Lemma 68.19.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

1. $Y$ is locally Noetherian,

2. $f$ is locally of finite type and quasi-separated,

3. for every commutative 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 discrete valuation ring and $K$ its fraction field, there is at most one dotted arrow making the diagram commute.

Then $f$ is separated.

Proof. We have to show that the diagonal $\Delta : X \to X \times _ Y X$ is a closed immersion. We already know $\Delta$ is representable, separated, a monomorphism, and locally of finite type, see Morphisms of Spaces, Lemma 66.4.1. Choose an affine scheme $U$ and an étale morphism $U \to X \times _ Y X$. Set $V = X \times _{\Delta , X \times _ Y X} U$. It suffices to show that $V \to U$ is a closed immersion (Morphisms of Spaces, Lemma 66.12.1). Since $X \times _ Y X$ is locally of finite type over $Y$ we see that $U$ is Noetherian (use Morphisms of Spaces, Lemmas 66.23.2, 66.23.3, and 66.23.5). Note that $V$ is a scheme as $\Delta$ is representable. Also, $V$ is quasi-compact because $f$ is quasi-separated. Hence $V \to U$ is of finite type. Consider a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & V \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & U }$

of morphisms of schemes where $A$ is a discrete valuation ring with fraction field $K$. We can interpret the composition $\mathop{\mathrm{Spec}}(A) \to U \to X \times _ Y X$ as a pair of morphisms $a, b : \mathop{\mathrm{Spec}}(A) \to X$ agreeing as morphisms into $Y$ and equal when restricted to $\mathop{\mathrm{Spec}}(K)$. Hence our assumption (3) guarantees $a = b$ and we find the dotted arrow in the diagram. By Limits, Lemma 32.15.3 we conclude that $V \to U$ is proper. In other words, $\Delta$ is proper. Since $\Delta$ is a monomorphism, we find that $\Delta$ is a closed immersion (Étale Morphisms, Lemma 41.7.2) as desired. $\square$

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