Lemma 26.22.1. Let $f : X \to S$ be a morphism of schemes. If $f$ is separated, then $f$ satisfies the uniqueness part of the valuative criterion.
26.22 Valuative criterion of separatedness
Proof. Let a diagram as in Definition 26.20.3 be given. Suppose there are two morphisms $a, b : \mathop{\mathrm{Spec}}(A) \to X$ fitting into the diagram. Let $Z \subset \mathop{\mathrm{Spec}}(A)$ be the equalizer of $a$ and $b$. By Lemma 26.21.5 this is a closed subscheme of $\mathop{\mathrm{Spec}}(A)$. By assumption it contains the generic point of $\mathop{\mathrm{Spec}}(A)$. Since $A$ is a domain this implies $Z = \mathop{\mathrm{Spec}}(A)$. Hence $a = b$ as desired. $\square$
Lemma 26.22.2 (Valuative criterion separatedness). Let $f : X \to S$ be a morphism. Assume
the morphism $f$ is quasi-separated, and
the morphism $f$ satisfies the uniqueness part of the valuative criterion.
Then $f$ is separated.
Proof. By assumption (1), Proposition 26.20.6, and Lemmas 26.21.2 and 26.10.4 we see that it suffices to prove the morphism $\Delta _{X/S} : X \to X \times _ S X$ satisfies the existence part of the valuative criterion. Let a solid commutative diagram
\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & X \times _ S X } \]
be given. The lower right arrow corresponds to a pair of morphisms $a, b : \mathop{\mathrm{Spec}}(A) \to X$ over $S$. By (2) we see that $a = b$. Hence using $a$ as the dotted arrow works. $\square$
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