The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 25.20.5. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

  1. Specializations lift along any base change of $f$

  2. The morphism $f$ satisfies the existence part of the valuative criterion.

Proof. Assume (1) holds. Let a solid diagram as in Definition 25.20.3 be given. In order to find the dotted arrow we may replace $X \to S$ by $X_{\mathop{\mathrm{Spec}}(A)} \to \mathop{\mathrm{Spec}}(A)$ since after all the assumption is stable under base change. Thus we may assume $S = \mathop{\mathrm{Spec}}(A)$. Let $x' \in X$ be the image of $\mathop{\mathrm{Spec}}(K) \to X$, so that we have $\kappa (x') \subset K$, see Lemma 25.13.3. By assumption there exists a specialization $x' \leadsto x$ in $X$ such that $x$ maps to the closed point of $S = \mathop{\mathrm{Spec}}(A)$. We get a local ring map $A \to \mathcal{O}_{X, x}$ and a ring map $\mathcal{O}_{X, x} \to \kappa (x')$, see Lemma 25.13.2 and the discussion following Lemma 25.13.3. The composition $A \to \mathcal{O}_{X, x} \to \kappa (x') \to K$ is the given injection $A \to K$. Since $A \to \mathcal{O}_{X, x}$ is local, the image of $\mathcal{O}_{X, x} \to K$ dominates $A$ and hence is equal to $A$, by Algebra, Definition 10.49.1. Thus we obtain a ring map $\mathcal{O}_{X, x} \to A$ and hence a morphism $\mathop{\mathrm{Spec}}(A) \to X$ (see Lemma 25.13.1 and discussion following it). This proves (2).

Conversely, assume (2) holds. It is immediate that the existence part of the valuative criterion holds for any base change $X_{S'} \to S'$ of $f$ by considering the following commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X_{S'} \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] \ar@{-->}[rru] & S' \ar[r] & S } \]

Namely, the more horizontal dotted arrow will lead to the other one by definition of the fibre product. OK, so it clearly suffices to show that specializations lift along $f$. Let $s' \leadsto s$ be a specialization in $S$, and let $x' \in X$ be a point lying over $s'$. Apply Lemma 25.20.4 to $s' \leadsto s$ and the extension of fields $\kappa (s') \subset \kappa (x') = K$. We get a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[rr] \ar[d] & & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[rru] & \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \ar[r] & S } \]

and by condition (2) we get the dotted arrow. The image $x$ of the closed point of $\mathop{\mathrm{Spec}}(A)$ in $X$ will be a solution to our problem, i.e., $x$ is a specialization of $x'$ and maps to $s$. $\square$


Comments (1)

Comment #3812 by Kestutis Cesnavicius on

I propose to strenghten the statement a little bit by including another equivalent condition:

(2') The morphism satisfies the existence part of the valuative criterion for those valuation rings that have an algebraically closed fraction field.

(Or even "that are strictly Henselian and have an algebraically closed fraction field.")

The proof needs virtually no changes to allow for this. This weaker condition is sometimes easier to verify when and are given through their functors of points.


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