Lemma 10.161.6. Let $R$ be a Noetherian domain. If $R[z, z^{-1}]$ is N-1, then so is $R$.

Proof. Let $R'$ be the integral closure of $R$ in its field of fractions $K$. Let $S'$ be the integral closure of $R[z, z^{-1}]$ in its field of fractions. Clearly $R' \subset S'$. Since $K[z, z^{-1}]$ is a normal domain we see that $S' \subset K[z, z^{-1}]$. Suppose that $f_1, \ldots , f_ n \in S'$ generate $S'$ as $R[z, z^{-1}]$-module. Say $f_ i = \sum a_{ij}z^ j$ (finite sum), with $a_{ij} \in K$. For any $x \in R'$ we can write

$x = \sum h_ i f_ i$

with $h_ i \in R[z, z^{-1}]$. Thus we see that $R'$ is contained in the finite $R$-submodule $\sum Ra_{ij} \subset K$. Since $R$ is Noetherian we conclude that $R'$ is a finite $R$-module. $\square$

There are also:

• 5 comment(s) on Section 10.161: Japanese rings

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).