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 Lemma 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

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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