Lemma 67.41.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent
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 67.41.1 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. $\square$
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