CharacterizationOfAbelianExtensionsOfExponentN 2009-01-05 22:42:41 2009-01-05 22:42:41 Theorem abelian extension % 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} \newcommand{\Order}[1]{\left\lvert #1 \right\rvert} % %% \theoremstyle{plain} %% This is the default \newtheorem{thm}{Theorem} \newtheorem{cor}[thm]{Corollary} \begin{thm} Let $K$ be a field containing the $n^{\mathrm{th}}$ roots of unity, with characteristic not dividing $n$. Let $L$ be a finite extension of $K$. Then the following are equivalent: \begin{enumerate} \item $L/K$ is Galois with abelian Galois group of exponent dividing $n$ \item $L=K(\sqrt[n]{a_1},\ \dotsc,\ \sqrt[n]{a_k})$ for some $a_i\in K$. \end{enumerate} \end{thm} \begin{proof} Let $\zeta\in K$ be a primitive $n^{\mathrm{th}}$ root of unity. $(2\Rightarrow 1):$ Choose $\alpha_i\in L$ such that $\alpha_i^n = a_i\in K$. Then for each $i$, the elements $\alpha_i,\ \zeta\alpha_i,\ \dotsc,\ \zeta^{n-1}\alpha_i$ are distinct and are all the roots of $x^n-a_i$ in $L$. Thus $x^n-a_i$ is separable over $K$ and splits in $L$, so that $L$ is the splitting field of the set of polynomials $\{x^n-a_i\ \mid\ 1\leq i\leq k\}$. Thus $L/K$ is Galois. Given $\sigma\in \Gal(L/K)$, for each $i$ we have $\sigma(\alpha_i) = \zeta^j\alpha_i$ for some $1\leq j\leq n$, so that $\sigma^k(\alpha_i) = \zeta^{kj}\alpha_i$. It follows that $\sigma^n$ is the identity for every $\sigma\in\Gal(L/K)$, so that the exponent of $\Gal(L/K)$ divides $n$. It remains to show that $\Gal(L/K)$ is abelian; this follows trivially from the simple definition of the Galois action as multiplication by some $n^{\mathrm{th}}$ root of unity: if $\sigma,\tau\in\Gal(L/K)$ with $\sigma(\alpha_i) = \zeta^r\alpha_i,\quad\tau(\alpha_i) = \zeta^s\alpha_i$, then \begin{align*} (\sigma\tau)(\alpha_i) &= \sigma(\zeta^s\alpha_i) = \zeta^s\zeta^r\alpha_i \\ (\tau\sigma)(\alpha_i) &= \tau(\zeta^r\alpha_i) = \zeta^r\zeta^s\alpha_i \end{align*} Thus $\sigma\tau=\tau\sigma$ on each $\alpha_i$. But the $\alpha_i$ generate $L/K$, so $\sigma\tau=\tau\sigma$ on $L$ and $\Gal(L/K)$ is abelian. $(1 \Rightarrow 2):$ Let $G=\Gal(L/K)$, and write $G=C_1\times \dots \times C_r$ where each $C_i$ is cyclic; $\Order{C_i}=m_i\mid n$ for each $i$. For each $i$, define a subgroup $H_i\leq G$ by \[ H_i = C_1 \times \dots \times C_{i-1} \times C_{i+1} \times \dots \times C_r \] Then $G/H_i \cong C_i$. Let $L_i$ be the fixed field of $H_i$. $L_i$ is normal over $K$ since $H_i$ is normal in $G$, and $\Gal(L_i/K) \cong G/H_i \cong C_i$ and thus $L_i/K$ is cyclic Galois of order $m_i$. $K$ contains the primitive $m_i^{\mathrm{th}}$ root of unity $\zeta^{n/m_i}$ and thus $L_i = K(\alpha_i)$ for some $\alpha_i\in L$ with $\alpha_i^{m_i}\in K$ (by Kummer theory). But then also $\alpha_i^n\in K$. Then \[ \Gal(L:K(\alpha_1,\dotsc,\alpha_r)) = H_1\cap \dots \cap H_r = \{1\} \] since any element of the left-hand group fixes each $\alpha_i$ and thus fixes $L_i$ so is the identity in $G/H_i$. Thus $L=K(\alpha_1,\dotsc,\alpha_r)$. \end{proof} \begin{cor} If $L/K$ is the maximal abelian extension of $K$ of exponent $n$, where $n$ is prime to the characteristic of $K$, then $L = K(\{\sqrt[n]{a}\})$ for some set of $a\in K$. \end{cor} \begin{proof} Clearly $K(\{\sqrt[n]{a}\mid a \in K^*)$ is an infinite abelian extension of exponent $n$. If $L$ is the maximal such extension, choose $b \in L$. Then $K(b)$ is a finite extension of exponent dividing $n$ and thus $K(b)$ is of the required form. Thus $L=\cup_{b\in L} K(b)$ is also of the required form; for example, if $S\subset K^*$ is a set of coset representatives for $K^*/(K^*)^n$, then $L=K(S)$. \end{proof} \begin{thebibliography}{10} \bibitem{bib:df} Morandi,~P., \emph{Field and Galois Theory}, Springer, 1996. \end{thebibliography}