A rank-two valuation ring that is not a DVR

Form the group algebra $k[t^{\mathrm{\Gamma }}]=\{{\sum }_{\gamma \in \mathrm{\Gamma }}{a}_{\gamma }{t}^{\gamma }:\text{only finitely many }{a}_{\gamma }\mathrm{\ne }0\}$, and let $K$ be its fraction field. For a nonzero element $f=\sum a_{\gamma }{t}^{\gamma }$, define $v(f)$ to be the smallest $\gamma$ in its finite support. Because the order on $\mathrm{\Gamma }$ is translation-invariant by [F1], the product of the two lowest terms is the unique lowest term of a product, so $v(fg)=v(f)+v(g)$. For sums, the minimum support can only stay the same or move upward, so $v(f+g)\ge \min \{v(f),v(g)\}$. Extending by $v(f\mathrm{/}g)=v(f)-v(g)$ and $v(0)=\mathrm{\infty }$ gives a valuation on $K$ in the sense of [F2], with value group all of $\mathrm{\Gamma }$.

Let $V:=\{x\in K:v(x)\ge 0\}$. Then $V$ is a valuation ring and is not a field because $t^{(0,1)}$ has positive value. It is not a DVR, since its value group is $\mathrm{\Gamma }$ rather than $\mathbb{Z}$: any cyclic subgroup of $\mathrm{\Gamma }$ is generated by one pair, so it cannot contain both $(1,0)$ and $(0,1)$. Hence [L1] implies that $V$ is not Noetherian.