Lemma 67.42.4. 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 67.30.4) 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 66.4.3). 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
Comments (0)