Lemma 66.42.4. Let $S$ be a scheme. Let $f : X \to Y$ be a flat morphism of algebraic spaces over $S$. Let $\mathop{\mathrm{Spec}}(A) \to Y$ be a morphism where $A$ is a valuation ring. If the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to a point of $|Y|$ in the image of $|X| \to |Y|$, then there exists a commutative diagram

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

where $A \to A'$ is an extension of valuation rings (More on Algebra, Definition 15.123.1).

Proof. The base change $X_ A \to \mathop{\mathrm{Spec}}(A)$ is flat (Lemma 66.30.4) and the closed point of $\mathop{\mathrm{Spec}}(A)$ is in the image of $|X_ A| \to |\mathop{\mathrm{Spec}}(A)|$ (Properties of Spaces, Lemma 65.4.3). Thus we may assume $Y = \mathop{\mathrm{Spec}}(A)$. Let $U \to X$ be a surjective étale morphism where $U$ is a scheme. Let $u \in U$ map to the closed point of $\mathop{\mathrm{Spec}}(A)$. Consider the flat local ring map $A \to B = \mathcal{O}_{U, u}$. By Algebra, Lemma 10.39.16 there exists a prime ideal $\mathfrak q \subset B$ such that $\mathfrak q$ lies over $(0) \subset A$. By Algebra, Lemma 10.50.2 we can find a valuation ring $A' \subset \kappa (\mathfrak q)$ dominating $B/\mathfrak q$. The induced morphism $\mathop{\mathrm{Spec}}(A') \to U \to X$ is a solution to the problem posed by the lemma. $\square$

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