## 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 8177–8183 (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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