The Stacks project

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 10.50.1. 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

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 $K = \kappa (x')/\kappa (s')$. We get a commutative diagram

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 $s$. $\square$