The Stacks project

26.22 Valuative criterion of separatedness

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.

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. $\square$

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01KY. Beware of the difference between the letter 'O' and the digit '0'.