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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. $\square$

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