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The Stacks project

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


Comments (2)

Comment #44 by Rankeya on

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

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