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Tag 032J

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Lemma 10.155.6. 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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42600–42604 (see updates for more information).

    \begin{lemma}
    \label{lemma-Laurent-ring-N-1}
    Let $R$ be a Noetherian domain.
    If $R[z, z^{-1}]$ is N-1, then so is $R$.
    \end{lemma}
    
    \begin{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.
    \end{proof}

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