# Group of units of a valuation

Let K be a field. Then a subring R of K is called a valuation ring if for all $x \in K^*,$ either $x \in R$ or $x^{-1} \in R$ (or both).

It can be shown that for any valuation $v$ on $K,$ the ring $\Theta_v = \{x \in K: v(x) \geq 0\}$ is a valuation ring of K. Conversely, given a valuation ring $\Theta$ of $K,$ there is a valuation $v$ on $K$ such that $\Theta = \Theta_v.$

My question is, whether it is possible to do the same for groups instead of rings. More precisely, given a subgroup $G \leq K^*,$ when is $G$ the group of units of a valuation on $K$ (which axioms does $G$ have to satisfy?), i.e. when is there a valuation $v$ on $K$ such that $G = \{x \in K: v(x) = 0\}.$

My attempt was to adapt the proof of the above result. So let $G \leq K^*.$ The value group of the looked-for valuation must be (isomorphic to) the quotient group $\Gamma:=K^*/G.$ One may rewrite this group additively (I'll do this because valuations are usually written additively) by defining $xG+yG:=xyG.$ The main problem is to make $\Gamma$ an ordered group (this is where the axiom $x \in K$ or $x^{-1}$ comes into play for the "ring case")

One may then proceed by defining $v: K \to \Gamma \cup \{\infty\}$ by setting $v(0):=\infty$ and $v(a) = aG$ for $a \neq 0.$ The remaining problem is then to show the triangle inequality, $v(x+y) \geq \min\{v(x),v(y)\}.$

A different approach might be studying the properties of group of units in general, in order to deduce some axioms. For instance, any such group $G$ must fulfill the axiom $x \in G \Rightarrow -x \in G,$ since $v(x) = v(-x)$ for any valuation $v$ and $x \in K.$

• Have you found something? One can prove that for such a group $G$, $\Theta_v = G + (G \cup \{0\})$, which should be enough to recover information from $v$ and translate it in the language of $(K,+,.,0,1,G)$. – nombre Nov 7 '16 at 18:03