Lemma 66.43.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is separated, then $f$ satisfies the uniqueness part of the valuative criterion.

## 66.43 Valuative criterion of separatedness

First we prove a converse and then we state the criterion.

**Proof.**
Let a diagram as in Definition 66.41.1 be given. Suppose there are two distinct 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 $b$. Then $Z = \mathop{\mathrm{Spec}}(A) \times _{(a, b), X \times _ Y X, \Delta } X$. If $f$ is separated, then $\Delta $ is a closed immersion, and 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 66.43.2 (Valuative criterion separatedness). Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

the morphism $f$ is quasi-separated, and

the morphism $f$ satisfies the uniqueness part of the valuative criterion.

Then $f$ is separated.

**Proof.**
Assumption (1) means $\Delta _{X/Y}$ is quasi-compact. We claim the morphism $\Delta _{X/Y} : X \to X \times _ Y X$ 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 $Y$. By assumption (2) we see that $a = b$. Hence using $a$ as the dotted arrow works. Hence Lemma 66.42.1 applies, and we see that $\Delta _{X/Y}$ is universally closed. Since always $\Delta _{X/Y}$ is locally of finite type and separated, we conclude from More on Morphisms, Lemma 37.44.1 that $\Delta _{X/Y}$ is a finite morphism (also, use the general principle of Spaces, Lemma 64.5.8). At this point $\Delta _{X/Y}$ is a representable, finite monomorphism, hence a closed immersion by Morphisms, Lemma 29.44.15. $\square$

Lemma 66.43.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-compact and quasi-separated. Then the following are equivalent

$f$ is separated and universally closed,

the valuative criterion holds as in Definition 66.41.1,

given any commutative solid diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & Y } \]where $A$ is a valuation ring with field of fractions $K$, there exists a unique dotted arrow, i.e., $f$ satisfies the valuative criterion as in Schemes, Definition 26.20.3.

**Proof.**
Since $f$ is quasi-separated, the uniqueness part of the valutative criterion implies $f$ is separated (Lemma 66.43.2). Conversely, if $f$ is separated, then it satisfies the uniqueness part of the valuative criterion (Lemma 66.43.1). Having said this, we see that in each of the three cases the morphism $f$ is separated and satisfies the uniqueness part of the valuative criterion. In this case the lemma is a formal consequence of Lemma 66.42.3.
$\square$

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