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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.

Johnny T.
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    No, it’s false. Consider for instance $R=k[[t]]$, $x=1$ and $y=t$. – Aphelli Oct 30 '19 at 21:05
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    Since $v(x^m) = m v(x)$ this is trivially true if $v(x) \ne 0$ and the valuation comes from an absolute value, ie. if it is $\Bbb{R}$ valued, not if it is $G$-valued for any ordered abelian group (try with $G=\Bbb{Z}+\epsilon \Bbb{Z}$ on $f[y]{(y)}+x f[x]{(x)}f(y)$) – reuns Oct 30 '19 at 21:05
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    Other counterexamples where the valuation is not trivial may be found in the realm of definable functions. You may wish to edit your post to clarify the scope of the question. – KReiser Oct 30 '19 at 21:07
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    With your update to ask about valuation rings having only two prime ideals, it should be noted that this is true in the case of discrete valuation rings, as seen in this MSE post. – KReiser Oct 30 '19 at 21:48
  • @KReiser I thought my first sentence was sufficient.. maybe it was too brief? – Johnny T. Oct 31 '19 at 05:46

2 Answers2

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$\Bbb{Z+\epsilon Z}$ 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.

reuns
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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*}

Arpan Dutta
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