The Stacks project

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

as in Definition 67.41.1, 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

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. $\square$

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

as in Definition 67.41.1 with $K \subset K'$ arbitrary. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Then

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

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

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

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. $\square$

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 67.4.4). Base changing all the morphisms of the diagram above we obtain

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 67.8.9. Let $Z \subset X$ be the scheme theoretic image of $\mathop{\mathrm{Spec}}(A') \to X$. Then $Z$ is a reduced (Lemma 67.16.4), quasi-compact (as a closed subspace of $X$), separated (as a closed subspace of $X$) algebraic space over $A$. Consider the base change

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 67.30.12 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 67.4.7). 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 66.6.3). 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 67.16.3. 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

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 65.9.1) and the proof is complete. $\square$

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

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

as in Definition 67.41.1 with $K \subset K'$ arbitrary. By Lemma 67.41.4 we can find a morphism $\mathop{\mathrm{Spec}}(A) \to X$ fitting into the diagram, i.e., (2) holds. $\square$

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

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

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

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

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

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 $\mathop{\mathrm{Spec}}(A'') \to X$. $\square$