A nondiscrete rank-one valuation from incommensurate values

As in the previous example, form the group algebra $k[t^{\mathrm{\Gamma }}]$ and its fraction field $K$. For a nonzero finite sum $f={\sum }_{\gamma \in \mathrm{\Gamma }}{a}_{\gamma }{t}^{\gamma }$, let $v(f)$ be the least element of its support. The same minimum-support argument as before gives $v(fg)=v(f)+v(g)$ and $v(f+g)\ge \min \{v(f),v(g)\}$, so extending by $v(f\mathrm{/}g)=v(f)-v(g)$ and $v(0)=\mathrm{\infty }$ yields a valuation on $K$ with value group $\mathrm{\Gamma }$. Its nonnegative locus is therefore a valuation ring.

The ordered group $\mathrm{\Gamma }$ has no least positive element. Indeed, $0<\sqrt{2}-1<1$, so any least positive element $\epsilon$ would satisfy $0<\epsilon <1$. Choose $n=\lceil 1\mathrm{/}\epsilon \rceil$, so $n>1$ and $0\le n\epsilon -1<\epsilon$. Equality $n\epsilon -1=0$ would give $\epsilon =1\mathrm{/}n$. But $1\mathrm{/}n$ cannot lie in $\mathrm{\Gamma }$: if $1\mathrm{/}n=a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$, then $nb\sqrt{2}=1-na$ is rational, so $b=0$ and then $na=1$, impossible for $n>1$. Thus $0<n\epsilon -1<\epsilon$, contradicting minimality. Hence the valuation is not discrete.