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.