## 26.20 Valuative criterion for universal closedness

In Topology, Section 5.17 there is a discussion of proper maps as closed maps of topological spaces all of whose fibres are quasi-compact, or as maps such that all base changes are closed maps. Here is the corresponding notion in algebraic geometry.

Definition 26.20.1. A morphism of schemes $f : X \to S$ is said to be universally closed if every base change $f' : X_{S'} \to S'$ is closed.

In fact the adjective “universally” is often used in this way. In other words, given a property $\mathcal{P}$ of morphisms then we say that “$X \to S$ is universally $\mathcal{P}$” if and only if every base change $X_{S'} \to S'$ has $\mathcal{P}$.

Please take a look at Morphisms, Section 29.41 for a more detailed discussion of the properties of universally closed morphisms. In this section we restrict the discussion to the relationship between universal closed morphisms and morphisms satisfying the existence part of the valuative criterion.

Lemma 26.20.2. Let $f : X \to S$ be a morphism of schemes.

1. If $f$ is universally closed then specializations lift along any base change of $f$, see Topology, Definition 5.19.4.

2. If $f$ is quasi-compact and specializations lift along any base change of $f$, then $f$ is universally closed.

Proof. Part (1) is a direct consequence of Topology, Lemma 5.19.7. Part (2) follows from Lemmas 26.19.8 and 26.19.3. $\square$

Definition 26.20.3. Let $f : X \to S$ be a morphism of schemes. We say $f$ satisfies the existence part of the valuative criterion if given any commutative solid diagram

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

where $A$ is a valuation ring with field of fractions $K$, the dotted arrow exists. We say $f$ satisfies the uniqueness part of the valuative criterion if there is at most one dotted arrow given any diagram as above (without requiring existence of course).

A valuation ring is a local domain maximal among the relation of domination in its fraction field, see Algebra, Definition 10.50.1. Hence the spectrum of a valuation ring has a unique generic point $\eta$ and a unique closed point $0$, and of course we have the specialization $\eta \leadsto 0$. The significance of valuation rings is that any specialization of points in any scheme is the image of $\eta \leadsto 0$ under some morphism from the spectrum of some valuation ring. Here is the precise result.

Lemma 26.20.4. Let $S$ be a scheme. Let $s' \leadsto s$ be a specialization of points of $S$. Then

1. there exists a valuation ring $A$ and a morphism $f : \mathop{\mathrm{Spec}}(A) \to S$ such that the generic point $\eta$ of $\mathop{\mathrm{Spec}}(A)$ maps to $s'$ and the special point maps to $s$, and

2. given a field extension $K/\kappa (s')$ we may arrange it so that the extension $\kappa (\eta )/\kappa (s')$ induced by $f$ is isomorphic to the given extension.

Proof. Let $s' \leadsto s$ be a specialization in $S$, and let $K/\kappa (s')$ be an extension of fields. By Lemma 26.13.2 and the discussion following Lemma 26.13.3 this leads to ring maps $\mathcal{O}_{S, s} \to \kappa (s') \to K$. Let $A \subset K$ be any valuation ring whose field of fractions is $K$ and which dominates the image of $\mathcal{O}_{S, s} \to K$, see Algebra, Lemma 10.50.2. The ring map $\mathcal{O}_{S, s} \to A$ induces the morphism $f : \mathop{\mathrm{Spec}}(A) \to S$, see Lemma 26.13.1. This morphism has all the desired properties by construction. $\square$

Lemma 26.20.5. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

1. Specializations lift along any base change of $f$

2. The morphism $f$ satisfies the existence part of the valuative criterion.

Proof. Assume (1) holds. Let a solid diagram as in Definition 26.20.3 be given. In order to find the dotted arrow we may replace $X \to S$ by $X_{\mathop{\mathrm{Spec}}(A)} \to \mathop{\mathrm{Spec}}(A)$ since after all the assumption is stable under base change. Thus we may assume $S = \mathop{\mathrm{Spec}}(A)$. Let $x' \in X$ be the image of $\mathop{\mathrm{Spec}}(K) \to X$, so that we have $\kappa (x') \subset K$, see Lemma 26.13.3. By assumption there exists a specialization $x' \leadsto x$ in $X$ such that $x$ maps to the closed point of $S = \mathop{\mathrm{Spec}}(A)$. We get a local ring map $A \to \mathcal{O}_{X, x}$ and a ring map $\mathcal{O}_{X, x} \to \kappa (x')$, see Lemma 26.13.2 and the discussion following Lemma 26.13.3. The composition $A \to \mathcal{O}_{X, x} \to \kappa (x') \to K$ is the given injection $A \to K$. Since $A \to \mathcal{O}_{X, x}$ is local, the image of $\mathcal{O}_{X, x} \to K$ dominates $A$ and hence is equal to $A$, by Algebra, Definition 10.50.1. Thus we obtain a ring map $\mathcal{O}_{X, x} \to A$ and hence a morphism $\mathop{\mathrm{Spec}}(A) \to X$ (see Lemma 26.13.1 and discussion following it). This proves (2).

Conversely, assume (2) holds. It is immediate that the existence part of the valuative criterion holds for any base change $X_{S'} \to S'$ of $f$ by considering the following commutative diagram

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

Namely, the more horizontal dotted arrow will lead to the other one by definition of the fibre product. OK, so it clearly suffices to show that specializations lift along $f$. Let $s' \leadsto s$ be a specialization in $S$, and let $x' \in X$ be a point lying over $s'$. Apply Lemma 26.20.4 to $s' \leadsto s$ and the extension of fields $K = \kappa (x')/\kappa (s')$. We get a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[rr] \ar[d] & & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[rru] & \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \ar[r] & S }$

and by condition (2) we get the dotted arrow. The image $x$ of the closed point of $\mathop{\mathrm{Spec}}(A)$ in $X$ will be a solution to our problem, i.e., $x$ is a specialization of $x'$ and maps to $s$. $\square$

Proof. This is a formal consequence of Lemmas 26.20.2 and 26.20.5 above. $\square$

Example 26.20.7. Let $k$ be a field. Consider the structure morphism $p : \mathbf{P}^1_ k \to \mathop{\mathrm{Spec}}(k)$ of the projective line over $k$, see Example 26.14.4. Let us use the valuative criterion above to prove that $p$ is universally closed. By construction $\mathbf{P}^1_ k$ is covered by two affine opens and hence $p$ is quasi-compact. Let a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_\xi \ar[d] & \mathbf{P}^1_ k \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r]^\varphi & \mathop{\mathrm{Spec}}(k) }$

be given, where $A$ is a valuation ring and $K$ is its field of fractions. Recall that $\mathbf{P}^1_ k$ is gotten by glueing $\mathop{\mathrm{Spec}}(k[x])$ to $\mathop{\mathrm{Spec}}(k[y])$ by glueing $D(x)$ to $D(y)$ via $x = y^{-1}$ (or more symmetrically $xy = 1$). To show there is a morphism $\mathop{\mathrm{Spec}}(A) \to \mathbf{P}^1_ k$ fitting diagonally into the diagram above we may assume that $\xi$ maps into the open $\mathop{\mathrm{Spec}}(k[x])$ (by symmetry). This gives the following commutative diagram of rings

$\xymatrix{ K & k[x] \ar[l]^{\xi ^\sharp } \\ A \ar[u] & k \ar[u] \ar[l]_{\varphi ^\sharp } }$

By Algebra, Lemma 10.50.4 we see that either $\xi ^\sharp (x) \in A$ or $\xi ^\sharp (x)^{-1} \in A$. In the first case we get a ring map

$k[x] \to A, \ \lambda \mapsto \varphi ^\sharp (\lambda ), \ x \mapsto \xi ^\sharp (x)$

fitting into the diagram of rings above, and we win. In the second case we see that we get a ring map

$k[y] \to A, \ \lambda \mapsto \varphi ^\sharp (\lambda ), \ y \mapsto \xi ^\sharp (x)^{-1}.$

This gives a morphism $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(k[y]) \to \mathbf{P}^1_ k$ which fits diagonally into the initial commutative diagram of this example (check omitted).

Comment #7479 by Fawzy Hegab on

In the paragraph after definition 01KB, there a typo in the statement that defines "Universally P" for a property P. Namely, the sentence says "given a property P of morphisms the we say that ". There is a typo: "THE" should be "Then".

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