The Stacks project

Proof. By Lemmas 67.8.4 and 67.41.7 properties (1) and (2) are preserved under any base change. By Lemma 67.9.5 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 66.6.3 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

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. $\square$

Proof. If (1) is true then combined with Lemma 67.42.1 we obtain (2). Assume $f$ is quasi-separated and universally closed. Assume given a diagram

as in Definition 67.41.1. A formal argument shows that the existence of the desired diagram

follows from existence in the case of the morphism $X_ A \to \mathop{\mathrm{Spec}}(A)$. Since being quasi-separated and universally closed are preserved by base change, the lemma follows from the result in the next paragraph.

Consider a solid diagram

where $A$ is a valuation ring with field of fractions $K$. By Lemma 67.8.9 and the fact that $f$ is quasi-separated we have that the morphism $x$ is quasi-compact. Since $f$ is universally closed, we have in particular that $|f|(\overline{\{ x\} })$ is closed in $\mathop{\mathrm{Spec}}(A)$. Since this image contains the generic point of $\mathop{\mathrm{Spec}}(A)$ there exists a point $x' \in |X|$ in the closure of $x$ mapping to the closed point of $\mathop{\mathrm{Spec}}(A)$. By Lemma 67.16.5 we can find a commutative diagram

such that the closed point of $\mathop{\mathrm{Spec}}(A')$ maps to $x' \in |X|$. It follows that $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ maps the closed point to the closed point, i.e., $A'$ dominates $A$ and this finishes the proof. $\square$

Proof. Since $f$ is separated parts (2) and (3) are equivalent by Lemma 67.41.5. The equivalence of (3) and (1) follows from Lemma 67.42.2. $\square$

Proof. The base change $X_ A \to \mathop{\mathrm{Spec}}(A)$ is flat (Lemma 67.30.4) and the closed point of $\mathop{\mathrm{Spec}}(A)$ is in the image of $|X_ A| \to |\mathop{\mathrm{Spec}}(A)|$ (Properties of Spaces, Lemma 66.4.3). Thus we may assume $Y = \mathop{\mathrm{Spec}}(A)$. Let $U \to X$ be a surjective étale morphism where $U$ is a scheme. Let $u \in U$ map to the closed point of $\mathop{\mathrm{Spec}}(A)$. Consider the flat local ring map $A \to B = \mathcal{O}_{U, u}$. By Algebra, Lemma 10.39.16 there exists a prime ideal $\mathfrak q \subset B$ such that $\mathfrak q$ lies over $(0) \subset A$. By Algebra, Lemma 10.50.2 we can find a valuation ring $A' \subset \kappa (\mathfrak q)$ dominating $B/\mathfrak q$. The induced morphism $\mathop{\mathrm{Spec}}(A') \to U \to X$ is a solution to the problem posed by the lemma. $\square$

Proof. Assume (1), (2), (3), and (4). We will verify the existence part of the valuative criterion for $f$ which will imply $f$ is universally closed by Lemma 67.42.1. To do this, consider a commutative diagram

where $A$ is a valuation ring and $K$ is the fraction field of $A$. Note that since valuation rings and fields are reduced, we may replace $U$, $X$, and $S$ by their respective reductions by Properties of Spaces, Lemma 66.12.4. In this case the assumption that $h(U)$ is dense means that the scheme theoretic image of $h : U \to X$ is $X$, see Lemma 67.16.4.

Reduction to the case $Y$ affine. Choose an étale morphism $\mathop{\mathrm{Spec}}(R) \to Y$ such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to an element of $\mathop{\mathrm{Im}}(|\mathop{\mathrm{Spec}}(R)| \to |Y|)$. By Lemma 67.42.4 we can find a local ring map $A \to A'$ of valuation rings and a morphism $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(R)$ fitting into a commutative diagram

Since in Definition 67.41.1 we allow for extensions of valuation rings it is clear that we may replace $A$ by $A'$, $Y$ by $\mathop{\mathrm{Spec}}(R)$, $X$ by $X \times _ Y \mathop{\mathrm{Spec}}(R)$ and $U$ by $U \times _ Y \mathop{\mathrm{Spec}}(R)$.

From now on we assume that $Y = \mathop{\mathrm{Spec}}(R)$ is an affine scheme. Let $\mathop{\mathrm{Spec}}(B) \to X$ be an étale morphism from an affine scheme such that the morphism $\mathop{\mathrm{Spec}}(K) \to X$ is in the image of $|\mathop{\mathrm{Spec}}(B)| \to |X|$. Since we may replace $K$ by an extension $K' \supset K$ and $A$ by a valuation ring $A' \subset K'$ dominating $A$ (which exists by Algebra, Lemma 10.50.2), we may assume the morphism $\mathop{\mathrm{Spec}}(K) \to X$ factors through $\mathop{\mathrm{Spec}}(B)$ (by definition of $|X|$). In other words, we may think of $K$ as a $B$-algebra. Choose a polynomial algebra $P$ over $B$ and a $B$-algebra surjection $P \to K$. Then $\mathop{\mathrm{Spec}}(P) \to X$ is flat as a composition $\mathop{\mathrm{Spec}}(P) \to \mathop{\mathrm{Spec}}(B) \to X$. Hence the scheme theoretic image of the morphism $U \times _ X \mathop{\mathrm{Spec}}(P) \to \mathop{\mathrm{Spec}}(P)$ is $\mathop{\mathrm{Spec}}(P)$ by Lemma 67.30.12. By Lemma 67.16.5 we can find a commutative diagram

where $A'$ is a valuation ring and $K'$ is the fraction field of $A'$ such that the closed point of $\mathop{\mathrm{Spec}}(A')$ maps to $\mathop{\mathrm{Spec}}(K) \subset \mathop{\mathrm{Spec}}(P)$. In other words, there is a $B$-algebra map $\varphi : K \to A'/\mathfrak m_{A'}$. Choose a valuation ring $A'' \subset A'/\mathfrak m_{A'}$ dominating $\varphi (A)$ with field of fractions $K'' = A'/\mathfrak m_{A'}$ (Algebra, Lemma 10.50.2). We set

which is a valuation ring by Algebra, Lemma 10.50.10. As $C$ is an $R$-algebra with fraction field $K'$, we obtain a solid commutative diagram

as in the statement of the lemma. Thus assumption (4) produces $C \to C_1$ and the dotted arrows making the diagram commute. Let $A_1' = (C_1)_\mathfrak p$ be the localization of $C_1$ at a prime $\mathfrak p \subset C_1$ lying over $\mathfrak m_{A'} \subset C$. Since $C \to C_1$ is flat by More on Algebra, Lemma 15.22.10 such a prime $\mathfrak p$ exists by Algebra, Lemmas 10.39.17 and 10.39.16. Note that $A'$ is the localization of $C$ at $\mathfrak m_{A'}$ and that $A'_1$ is a valuation ring (Algebra, Lemma 10.50.9). In other words, $A' \to A'_1$ is a local ring map of valuation rings. Assumption (3) implies

commutes. Hence the restriction of the morphism $\mathop{\mathrm{Spec}}(C_1) \to X$ to $\mathop{\mathrm{Spec}}(C_1/\mathfrak p)$ restricts to the composition

on the generic point of $\mathop{\mathrm{Spec}}(C_1/\mathfrak p)$. Moreover, $C_1/\mathfrak p$ is a valuation ring (Algebra, Lemma 10.50.9) dominating $A''$ which dominates $A$. Thus the morphism $\mathop{\mathrm{Spec}}(C_1/\mathfrak p) \to X$ witnesses the existence part of the valuative criterion for the diagram (67.42.5.1) as desired.

Next, suppose that (5) is satisfied as well, i.e., the morphism $\Delta : X \to X \times _ S X$ is quasi-compact. In this case assumptions (1) – (4) hold for $h$ and $\Delta $. Hence the first part of the proof shows that $\Delta $ is universally closed. By Lemma 67.40.9 we conclude that $f$ is separated. $\square$