Lemma 10.49.18. A valuation ring is Noetherian if and only if it is a discrete valuation ring or a field.

**Proof.**
Suppose $A$ is a discrete valuation ring with valuation $v : A \setminus \{ 0\} \to \mathbf{Z}$ normalized so that $\mathop{\mathrm{Im}}(v) \subset \mathbf{Z}_{\geq 0}$. By Lemma 10.49.17 the ideals of $A$ are the subsets $I_ n = \{ 0\} \cup v^{-1}(\mathbf{Z}_{\geq n})$. It is clear that any element $x \in A$ with $v(x) = n$ generates $I_ n$. Hence $A$ is a PID so certainly Noetherian.

Suppose $A$ is a Noetherian valuation ring with value group $\Gamma $. By Lemma 10.49.17 we see the ascending chain condition holds for ideals in $\Gamma $. We may assume $A$ is not a field, i.e., there is a $\gamma \in \Gamma $ with $\gamma > 0$. Applying the ascending chain condition to the subsets $\gamma + \Gamma _{\geq 0}$ with $\gamma > 0$ we see there exists a smallest element $\gamma _0$ which is bigger than $0$. Let $\gamma \in \Gamma $ be an element $\gamma > 0$. Consider the sequence of elements $\gamma $, $\gamma - \gamma _0$, $\gamma - 2\gamma _0$, etc. By the ascending chain condition these cannot all be $> 0$. Let $\gamma - n \gamma _0$ be the last one $\geq 0$. By minimality of $\gamma _0$ we see that $0 = \gamma - n \gamma _0$. Hence $\Gamma $ is a cyclic group as desired. $\square$

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