Bdtyktl

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?

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

• $Phineqemptyset$


• 
  $forall a,binPhi exists cinPhi: cleq a wedge cleq b$.


• $forall ain Phiforall binLambda: aleq b implies binPhi$

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

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 $Lambdasqcup{infty}$ (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 $PhisubseteqLambdasqcup{infty}$.

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

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.