# The Stacks Project

## Tag 0BI0

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 7825–7829 (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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