The Stacks project

Example 10.162.17. A discrete valuation ring is Nagata if and only if it is N-2 (because the quotient by the maximal ideal is a field and hence N-2). The discrete valuation ring $A$ of Example 10.119.5 is not Nagata, i.e., it is not N-2. Namely, the finite extension $A \subset R = A[f]$ is not N-1. To see this say $f = \sum a_ i x^ i$. For every $n \geq 1$ set $g_ n = \sum _{i < n} a_ i x^ i \in A$. Then $h_ n = (f - g_ n)/x^ n$ is an element of the fraction field of $R$ and $h_ n^ p \in k^ p[[x]] \subset A$. Hence the integral closure $R'$ of $R$ contains $h_1, h_2, h_3, \ldots $. Now, if $R'$ were finite over $R$ and hence $A$, then $f = x^ n h_ n + g_ n$ would be contained in the submodule $A + x^ nR'$ for all $n$. By Artin-Rees this would imply $f \in A$ (Lemma 10.51.4), a contradiction.


Comments (0)

There are also:

  • 2 comment(s) on Section 10.162: Nagata 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Your email address will not be published.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 09E1. Beware of the difference between the letter 'O' and the digit '0'.