# Analogue of a ring extension splitting in the Kummer case

Background (the Kummer extension case)

Let $R$ be a complete regular local ring (it follws that it's a UFD) with a prime integer $p$ contained in the maximal ideal of $R$ (I'm mostly interested in $R=V[[x_1\ldots x_n]]$ with V a DVR). Consider a finite extension of rings $R\hookrightarrow A$ with $A = R[u^{\frac{1}{p}}]$.

Let $K$ and $F$ be the fraction fields of $R$ and $A$ respectively. Let $\xi$ be a $p^{th}$ primitive root of 1 and consider the following commutative diagrams where horizontally displayed extensions are of order $p-1$ and vertical ones are of order $p$.

$\begin{array}{ccc} K & \hookrightarrow & K[\xi]\\ \downarrow & & \downarrow \\ L &\hookrightarrow & L[\xi]\\ \end{array}$ where $K[\xi]\subset L[\xi]$ is a Kummer extension with $Gal((L[\xi]/K[\xi])=<\sigma>\simeq \mathbf{Z}/p\mathbf{Z}$

$\begin{array}{ccc} R & \hookrightarrow & R' \\ \downarrow & & \downarrow \\ A &\hookrightarrow & A' \\ \end{array}$ where $R'$ is the integral closure of $R$ in $R[\xi]$ and $A'$ is the integral closure of $A$ in $L[\xi]$.

Now $L[\xi]$ has an eigenspace decomposition $L[\xi]=L_1\bigoplus \ldots \bigoplus L_{p}$, where $L_i= \{ x \in L[\xi]| \sigma(x)=\xi^i *x \}$.

Here the contractions $S_i=A'\cap L_i$ are rank one $R'$-modules (in fact $S_i= \{x \in A' |\sigma(x)=\xi^i*x\}$).

Question

Suppose that we DROP the hypothesis $A = R[u^{\frac{1}{p}}]$ and only assume $[L:K]=p$. Let $L'$ be such that $L'/K$ is Galois and consider the analogous diagrams $\begin{array}{ccc} K & \hookrightarrow & K'\\ \downarrow & & \downarrow \\ L &\hookrightarrow & L'\\ \end{array}$ and $\begin{array}{ccc} R & \hookrightarrow & R'\\ \downarrow & & \downarrow \\ A &\hookrightarrow & A'\\ \end{array}$.

$L'$ still has an eigenspace decomposition of the type $L'=L_1\bigoplus \ldots \bigoplus L_{p}$, where $L_i\simeq L$.

What conditions would ensure that the contractions $S_i=L_i\cap A'$ are still rank one $R'$-modules?

I am guessing that we need $[L':L]$ not to be divisible by $p$, but would be happy to see a proof even under more restrictive conditions (such as $Gal(L'/L)$ commutes with $Gal(L'/R')$ perhaps).

Any help is much appreciated.

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While I cannot help with the question, I wanted to commend you on the exemplary presentation! – Victor Protsak Jun 23 '10 at 4:58