Lemma 26.20.5. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

Specializations lift along any base change of $f$

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

Lemma 26.20.5. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

Specializations lift along any base change of $f$

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 10.50.1. 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

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

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 $K = \kappa (x')/\kappa (s')$. 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 $s$. $\square$

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