The Stacks project

Lemma 33.39.4. Let $A$ be a reduced Nagata local ring of dimension $1$. The $\delta $-invariant of $A$ is $0$ if and only if $A$ is a discrete valuation ring.

Proof. If $A$ is a discrete valuation ring, then $A$ is normal and the ring $A'$ is equal to $A$. Conversely, if the $\delta $-invariant of $A$ is $0$, then $A$ is integrally closed in its total ring of fractions which implies that $A$ is normal (Algebra, Lemma 10.37.16) and this forces $A$ to be a discrete valuation ring by Algebra, Lemma 10.119.7. $\square$

Comments (0)

There are also:

  • 3 comment(s) on Section 33.39: The delta invariant

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Your email address will not be published.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 0C3U. Beware of the difference between the letter 'O' and the digit '0'.