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Tag 030A

Chapter 10: Commutative Algebra > Section 10.36: Normal rings

Lemma 10.36.8. Let $R$ be a normal domain. Then $R[x]$ is a normal domain.

Proof. The result is true if $R$ is a field $K$ because $K[x]$ is a euclidean domain and hence a principal ideal domain and hence normal by Lemma 10.36.6. Let $g$ be an element of the fraction field of $R[x]$ which is integral over $R[x]$. Because $g$ is integral over $K[x]$ where $K$ is the fraction field of $R$ we may write $g = \alpha_d x^d + \alpha_{d-1}x^{d-1} + \ldots + \alpha_0$ with $\alpha_i \in K$. By Lemma 10.36.7 the elements $\alpha_i$ are integral over $R$ and hence are in $R$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 7805–7809 (see updates for more information).

    \begin{lemma}
    \label{lemma-polynomial-domain-normal}
    Let $R$ be a normal domain.
    Then $R[x]$ is a normal domain.
    \end{lemma}
    
    \begin{proof}
    The result is true if $R$ is a field $K$ because
    $K[x]$ is a euclidean domain and hence a principal ideal
    domain and hence normal by Lemma \ref{lemma-PID-normal}.
    Let $g$ be an element of the fraction field of
    $R[x]$ which is integral over $R[x]$. Because $g$
    is integral over $K[x]$ where $K$ is the fraction
    field of $R$ we may write $g = \alpha_d x^d + \alpha_{d-1}x^{d-1} +
    \ldots + \alpha_0$ with $\alpha_i \in K$.
    By Lemma \ref{lemma-prepare-polynomial-ring-normal}
    the elements $\alpha_i$ are integral over $R$ and
    hence are in $R$.
    \end{proof}

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