Lemma 10.50.18. 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 $\mathop{\mathrm{Im}}(v) = \mathbf{Z}_{\geq 0}$. By Lemma 10.50.17 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 10.50.17 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. $\square$

Comment #44 by Rankeya on

In the first line of the proof, 'valutation ring' should be valuation ring.

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