The Stacks project

Lemma 10.161.11. A Noetherian domain whose fraction field has characteristic zero is N-1 if and only if it is N-2 (i.e., Japanese).

Proof. This is clear from Lemma 10.161.8 since every field extension in characteristic zero is separable. $\square$

Comments (4)

Comment #4965 by Rankeya on

When you say a "Noetherian domain of characteristic 0" do you mean the domain contains the rationals or that its fraction field has characteristic 0? I think this result holds as long as the fraction field of the domain has characteristic 0, so may be it is good to specify this in the statement?

Comment #7448 by Haohao Liu on

From the proof we see that the condition "fraction field has characteristic zero " can be relaxed to "fraction field is perfect".

Comment #7449 by Laurent Moret-Bailly on

@ #7448: True, but if is a noetherian domain with perfect fraction field of char. , then . Otherwise, by Krull-Akizuki, there is a discrete valuation ring between and , and a uniformizer of cannot be a -th power in . (There may be simpler arguments).

There are also:

  • 7 comment(s) on Section 10.161: Japanese 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Your email address will not be published.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 032M. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 10.161.11 as pdf