The Stacks project

Specializations are witnessed by valuation rings.

Lemma 26.20.4. Let $S$ be a scheme. Let $s' \leadsto s$ be a specialization of points of $S$. Then

  1. there exists a valuation ring $A$ and a morphism $f : \mathop{\mathrm{Spec}}(A) \to S$ such that the generic point $\eta $ of $\mathop{\mathrm{Spec}}(A)$ maps to $s'$ and the special point maps to $s$, and

  2. given a field extension $K/\kappa (s')$ we may arrange it so that the extension $\kappa (\eta )/\kappa (s')$ induced by $f$ is isomorphic to the given extension.

Proof. Let $s' \leadsto s$ be a specialization in $S$, and let $K/\kappa (s')$ be an extension of fields. By Lemma 26.13.2 and the discussion following Lemma 26.13.3 this leads to ring maps $\mathcal{O}_{S, s} \to \kappa (s') \to K$. Let $A \subset K$ be any valuation ring whose field of fractions is $K$ and which dominates the image of $\mathcal{O}_{S, s} \to K$, see Algebra, Lemma 10.50.2. The ring map $\mathcal{O}_{S, s} \to A$ induces the morphism $f : \mathop{\mathrm{Spec}}(A) \to S$, see Lemma 26.13.1. This morphism has all the desired properties by construction. $\square$

Comments (4)

Comment #4996 by Jérôme Poineau on

In (1), the morphism Spec should be called , so that we know what refers to in (2).

Comment #10743 by Anon on

suggested slogan Specializations are witnessed by valuation rings.

There are also:

  • 2 comment(s) on Section 26.20: Valuative criterion for universal closedness

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