Lemma 110.12.5. 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
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
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. $\square$
Post a comment
Your email address will not be published.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 toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #8815 by Laurent Moret-Bailly on
Comment #9274 by Stacks project on
There are also: