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Tag 0BI0

Chapter 10: Commutative Algebra > Section 10.36: Normal rings

Lemma 10.36.9. Let $R$ be a Noetherian normal domain. Then $R[[x]]$ is a Noetherian normal domain.

Proof. The power series ring is Noetherian by Lemma 10.30.2. Let $f, g \in R[[x]]$ be nonzero elements such that $w = f/g$ is integral over $R[[x]]$. Let $K$ be the fraction field of $R$. Since the ring of Laurent series $K((x)) = K[[x]][1/x]$ is a field, we can write $w = a_n x^n + a_{n + 1} x^{n + 1} + \ldots)$ for some $n \in \mathbf{Z}$, $a_i \in K$, and $a_n \not = 0$. By Lemma 10.36.4 we see there exists a nonzero element $h = b_m x^m + b_{m + 1} x^{m + 1} + \ldots$ in $R[[x]]$ with $b_m \not = 0$ such that $w^e h \in R[[x]]$ for all $e \geq 1$. We conclude that $n \geq 0$ and that $b_m a_n^e \in R$ for all $e \geq 1$. Since $R$ is Noetherian this implies that $a_n \in R$ by the same lemma. Now, if $a_n, a_{n + 1}, \ldots, a_{N - 1} \in R$, then we can apply the same argument to $w - a_n x^n - \ldots - a_{N - 1} x^{N - 1} = a_N x^N + \ldots$. In this way we see that all $a_i \in R$ and the lemma is proved. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 7796–7800 (see updates for more information).

    \begin{lemma}
    \label{lemma-power-series-over-Noetherian-normal-domain}
    Let $R$ be a Noetherian normal domain. Then $R[[x]]$ is
    a Noetherian normal domain.
    \end{lemma}
    
    \begin{proof}
    The power series ring is Noetherian by
    Lemma \ref{lemma-Noetherian-power-series}.
    Let $f, g \in R[[x]]$ be nonzero elements such that
    $w = f/g$ is integral over $R[[x]]$.
    Let $K$ be the fraction field of $R$. Since the ring of Laurent series
    $K((x)) = K[[x]][1/x]$ is a field, we can write
    $w = a_n x^n + a_{n + 1} x^{n + 1} + \ldots)$
    for some $n \in \mathbf{Z}$, $a_i \in K$, and $a_n \not = 0$.
    By Lemma \ref{lemma-almost-integral} we see there exists a
    nonzero element $h = b_m x^m + b_{m + 1} x^{m + 1} + \ldots$
    in $R[[x]]$ with $b_m \not = 0$ such that
    $w^e h \in R[[x]]$ for all $e \geq 1$. We conclude that $n \geq 0$ and that
    $b_m a_n^e \in R$ for all $e \geq 1$.
    Since $R$ is Noetherian this implies that $a_n \in R$ by
    the same lemma. Now, if $a_n, a_{n + 1}, \ldots, a_{N - 1} \in R$,
    then we can apply the same argument to
    $w - a_n x^n - \ldots - a_{N - 1} x^{N - 1} = a_N x^N + \ldots$.
    In this way we see that all $a_i \in R$ and the lemma is proved.
    \end{proof}

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