Valuation rings

Question

What's the spectrum of a valuation ring? How to describe morphisms from it to a scheme? Is it enough to set the image of generic point and of a maximal ideal and correspondent map of local rings?

Answer 1

If $R$ is a valuation ring with value group $\Gamma$, then the prime ideals of $R$ are totally ordered by inclusion and correspond 1:1 to the convex subgroups of $\Gamma$ (see here). In particular $\mathrm{dim}(R)=\mathrm{height}(\Gamma)$. Here, every cardinal is possible. See also Exercise 3.3.26 in Liu's book. This illstrates that $\mathrm{Spec}(R)$ can be quite complicated.

If $R$ is a local ring, then morphisms $\mathrm{Spec}(R) \to X$ (where $X$ is a scheme) correspond 1:1 to points $x \in X$ together with a local homomorphism $\mathcal{O}_{X,x} \to R$ (EGA I, 2.5.3). Here, $x$ is the image of the unique closed point of $\mathrm{Spec}(R)$. Formulated more abstractly, the functor $(X,x) \mapsto \mathrm{Spec}(\mathcal{O}_{X,x})$ from "pointed" schemes to local schemes is right adjoint to the forgetful functor.

This cannot be simplified when $R$ is a valuation ring, even for discrete valuation rings (just consider the case that $X$ is affine). However, note the the following "converse", which is the main ingredient for the valuative criteria:

If $X$ is a scheme and $x,y \in X$ satisfy $x \in \overline{\{y\}}$, $K$ is a field and $y$ extends to a $K$-valued point of $X$, then there is a valuation ring $R$ with field of fractions $K$ together with an $R$-valued point of $X$ extending $x$, such that its restriction to $K$ is the given $K$-valued point.