## 66.42 Valuative criterion for universal closedness

The existence part of the valuative criterion implies universal closedness for quasi-compact morphisms, see Lemma 66.42.1. In the case of schemes, this is an “if and only if” statement, but for morphisms of algebraic spaces this is wrong. Example 66.9.6 shows that $\mathbf{A}^1_ k/\mathbf{Z} \to \mathop{\mathrm{Spec}}(k)$ is universally closed, but it is easy to see that the existence part of the valuative criterion fails. We revisit this topic in Decent Spaces, Section 67.16 and show the converse holds if the source of the morphism is a decent space (see also Decent Spaces, Lemma 67.17.11 for a relative version).

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

1. $f$ is quasi-compact, and

2. $f$ 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. $\square$

The following lemma will be generalized in Decent Spaces, Lemma 67.17.11.

Lemma 66.42.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

1. If $f$ is quasi-separated and universally closed, then $f$ satisfies the existence part of the valuative criterion.

2. If $f$ is quasi-compact and quasi-separated, then $f$ is universally closed if and only if the existence part of the valuative criterion holds.

Proof. If (1) is true then combined with Lemma 66.42.1 we obtain (2). Assume $f$ is quasi-separated and universally closed. Assume given a diagram

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

as in Definition 66.41.1. A formal argument shows that the existence of the desired diagram

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

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

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-x \ar[d] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar@{=}[r] \ar@{..>}[ru] & \mathop{\mathrm{Spec}}(A) }$

where $A$ is a valuation ring with field of fractions $K$. By Lemma 66.8.9 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 66.16.5 we can find a commutative diagram

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

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. $\square$

Lemma 66.42.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-compact and separated. Then the following are equivalent

1. $f$ is universally closed,

2. the existence part of the valuative criterion holds as in Definition 66.41.1, and

3. given any commutative solid diagram

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

where $A$ is a valuation ring with field of fractions $K$, there exists a dotted arrow, i.e., $f$ satisfies the existence part of the valuative criterion as in Schemes, Definition 26.20.3.

Proof. Since $f$ is separated parts (2) and (3) are equivalent by Lemma 66.41.5. The equivalence of (3) and (1) follows from Lemma 66.42.2. $\square$

Lemma 66.42.4. Let $S$ be a scheme. Let $f : X \to Y$ be a flat morphism of algebraic spaces over $S$. Let $\mathop{\mathrm{Spec}}(A) \to Y$ be a morphism where $A$ is a valuation ring. If the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to a point of $|Y|$ in the image of $|X| \to |Y|$, then there exists a commutative diagram

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

where $A \to A'$ is an extension of valuation rings (More on Algebra, Definition 15.123.1).

Proof. The base change $X_ A \to \mathop{\mathrm{Spec}}(A)$ is flat (Lemma 66.30.4) 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 65.4.3). 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. $\square$

Lemma 66.42.5. Let $S$ be a scheme. Let $f : X \to Y$ and $h : U \to X$ be morphisms of algebraic spaces over $S$. If

1. $f$ and $h$ are quasi-compact,

2. $|h|(|U|)$ is dense in $|X|$, and

given any commutative solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & U \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[rr] \ar@{-->}[rru] & & Y }$

where $A$ is a valuation ring with field of fractions $K$

1. there exists at most one dotted arrow making the diagram commute, and

2. there exists an extension $K'/K$ of fields, a valuation ring $A' \subset K'$ dominating $A$ and a morphism $\mathop{\mathrm{Spec}}(A') \to X$ such that the following diagram commutes

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

then $f$ is universally closed. If moreover

1. $f$ is quasi-separated

then $f$ is separated and universally closed.

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 66.42.1. To do this, consider a commutative diagram

66.42.5.1
$$\label{spaces-morphisms-equation-start-with} \vcenter { \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y } }$$

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 65.12.4. 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 66.16.4.

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 66.42.4 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

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

Since in Definition 66.41.1 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 66.30.12. By Lemma 66.16.5 we can find a commutative diagram

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

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

$C = \{ \lambda \in A' \mid \lambda \bmod \mathfrak m_{A'} \in A''\} .$

which is a valuation ring by Algebra, Lemma 10.50.9. As $C$ is an $R$-algebra with fraction field $K'$, we obtain a solid commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K'_1) \ar@{-->}[r] \ar@{-->}[d] & \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & U \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(C_1) \ar@{-->}[r] \ar@{-->}[rrru] & \mathop{\mathrm{Spec}}(C) \ar[rr] & & Y }$

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 10.50.8). In other words, $A' \to A'_1$ is a local ring map of valuation rings. Assumption (3) implies

$\xymatrix{ \mathop{\mathrm{Spec}}(A'_1) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(C_1) \ar[r] & X \\ \mathop{\mathrm{Spec}}(A') \ar[r] & \mathop{\mathrm{Spec}}(P) \ar[r] & \mathop{\mathrm{Spec}}(B) \ar[u] }$

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

$\mathop{\mathrm{Spec}}(\kappa (\mathfrak p)) \to \mathop{\mathrm{Spec}}(A'/\mathfrak m_{A'}) = \mathop{\mathrm{Spec}}(K'') \to \mathop{\mathrm{Spec}}(K) \to X$

on the generic point of $\mathop{\mathrm{Spec}}(C_1/\mathfrak p)$. Moreover, $C_1/\mathfrak p$ is a valuation ring (Algebra, Lemma 10.50.8) 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 (66.42.5.1) 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 66.40.9 we conclude that $f$ is separated. $\square$

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