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. $\square$

Comment #2101 by Kestutis Cesnavicius on

In (1), "valuation criterion" ---> "valuative criterion", and similarly later in the section.

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