See my post here: Valuation Rings and Ultrafilters

Let $K$ be a field, and let $\mathcal{S}$ be the set of pairs $(R, \mathfrak{p})$ of subrings $R$ of $K$ with designated prime ideals $\mathfrak{p}$ of $R$. We order this set where $(R , \mathfrak{p}) \leq (S, \mathfrak{q})$ when $R \subset S$ and $\mathfrak{q} \cap R = \mathfrak{p}$. One of the elementary results about valuations is that valuations with fraction field $K$ are precisely the maximal elements of this set. This is sometimes even given as the definition.

The usual proof works by showing that we always have either $(R, \mathfrak{p}) \leq (R[a], \mathfrak{p}R[a])$ or $(R, \mathfrak{p}) \leq (R[a^{-1}], \mathfrak{p}R[a^{-1}])$ for each pair $(R, \mathfrak{p})$ in $\mathcal{S}$. Then, as valuations are those rings which always contain either $a$ or $a^{-1}$, the result follows.

I am trying to give an alternative proof of this fact, using a near correspondence between subrings of a ring $R$ and partially ordered abelian groups, but I can't seem to see how it works out, so I want some help.

See here for how, for a subring $R$ of a field $K$,

1) certain $R$ submodules of $K$ correspond to filters on $\Lambda: = K^\times / R^\times$

2) certain subrings of $K$ contianing $R$ correspond to filters on the partially ordered abelian group $K^\times / R^\times$ which are submonoids.

3) filters on $K^\times / R^\times$ which are submonoids correspond to extensions of the partial order on $K^\times / R^\times$.

This is a full correspondence when $R$ is a UFD, but for the general case, I think this could be close to an elegant proof that valuation rings are maximal in $\mathcal{S}$, just like the maximal extensions of the order on $\Lambda$ are total orders.

But things don't quite match up, since not all rings correspond to filters which are submonoids.

So, can anyone fix up this near-correspondence enough to get a clean proof that valuation rings are maximal elements of $\mathcal{S}$ in the general case?