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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?

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

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  • $\begingroup$ 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? $\endgroup$ – Dean Young Jan 29 at 0:47
  • $\begingroup$ See here: math.stackexchange.com/questions/3091610/… $\endgroup$ – Dean Young Jan 29 at 1:00
  • $\begingroup$ Also, is there a place where I could read about this in more detail? $\endgroup$ – Dean Young Jan 29 at 1:04

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