Valuation rings only have two prime ideals?

Question

If $R$ is a DVR then we know that $R$ has only two prime ideals. Does this still hold true for valuation ring $R$?

I was trying to prove this by showing: Let $R$ be a valuation ring. I wanted to prove given $x,y \in R$ if $0 < v(x) < v(y)$ then $v(x^m)> v(y)$ for some $m \in \mathbb{Z}$.

But as I have been pointed out by the comments this does not hold in general.

Are there examples of valuation ring $R$ with more than two prime ideals? Any comments/examples would be appreciated. Thank you.

ps I have changed my question from asking how to show the above inequality to this since the question I was asking is not true in general as pointed out in the comments.

Answer 1

$\Bbb{Z+\epsilon Z}$ is an ordered abelian group. On $k[x,y]$ we have the valuation $$v(x)=1,v(y)=\epsilon, \qquad v(\sum_{i,j} c_{i,j} x^i y^j) = \inf_{c_{i,j}\ne 0} v(x^i y^j)\in \Bbb{Z+\epsilon Z}$$ $v^{-1}(\infty)=\{0\}$ so $v$ extends to the fraction field and the valuation ring is $$O_v = \{ u\in Frac(k[x,y]), v(u)\ge 0\}=k[y]_{(y)}+x k(y)[x]_{(x)}$$ you'll obtain 3 prime ideals : $(0), (xk[y^{-1}]),(y)$

This valuation is not $\Bbb{R}$-valued : it doesn't correspond to an absolute value.

Answer 2

Let $\nu$ be the valuation corresponding to the valuation ring $R$. Then the rank of $\nu$ is equal to the Krull dimension of $R$. A valuation has rank 1 if and only if the value group is a subgroup of $\mathbb{R}$. One easy recipe to construct valuations of arbitrary rank is as follows :

Let $\Gamma$ be an ordered abelian group and $\mathtt{k}$ be any field. Then we can construct a valuation with value group $\Gamma$ and residue field $\mathtt{k}$ in the generalized power series field $\mathtt{k}((t^{\Gamma}))$ by defining \begin{equation*} \nu (\sum a_{\gamma} t^{\gamma}) = \text{min} \{ \gamma \mid a_{\gamma} \neq 0 \}. \end{equation*}