# Valuation Rings and Ultrafilters

I notice there is a certain similarity between the definition of a valuation ring and the definition of an ultrafilter.

To begin, take a field $$K$$ and let $$\mathcal{A}$$ be the set of subrings of $$K$$. Let $$\mathcal{B}'$$ be the class of pairs $$(\nu, \Lambda)$$, where $$\Lambda$$ is a partially ordered abelian group, and $$\nu : K^\times \rightarrow \Lambda$$ is a surjective map of abelian groups such that $$\nu(a),\nu (b) \geq 0 \implies \nu(a+b) \geq 0$$. Note that $$\nu$$ does necessarily respect the partial order of $$\Lambda$$. We form the set $$\mathcal{B}$$ of equivalence classes of elements in $$\mathcal{B}'$$, where $$(\nu, \Lambda) \sim (\nu', \Lambda')$$ when $$\nu$$ factors through $$\nu'$$ by an isomorphism of partially ordered abelian groups.

There is a $$1$$-to-$$1$$ correspondence between $$\mathcal{A}$$ and $$\mathcal{B}$$. We send a subring $$R$$ of $$K$$ to the abelian group $$K^\times / R^\times$$, with the smallest admissible partial order generated by declaring elements $$r R^\times$$ to be non-negative, paried with the natural map $$K^\times \rightarrow K^\times /R^\times$$. We send a pair $$(\nu, \Lambda)$$ in $$\mathcal{B}$$ to $$\{ r \in K : \nu(r) \geq 0 \}$$.

To see the similarity, take a boolean algebra $$A$$ with filter $$F$$ and a field $$K$$ with subring $$R$$ inducing a pair $$(\nu, \Lambda)$$. To make the similarity more clear, I want to change the notation a bit for the field $$K$$: for $$a, b \in K$$, write $$a \leq b$$ when $$\nu(a) \leq \nu(b)$$. Write $$a \wedge b$$ for $$a + b$$. Write $$a^c$$ for $$a^{-1}$$ ($$c$$ for complement). Then we have

1) $$a, b \in R \implies a \wedge b \in R \forall a,b \in K^\times$$, just as $$a, b \in F \implies a \wedge b \in F \forall a, b \in A$$.

2) $$1 \in R$$, just as $$1 \in F$$.

3) $$a \in R, a \leq b \implies b \in R$$, just as $$a \in F, a \leq b \implies b \in F$$.

4) $$R$$ is a valuation ring when $$a \in R$$ or $$a^c \in R$$ for all $$a \in K^\times$$, just as $$F$$ is an ultrafilter when $$a \in F$$ or $$a^c \in F$$ forall $$a \in A$$.

Can anyone illuminate the similarity going on here? How is $$K^\times$$ formally like a boolean algebra?

• $\mathcal{B}$ is not a set. You want to mod out by some equivalence relation. – YCor Jan 27 at 21:19
• Good point. I fixed it. – Dean Young Jan 27 at 21:40

The similarity has nothing to do with boolean algebras, but with orders in general. Filters can be defined for every partial order: A subset $$\Phi$$ of a poset $$\Lambda$$ is a filter if

• $$\Phi\neq\emptyset$$
• $$\forall a,b\in\Phi \exists c\in\Phi: c\leq a \wedge c\leq b$$.
• $$\forall a\in \Phi\forall b\in\Lambda: a\leq b \implies b\in\Phi$$

If $$\Lambda$$ has a greatest element $$\infty$$, then the first condition can be replaced by $$\infty\in\Phi$$. If $$\Lambda$$ has meets, then the second condition can be replaced by $$\forall a,b\in\Phi: a\wedge b \in \Phi$$.

What you observe is simply that the order on $$\Lambda:=K^\times / R^\times$$ is defined in such a way that $$\nu(R)$$ is a filter in $$\Lambda\sqcup\{\infty\}$$ (where we define $$\nu(0) := \infty$$ as usual), namely the filter of all elements which are greater or equal $$0=\nu(1)$$.

In fact you can do this more generally find that $$\nu^{-1}(\Phi)$$ is an $$R$$-submodule of $$K$$ for every filter $$\Phi\subseteq\Lambda\sqcup\{\infty\}$$.

At least for some rings, say UFD rings $$R$$ and their quotient fields $$K$$, the reverse is also true and we get a bijection $$\begin{array}{rcl} \{L\subseteq K \;R\text{-submodule}\} & \overset{\cong}{\leftrightarrow} & \{\Phi \subseteq \Lambda\sqcup\{\infty\} \;\text{filter}\} \\ L &\mapsto& \nu(L) \\ \nu^{-1}(\Phi) &\leftarrow& \Phi \end{array}$$

If the filter is also a submonoid of $$(\Lambda,+)$$, then $$\nu^{-1}(\Phi)$$ is also multiplicatively closed, i.e. a $$R$$-algebra.

Now what a submonoid that is a filter must contain $$0$$ and therefore all larger elements, i.e. they contain all of $$\Lambda_{\geq 0}$$. They are also additively closed. In other words, they are a positive cone for an extension of the partial order on $$\Lambda$$. The maximal submonoid-filters therefore correspond to maximal extensions of the partial order, i.e. total orders. And $$K^\times / R^\times$$ is totally ordered iff $$R$$ is a valuation ring.

• Would it be possible to use this correspondence to give an alternative proof that valuation rings are maximal in general? Also, could you elaborate a bit on how submonoid filters of $\Lambda$ are in correspondence with extensions of the partial order? – Dean Young Jan 29 at 0:47
• – Dean Young Jan 29 at 1:00