
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. $\square$

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