Lemma 26.20.5. 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 26.20.3 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 26.13.3. 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 26.13.2 and the discussion following Lemma 26.13.3. 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 26.13.1 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 26.20.4 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 $s$. $\square$

Comment #3812 by Kestutis Cesnavicius on

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

(2') The morphism $f$ 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 $X$ and $S$ are given through their functors of points.

Comment #3921 by on

OK, I agree you can do this and even that we should do this. But then I think we should go whole hog and make this modification consistently throughout the stacks project with all the different valuative criteria in the stacks project. See Section 103.11 for a very long list.

Maybe the solution is to add a remark saying you can do this... In other words, make a remark of the form: if you can find the dotted arrow after doing an extension of valuation rings, then... For example, in Noetherian cases it is also often useful to know you can use complete discrete valuation rings and in geometric situations it can be useful to use actual curves or local rings on curves. Etc, etc, etc. I think I will add this if I need it for some particular argument.

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