] }$

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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