Two dimensional valuation domain with value group $\Bbb Z \oplus \Bbb Q$

Question

Let $R$ be a two-dimensional valuation domain with prime ideals $0 \subset P \subset M$ and value group $G=\Bbb Z \oplus \Bbb Q$. Then $M^2=M$ and $P^2 \neq P$. Can we say $P^n \neq P^{n+1}$ for all $n$?

Why $M^2=M$ and $P^2\neq P$?

Answer 1

I suppose the order on $\mathbb Z\oplus\mathbb Q$ is the lexicographical order.

Denote by $v$ a (surjective) valuation whose value group is $\mathbb Z\oplus\mathbb Q$, and set $R=R_v$.

Then $(m,q)\in\mathbb Z\oplus\mathbb Q$ is such that $(m,q)>(0,0)$ iff $m\ge1$, or $m=0$ and $q>0$. In the first case $(m,q)=(m,q-1)+(0,1)$, while in the second $(0,q)=(0,q/2)+(0,q/2)$. This shows that $M=M^2$.

We can identify $P$ with the set $\{x\in R:v(x)\in\mathbb Z_{>0}\oplus\mathbb Q\}$, and its easily seen that if $v(x_0)=(1,0)$ then $x_0\in P-P^2$. Moreover, $x_0^n\in P^n-P^{n+1}$.