Lemma 70.22.2. Let $S$ be a scheme. Let $f : X \to Y$ and $h : U \to X$ be morphisms of algebraic spaces over $S$. Assume that $Y$ is locally Noetherian, that $f$ is locally of finite type and quasi-separated, that $h$ is of finite type, and that the image of $|h| : |U| \to |X|$ is dense in $|X|$. If given any commutative solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & U \ar[r]^ h & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[rr] \ar@{-->}[rru] & & Y }$

where $A$ is a discrete valuation ring with field of fractions $K$, there exists at most one dotted arrow making the diagram commute, then $f$ is separated.

Proof. We will apply Lemma 70.22.1 to the morphisms $U \to X$ and $\Delta : X \to X \times _ Y X$. We check the conditions. Observe that $\Delta$ is quasi-compact because $f$ is quasi-separated. Of course $\Delta$ is locally of finite type and separated (true for any diagonal morphism). Finally, suppose given a commutative solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & U \ar[r]^ h & X \ar[d]^\Delta \\ \mathop{\mathrm{Spec}}(A) \ar[rr]^{(a, b)} \ar@{-->}[rru] & & X \times _ Y X }$

where $A$ is a discrete valuation ring with field of fractions $K$. Then $a$ and $b$ give two dotted arrows in the diagram of the lemma and have to be equal. Hence as dotted arrow we can use $a = b$ which gives existence. This finishes the proof. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).