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 set 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 $\Lambda$.
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
$a, b \in R \implies a \wedge b \in R \forall a,b \in K^\times$, just as $a, b \in F \implies a \wedge b \in F \forall a, b \in A$.
$1 \in R$, just as $1 \in F$.
$a \in R, a \leq b \implies b \in R$, just as $a \in F, a \leq b \implies b \in F$.
$R$ 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?