Let $(R, \mathfrak m)$ be a valuation ring such that $\mathfrak m$ and every ideal of finite height is principal. Then is $R$ Noetherian , i.e. a discrete valuation ring ?

## 1 Answer

This is false. For example, there can be no nonzero ideals of finite height.

**Remark.** Note that for every totally ordered abelian group $\Gamma$, there exists a valuation ring with value group $\Gamma$. For example, we may extend the valuation
\begin{align*}
v \colon \mathbb C[\Gamma] &\to \Gamma \cup \{\infty\}\\
\sum \alpha_\gamma \gamma &\mapsto \min\{\gamma\ |\ \alpha_\gamma \neq 0\}
\end{align*}
to the fraction field $K = \operatorname{Frac}(\mathbb C[\Gamma])$, and then take $R$ to be the valuation ring $\{x \in K\ |\ v(x) \geq 0\}$.
Thus, the question is really one about totally ordered abelian groups. In this language, an *ideal* of $\Gamma$ is a set $I \subseteq \Gamma_{\geq 0}$ such that $i \in I$ and $\gamma \geq 0$ implies $i + \gamma \in I$, and an ideal $I$ is *prime* if $x+y \in I$ implies $x \in I$ or $y \in I$.

**Example.** Let $\Gamma = \mathbb Z[x]$, with the ordering given by $f \leq g$ if and only if $f(x) \leq g(x)$ for $x \gg 0$. Equivalently, $\Gamma$ is a free $\mathbb Z$-module with countably infinite basis $\{x^i\ |\ i \in \mathbb Z_{\geq 0}\}$, equipped with the lexicographical ordering such that
$$1 < x < x^2 < \ldots.$$
The principal ideals $\mathfrak p_i = \{f \ \big|\ f \geq x^i\} \subseteq \Gamma_{\geq 0}$ are prime for all $i \in \mathbb Z_{\geq 0}$, giving an infinite chain
$$\ldots \subsetneq \mathfrak p_2 \subsetneq \mathfrak p_1 \subsetneq \mathfrak p_0 = \mathfrak m.$$
Moreover, if $\mathfrak q$ is a nonempty prime, then there exists $f \in \mathfrak q$. If $\deg f = n$, then $f < x^{n+1}$, so $\mathfrak p_{n+1} \subseteq \mathfrak q$ since $\mathfrak q$ is an ideal. Hence, $\mathfrak q$ has infinite height. On the other hand, $\mathfrak m$ is principal (in fact, so is every $\mathfrak p_i$, and one can show that these are the only prime ideals).

Since the value group is larger than $\mathbb Z$, this is a counterexample to your question.