The Stacks project

Lemma 10.50.5. Let $A \subset K$ be a subring of a field $K$ such that for all $x \in K$ either $x \in A$ or $x^{-1} \in A$ or both. Then $A$ is a valuation ring with fraction field $K$.

Proof. If $A$ is not $K$, then $A$ is not a field and there is a nonzero maximal ideal $\mathfrak m$. If $\mathfrak m'$ is a second maximal ideal, then choose $x, y \in A$ with $x \in \mathfrak m$, $y \not\in \mathfrak m$, $x \not\in \mathfrak m'$, and $y \in \mathfrak m'$. Then neither $x/y \in A$ nor $y/x \in A$ contradicting the assumption of the lemma. Thus we see that $A$ is a local ring. Suppose that $A'$ is a local ring contained in $K$ which dominates $A$. Let $x \in A'$. We have to show that $x \in A$. If not, then $x^{-1} \in A$, and of course $x^{-1} \in \mathfrak m_ A$. But then $x^{-1} \in \mathfrak m_{A'}$ which contradicts $x \in A'$. $\square$


Comments (2)

Comment #8361 by Et on

There's no need to quote lemma 10.15.2

There are also:

  • 4 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Your email address will not be published.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 052K. Beware of the difference between the letter 'O' and the digit '0'.