Lemma 10.161.13. Let $R$ be a Noetherian domain. If $R$ is N-1 then $R[x]$ is N-1. If $R$ is N-2 then $R[x]$ is N-2.

**Proof.**
Assume $R$ is N-1. Let $R'$ be the integral closure of $R$ which is finite over $R$. Hence also $R'[x]$ is finite over $R[x]$. The ring $R'[x]$ is normal (see Lemma 10.37.8), hence N-1. This proves the first assertion.

For the second assertion, by Lemma 10.161.7 it suffices to show that $R'[x]$ is N-2. In other words we may and do assume that $R$ is a normal N-2 domain. In characteristic zero we are done by Lemma 10.161.11. In characteristic $p > 0$ we have to show that the integral closure of $R[x]$ is finite in any finite purely inseparable extension of $L/K(x)$ where $K$ is the fraction field of $R$. There exists a finite purely inseparable field extension $L'/K$ and $q = p^ e$ such that $L \subset L'(x^{1/q})$; some details omitted. As $R[x]$ is Noetherian it suffices to show that the integral closure of $R[x]$ in $L'(x^{1/q})$ is finite over $R[x]$. And this integral closure is equal to $R'[x^{1/q}]$ with $R \subset R' \subset L'$ the integral closure of $R$ in $L'$. Since $R$ is N-2 we see that $R'$ is finite over $R$ and hence $R'[x^{1/q}]$ is finite over $R[x]$. $\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: