## 66.41 Valuative criteria

The section introduces the basics on valuative criteria for morphisms of algebraic spaces. Here is a list of references to further results

1. the valuative criterion for universal closedness can be found in Section 66.42,

2. the valuative criterion of separatedness can be found in Section 66.43,

3. the valuative criterion for properness can be found in Section 66.44,

4. additional converse statements can be found in Decent Spaces, Section 67.16 and Decent Spaces, Lemma 67.17.11, and

5. in the Noetherian case it is enough to check the criterion for discrete valuation rings as is shown in Cohomology of Spaces, Section 68.19 and Limits of Spaces, Section 69.21, and

6. refined versions of the valuative criteria in the Noetherian case can be found in Limits of Spaces, Section 69.22.

We first formally state the definition and then we discuss how this differs from the case of morphisms of schemes.

Definition 66.41.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. We say $f$ satisfies the uniqueness part of the valuative criterion if given any commutative solid diagram

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

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

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

We say $f$ satisfies the valuative criterion if $f$ satisfies both the existence and uniqueness part.

The formulation of the existence part of the valuative criterion is slightly different for morphisms of algebraic spaces, since it may be necessary to extend the fraction field of the valuation ring. In practice this difference almost never plays a role.

1. Checking the uniqueness part of the valuative criterion never involves any fraction field extensions, hence this is exactly the same as in the case of schemes.

2. It is necessary to allow for field extensions in general, see Example 66.41.6.

3. For morphisms of algebraic spaces it always suffices to take a finite separable extensions $K'/K$ in the existence part of the valuative criterion, see Lemma 66.41.3.

4. If $f : X \to Y$ is a separated morphism of algebraic spaces, then we can always take $K = K'$ when we check the existence part of the valuative criterion, see Lemma 66.41.5.

5. For a quasi-compact and quasi-separated morphism $f : X \to Y$, we get an equivalence between “$f$ is separated and universally closed” and “$f$ satisfies the usual valuative criterion”, see Lemma 66.43.3. The valuative criterion for properness is the usual one, see Lemma 66.44.1.

As a first step in the theory, we show that the criterion is identical to the criterion as formulated for morphisms of schemes in case the morphism of algebraic spaces is representable.

Lemma 66.41.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is representable. The following are equivalent

1. $f$ satisfies the existence part of the valuative criterion as in Definition 66.41.1,

2. 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. It suffices to show that given a commutative diagram of the form

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

as in Definition 66.41.1, then we can find a morphism $\mathop{\mathrm{Spec}}(A) \to X$ fitting into the diagram too. Set $X_ A = \mathop{\mathrm{Spec}}(A) \times _ Y Y$. As $f$ is representable we see that $X_ A$ is a scheme. The morphism $\varphi$ gives a morphism $\varphi ' : \mathop{\mathrm{Spec}}(A') \to X_ A$. Let $x \in X_ A$ be the image of the closed point of $\varphi ' : \mathop{\mathrm{Spec}}(A') \to X_ A$. Then we have the following commutative diagram of rings

$\xymatrix{ K' & K \ar[l] & \mathcal{O}_{X_ A, x} \ar[l] \ar[lld] \\ A' \ar[u] & A \ar[l] & A \ar[l] \ar[u] }$

Since $A$ is a valuation ring, and since $A'$ dominates $A$, we see that $K \cap A' = A$. Hence the ring map $\mathcal{O}_{X_ A, x} \to K$ has image contained in $A$. Whence a morphism $\mathop{\mathrm{Spec}}(A) \to X_ A$ (see Schemes, Section 26.13) as desired. $\square$

Lemma 66.41.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent

1. $f$ satisfies the existence part of the valuative criterion as in Definition 66.41.1,

2. $f$ satisfies the existence part of the valuative criterion as in Definition 66.41.1 modified by requiring the extension $K'/K$ to be finite separable.

Proof. We have to show that (1) implies (2). Suppose given a diagram

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

as in Definition 66.41.1 with $K \subset K'$ arbitrary. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Then

$\mathop{\mathrm{Spec}}(A') \times _ X U \longrightarrow \mathop{\mathrm{Spec}}(A')$

is surjective étale. Let $p$ be a point of $\mathop{\mathrm{Spec}}(A') \times _ X U$ mapping to the closed point of $\mathop{\mathrm{Spec}}(A')$. Let $p' \leadsto p$ be a generalization of $p$ mapping to the generic point of $\mathop{\mathrm{Spec}}(A')$. Such a generalization exists because generalizations lift along flat morphisms of schemes, see Morphisms, Lemma 29.25.9. Then $p'$ corresponds to a point of the scheme $\mathop{\mathrm{Spec}}(K') \times _ X U$. Note that

$\mathop{\mathrm{Spec}}(K') \times _ X U = \mathop{\mathrm{Spec}}(K') \times _{\mathop{\mathrm{Spec}}(K)} (\mathop{\mathrm{Spec}}(K) \times _ X U)$

Hence $p'$ maps to a point $q' \in \mathop{\mathrm{Spec}}(K) \times _ X U$ whose residue field is a finite separable extension of $K$. Finally, $p' \leadsto p$ maps to a specialization $u' \leadsto u$ on the scheme $U$. With all this notation we get the following diagram of rings

$\xymatrix{ \kappa (p') & & \kappa (q') \ar[ll] & \kappa (u') \ar[l] \\ & \mathcal{O}_{\mathop{\mathrm{Spec}}(A') \times _ X U, p} \ar[lu] & & \mathcal{O}_{U, u} \ar[ll] \ar[u] \\ K' \ar[uu] & A' \ar[l] \ar[u] & A \ar[l] \ar '[u][uu] }$

This means that the ring $B \subset \kappa (q')$ generated by the images of $A$ and $\mathcal{O}_{U, u}$ maps to a subring of $\kappa (p')$ contained in the image $B'$ of $\mathcal{O}_{\mathop{\mathrm{Spec}}(A') \times _ X U, p} \to \kappa (p')$. Note that $B'$ is a local ring. Let $\mathfrak m \subset B$ be the maximal ideal. By construction $A \cap \mathfrak m$, (resp. $\mathcal{O}_{U, u} \cap \mathfrak m$, resp. $A' \cap \mathfrak m$) is the maximal ideal of $A$ (resp. $\mathcal{O}_{U, u}$, resp. $A'$). Set $\mathfrak q = B \cap \mathfrak m$. This is a prime ideal such that $A \cap \mathfrak q$ is the maximal ideal of $A$. Hence $B_{\mathfrak q} \subset \kappa (q')$ is a local ring dominating $A$. By Algebra, Lemma 10.50.2 we can find a valuation ring $A_1 \subset \kappa (q')$ with field of fractions $\kappa (q')$ dominating $B_{\mathfrak q}$. The (local) ring map $\mathcal{O}_{U, u} \to A_1$ gives a morphism $\mathop{\mathrm{Spec}}(A_1) \to U \to X$ such that the diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(\kappa (q')) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A_1) \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$

is commutative. Since the fraction field of $A_1$ is $\kappa (q')$ and since $\kappa (q')/K$ is finite separable by construction the lemma is proved. $\square$

Lemma 66.41.4. Let $S$ be a scheme. Let $f : X \to Y$ be a separated morphism of algebraic spaces over $S$. Suppose given a diagram

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

as in Definition 66.41.1 with $K \subset K'$ arbitrary. Then the dotted arrow exists making the diagram commute.

Proof. We have to show that we can find a morphism $\mathop{\mathrm{Spec}}(A) \to X$ fitting into the diagram.

Consider the base change $X_ A = \mathop{\mathrm{Spec}}(A) \times _ Y X$ of $X$. Then $X_ A \to \mathop{\mathrm{Spec}}(A)$ is a separated morphism of algebraic spaces (Lemma 66.4.4). Base changing all the morphisms of the diagram above we obtain

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

Thus we may replace $X$ by $X_ A$, assume that $Y = \mathop{\mathrm{Spec}}(A)$ and that we have a diagram as above. We may and do replace $X$ by a quasi-compact open subspace containing the image of $|\mathop{\mathrm{Spec}}(A')| \to |X|$.

The morphism $\mathop{\mathrm{Spec}}(A') \to X$ is quasi-compact by Lemma 66.8.9. Let $Z \subset X$ be the scheme theoretic image of $\mathop{\mathrm{Spec}}(A') \to X$. Then $Z$ is a reduced (Lemma 66.16.4), quasi-compact (as a closed subspace of $X$), separated (as a closed subspace of $X$) algebraic space over $A$. Consider the base change

$\mathop{\mathrm{Spec}}(K') = \mathop{\mathrm{Spec}}(A') \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(K) \to X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(K) = X_ K$

of the morphism $\mathop{\mathrm{Spec}}(A') \to X$ by the flat morphism of schemes $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(A)$. By Lemma 66.30.12 we see that the scheme theoretic image of this morphism is the base change $Z_ K$ of $Z$. On the other hand, by assumption (i.e., the commutative diagram above) this morphism factors through a morphism $\mathop{\mathrm{Spec}}(K) \to Z_ K$ which is a section to the structure morphism $Z_ K \to \mathop{\mathrm{Spec}}(K)$. As $Z_ K$ is separated, this section is a closed immersion (Lemma 66.4.7). We conclude that $Z_ K = \mathop{\mathrm{Spec}}(K)$.

Let $V \to Z$ be a surjective étale morphism with $V$ an affine scheme (Properties of Spaces, Lemma 65.6.3). Say $V = \mathop{\mathrm{Spec}}(B)$. Then $V \times _ Z \mathop{\mathrm{Spec}}(A') = \mathop{\mathrm{Spec}}(C)$ is affine as $Z$ is separated. Note that $B \to C$ is injective as $V$ is the scheme theoretic image of $V \times _ Z \mathop{\mathrm{Spec}}(A') \to V$ by Lemma 66.16.3. On the other hand, $A' \to C$ is étale as corresponds to the base change of $V \to Z$. Since $A'$ is a torsion free $A$-module, the flatness of $A' \to C$ implies $C$ is a torsion free $A$-module, hence $B$ is a torsion free $A$-module. Note that being torsion free as an $A$-module is equivalent to being flat (More on Algebra, Lemma 15.22.10). Next, we write

$V \times _ Z V = \mathop{\mathrm{Spec}}(B')$

Note that the two ring maps $B \to B'$ are étale as $V \to Z$ is étale. The canonical surjective map $B \otimes _ A B \to B'$ becomes an isomorphism after tensoring with $K$ over $A$ because $Z_ K = \mathop{\mathrm{Spec}}(K)$. However, $B \otimes _ A B$ is torsion free as an $A$-module by our remarks above. Thus $B' = B \otimes _ A B$. It follows that the base change of the ring map $A \to B$ by the faithfully flat ring map $A \to B$ is étale (note that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is surjective as $X \to \mathop{\mathrm{Spec}}(A)$ is surjective). Hence $A \to B$ is étale (Descent, Lemma 35.23.29), in other words, $V \to X$ is étale. Since we have $V \times _ Z V = V \times _{\mathop{\mathrm{Spec}}(A)} V$ we conclude that $Z = \mathop{\mathrm{Spec}}(A)$ as algebraic spaces (for example by Spaces, Lemma 64.9.1) and the proof is complete. $\square$

Lemma 66.41.5. Let $S$ be a scheme. Let $f : X \to Y$ be a separated morphism of algebraic spaces over $S$. The following are equivalent

1. $f$ satisfies the existence part of the valuative criterion as in Definition 66.41.1,

2. 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. We have to show that (1) implies (2). Suppose given a commutative diagram

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

as in part (2). By (1) there exists a commutative diagram

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

as in Definition 66.41.1 with $K \subset K'$ arbitrary. By Lemma 66.41.4 we can find a morphism $\mathop{\mathrm{Spec}}(A) \to X$ fitting into the diagram, i.e., (2) holds. $\square$

Example 66.41.6. Consider the algebraic space $X$ constructed in Spaces, Example 64.14.2. Recall that it is Galois twist of the affine line with zero doubled. The Galois twist is with respect to a degree two Galois extension $k'/k$ of fields. As such it comes with a morphism

$\pi : X \longrightarrow S = \mathbf{A}^1_ k$

which is quasi-compact. We claim that $\pi$ is universally closed. Namely, after base change by $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ the morphism $\pi$ is identified with the morphism

$\text{affine line with zero doubled} \longrightarrow \text{affine line}$

which is universally closed (some details omitted). Since the morphism $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ is universally closed and surjective, a diagram chase shows that $\pi$ is universally closed. On the other hand, consider the diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(k((x))) \ar[r] \ar[d] & X \ar[d]^\pi \\ \mathop{\mathrm{Spec}}(k[[x]]) \ar[r] \ar@{..>}[ru] & \mathbf{A}^1_ k }$

Since the unique point of $X$ above $0 \in \mathbf{A}^1_ k$ corresponds to a monomorphism $\mathop{\mathrm{Spec}}(k') \to X$ it is clear there cannot exist a dotted arrow! This shows that a finite separable field extension is needed in general.

Lemma 66.41.7. The base change of a morphism of algebraic spaces which satisfies the existence part of (resp. uniqueness part of) the valuative criterion by any morphism of algebraic spaces satisfies the existence part of (resp. uniqueness part of) the valuative criterion.

Proof. Let $f : X \to Y$ be a morphism of algebraic spaces over the scheme $S$. Let $Z \to Y$ be any morphism of algebraic spaces over $S$. Consider a solid commutative diagram of the following shape

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

Then the set of north-west dotted arrows making the diagram commute is in 1-1 correspondence with the set of west-north-west dotted arrows making the diagram commute. This proves the lemma in the case of “uniqueness”. For the existence part, assume $f$ satisfies the existence part of the valuative criterion. If we are given a solid commutative diagram as above, then by assumption there exists an extension $K'/K$ of fields and a valuation ring $A' \subset K'$ dominating $A$ and a morphism $\mathop{\mathrm{Spec}}(A') \to X$ fitting into the following commutative diagram

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

And by the remarks above the skew arrow corresponds to an arrow $\mathop{\mathrm{Spec}}(A') \to Z \times _ Y X$ as desired. $\square$

Lemma 66.41.8. The composition of two morphisms of algebraic spaces which satisfy the (existence part of, resp. uniqueness part of) the valuative criterion satisfies the (existence part of, resp. uniqueness part of) the valuative criterion.

Proof. Let $f : X \to Y$, $g : Y \to Z$ be morphisms of algebraic spaces over the scheme $S$. Consider a solid commutative diagram of the following shape

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

If we have the uniqueness part for $g$, then there exists at most one north-west dotted arrow making the diagram commute. If we also have the uniqueness part for $f$, then we have at most one north-north-west dotted arrow making the diagram commute. The proof in the existence case comes from contemplating the following diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K'') \ar[r] \ar[dd] & \mathop{\mathrm{Spec}}(K') \ar[r] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d]^ f \\ & & & Y \ar[d]^ g \\ \mathop{\mathrm{Spec}}(A'') \ar[r] \ar[rrruu] & \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Z }$

Namely, the existence part for $g$ gives us the extension $K'$, the valuation ring $A'$ and the arrow $\mathop{\mathrm{Spec}}(A') \to Y$, whereupon the existence part for $f$ gives us the extension $K''$, the valuation ring $A''$ and the arrow $\mathop{\mathrm{Spec}}(A'') \to X$. $\square$

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