The Stacks project

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. $\square$

Comments (0)

There are also:

  • 3 comment(s) on Section 109.12: Nonflat completions

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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F1Y. Beware of the difference between the letter 'O' and the digit '0'.