Lemma 101.25.6 (0DN5)—The Stacks project

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Table of contents
Part 7: Algebraic Stacks
Chapter 101: Morphisms of Algebraic Stacks
Section 101.25: Flat morphisms
Lemma 101.25.6 (cite)

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Lemma 101.25.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a flat morphism of algebraic stacks. Let $\mathop{\mathrm{Spec}}(A) \to \mathcal{Y}$ be a morphism where $A$ is a valuation ring. If the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to a point of $|\mathcal{Y}|$ in the image of $|\mathcal{X|} \to |\mathcal{Y}|$ , then there exists a commutative diagram

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

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

Proof. The base change $\mathcal{X}_ A \to \mathop{\mathrm{Spec}}(A)$ is flat (Lemma 101.25.3) and the closed point of $\mathop{\mathrm{Spec}}(A)$ is in the image of $|\mathcal{X}_ A| \to |\mathop{\mathrm{Spec}}(A)|$ (Properties of Stacks, Lemma 100.4.3). Thus we may assume $\mathcal{Y} = \mathop{\mathrm{Spec}}(A)$ . Let $U \to \mathcal{X}$ be a surjective smooth morphism where $U$ is a scheme. Then we can apply Morphisms of Spaces, Lemma 67.42.4 to the morphism $U \to \mathop{\mathrm{Spec}}(A)$ to conclude. $\square$

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