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.

There are also:

  • 3 comment(s) on Section 10.50: Valuation rings

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00II. Beware of the difference between the letter 'O' and the digit '0'.