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Tag 00H7

Chapter 10: Commutative Algebra > Section 10.37: Going down for integral over normal

Lemma 10.37.6. Let $R \subset S$ be an inclusion of domains. Assume $R$ is normal. Let $g \in S$ be integral over $R$. Then the minimal polynomial of $g$ has coefficients in $R$.

Proof. Let $P = x^m + b_{m-1} x^{m-1} + \ldots + b_0$ be a polynomial with coefficients in $R$ such that $P(g) = 0$. Let $Q = x^n + a_{n-1}x^{n-1} + \ldots + a_0$ be the minimal polynomial for $g$ over the fraction field $K$ of $R$. Then $Q$ divides $P$ in $K[x]$. By Lemma 10.37.5 we see the $a_i$ are integral over $R$. Since $R$ is normal this means they are in $R$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 8174–8180 (see updates for more information).

    \begin{lemma}
    \label{lemma-minimal-polynomial-normal-domain}
    Let $R \subset S$ be an inclusion of domains.
    Assume $R$ is normal. Let $g \in S$ be integral
    over $R$. Then the minimal polynomial of $g$
    has coefficients in $R$.
    \end{lemma}
    
    \begin{proof}
    Let $P = x^m + b_{m-1} x^{m-1} + \ldots + b_0$
    be a polynomial with coefficients in $R$
    such that $P(g) = 0$. Let $Q = x^n + a_{n-1}x^{n-1} + \ldots + a_0$
    be the minimal polynomial for $g$ over the fraction field
    $K$ of $R$. Then $Q$ divides $P$ in $K[x]$. By Lemma
    \ref{lemma-polynomials-divide} we see the $a_i$ are
    integral over $R$. Since $R$ is normal this
    means they are in $R$.
    \end{proof}

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