Lemma 109.12.5. Let $R$ be a domain with fraction field $K$. If $R[[x]]$ is flat over $R[x]$, then $R$ is normal if and only if $R$ is completely normal (Algebra, Definition 10.37.3).

Proof. Suppose we have $\alpha \in K$ and a nonzero $r \in R$ such that $r \alpha ^ n \in R$ for all $n \geq 1$. Then we consider $f = \sum r \alpha ^{n - 1} x^ n$ in $R[[x]]$. Write $\alpha = a/b$ for $a, b \in R$ with $b$ nonzero. Then we see that $(a x - b)f = -rb$. It follows that $rb$ is in the ideal $(ax - b)R[[x]]$. Let $S = \{ h \in R[x] : h(0) = 1\}$. This is a multiplicative subset and flatness of $R[x] \to R[[x]]$ implies that $S^{-1}R[x] \to R[[x]]$ is faithfully flat (details omitted; hint: use Algebra, Lemma 10.39.16). Hence

$S^{-1}R/(ax - b)S^{-1}R \to R[[x]]/(ax - b)R[[x]]$

is injective. We conclude that $h rb = (ax - b) g$ for some $h \in S$ and $g \in R[x]$. Writing $h = 1 + h_1 x + \ldots + h_ d x^ d$ shows that we obtain

$1 + h_1 x + \ldots + h_ d x^ d = (1/r)(\alpha x - 1)g$

This factorization in $K[x]$ gives a corresponding factorization in $K[x^{-1}]$ which shows that $\alpha$ is the root of a monic polynomial with coefficients in $R$ as desired. $\square$

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