ProofOfKummerTheory 2009-01-05 20:11:32 2009-01-06 15:51:08 Proof Kummer theory 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\N}{N} % %% \theoremstyle{plain} %% This is the default \newtheorem{thm}{Theorem} \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \begin{proof} Let $\zeta\in K$ be a primitive $n^{\mathrm{th}}$ root of unity, and denote by $\boldsymbol{\mu}_n$ the subgroup of $K^{\star}$ generated by $\zeta$. (1) Let $L=K(\sqrt[n]{a})$; then $L/K$ is Galois since $K$ contains all $n^{\mathrm{th}}$ roots of unity and thus is a splitting field for $x^n-a$, which is separable since $n\neq 0$ in $K$. Thus the elements of $\Gal(L/K)$ permute the roots of $x^n-a$, which are \[\sqrt[n]{a},\ \zeta \sqrt[n]{a},\ \zeta^2 \sqrt[n]{a},\ \dotsc,\ \zeta^{n-1}\sqrt[n]{a}\] and thus for $\sigma\in \Gal(L/K)$, we have $\sigma(\sqrt[n]{a}) = \zeta_{\sigma} \sqrt[n]{a}$ for some $\zeta_{\sigma}\in\boldsymbol{\mu}_n$. Define a map \[p:\Gal(L/K)\to \boldsymbol{\mu}_n:\sigma\mapsto\zeta_{\sigma}\] We will show that $p$ is an injective homomorphism, which proves the result. Since $\boldsymbol{\mu}_n\subset K$, each $n^{\mathrm{th}}$ root of unity is fixed by $\Gal(L/K)$. Then for $\sigma,\tau\in\Gal(L/K)$, \[ \zeta_{\sigma\tau}\sqrt[n]{a} =\sigma\tau(\sqrt[n]{a}) = \sigma(\zeta_{\tau}\sqrt[n]{a}) = \zeta_{\tau}(\sigma(\sqrt[n]{a})) = \zeta_{\sigma}\zeta_{\tau}\sqrt[n]{a} \] so that $\zeta_{\sigma\tau} = \zeta_{\sigma}\zeta_{\tau}$ and $p$ is a homomorphism. The kernel of the map consists of all elements of $\Gal(L/K)$ which fix $\sqrt[n]{a}$, so that $p$ is injective and we are done. (2) Note that $\N_{L/K}(\zeta) = 1$ since $\zeta$ is a root of $x^n-1$, so that by Hilbert's Theorem 90, \[\zeta = \sigma(u)/u,\quad\text{for some }u\in L\] But then $\sigma(u) = \zeta u$ so that $\sigma(u^n) = \sigma(u)^n = \zeta^n u^n = u^n$ and $a=u^n\in K$ since it is fixed by a generator of $\Gal(L/K)$. Then clearly $K(u)$ is a splitting field of $x^n-a$, and the elements of $\Gal(L/K)$ send $u$ into distinct elements of $K(u)$. Thus $K(u)$ admits at least $n$ automorphisms over $K$, so that $[K(u):K]\geq n = [L:K]$. But $K(u)\subset L$, so $K(\sqrt[n]{a})=K(u)=L$. \end{proof} \begin{thebibliography}{10} \bibitem{bib:df} Dummit,~D.,~Foote,~R.M., \emph{Abstract Algebra, Third Edition}, Wiley, 2004. \bibitem{bib:kap} Kaplansky,~I., \emph{Fields and Rings}, University of Chicago Press, 1969. \end{thebibliography}