Analogue of a ring extension splitting in the Kummer case - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T14:35:28Z http://mathoverflow.net/feeds/question/29156 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29156/analogue-of-a-ring-extension-splitting-in-the-kummer-case Analogue of a ring extension splitting in the Kummer case Alexandra Seceleanu 2010-06-22T22:10:08Z 2010-06-22T22:10:08Z <blockquote> <p>Background (the Kummer extension case)</p> </blockquote> <p>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 $A = R[u^{\frac{1}{p}}]$.</p> <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 $p$.</p> <p><code>$\begin{array}{ccc} K &amp; \hookrightarrow &amp; K[\xi]\\ \downarrow &amp; &amp; \downarrow \\ L &amp;\hookrightarrow &amp; L[\xi]\\ \end{array}$</code> where $K[\xi]\subset L[\xi]$ is a Kummer extension with $Gal((L[\xi]/K[\xi])=&lt;\sigma>\simeq \mathbf{Z}/p\mathbf{Z}$ </p> <p><code>$\begin{array}{ccc} R &amp; \hookrightarrow &amp; R' \\ \downarrow &amp; &amp; \downarrow \\ A &amp;\hookrightarrow &amp; A' \\ \end{array}$</code> where $R'$ is the integral closure of $R$ in $R[\xi]$ and $A'$ is the integral closure of $A$ in $L[\xi]$.</p> <p>Now $L[\xi]$ has an eigenspace decomposition $L[\xi]=L_1\bigoplus \ldots \bigoplus L_{p}$, where <code>$L_i= \{ x \in L[\xi]| \sigma(x)=\xi^i *x \}$</code>. </p> <p>Here the contractions $S_i=A'\cap L_i$ are rank one $R'$-modules (in fact <code>$S_i= \{x \in A' |\sigma(x)=\xi^i*x\}$</code>).</p> <blockquote> <p>Question</p> </blockquote> <p>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 <code>$\begin{array}{ccc} K &amp; \hookrightarrow &amp; K'\\ \downarrow &amp; &amp; \downarrow \\ L &amp;\hookrightarrow &amp; L'\\ \end{array}$</code> and <code>$\begin{array}{ccc} R &amp; \hookrightarrow &amp; R'\\ \downarrow &amp; &amp; \downarrow \\ A &amp;\hookrightarrow &amp; A'\\ \end{array}$</code>.</p> <p>$L'$ still has an eigenspace decomposition of the type $L'=L_1\bigoplus \ldots \bigoplus L_{p}$, where $L_i\simeq L$. </p> <blockquote> <p>What conditions would ensure that the contractions $S_i=L_i\cap A'$ are still rank one $R'$-modules? </p> </blockquote> <p>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). </p> <p>Any help is much appreciated.</p>