## 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 7775–7779 (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}
```

## Comments (0)

## Add a comment on tag `030A`

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

There are no comments yet for this tag.