GaloisSubfieldsOfRealRootExtensionsAreAtMostQuadratic 2007-12-30 10:49:06 2007-12-30 15:52:02 Theorem % 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 \newcommand{\Reals}{\mathbb{R}} \DeclareMathOperator{\Gal}{Gal} \newtheorem{thm}{Theorem} \begin{thm} Suppose $F\subset L\subset K=F(\sqrt[n]\alpha)\subset\Reals$ are fields with $\alpha\in F$ and $L$ Galois over $F$. Then $[L:F]\leq 2$. \end{thm} \textbf{Proof. } Let $\zeta_n$ be a primitive $n^{\mathrm{th}}$ root of unity, and define $F'=F(\zeta_n)$, $L'=L(\zeta_n)$, and $K'=K(\zeta_n)=F'(\sqrt[n]\alpha)$. \[\xymatrix @R1pc@C.3pc{ & K'=K(\zeta_n)=F'(\sqrt[n]\alpha) \ar@{-}[dl]\ar@{-}[dr] & & \\ K=F(\sqrt[n]\alpha) \ar@{-}[dr] & & L'=L(\zeta_n) \ar@{-}[dl]\ar@{-}[dr] & \\ & L \ar@{-}[dr] & & F'=F(\zeta_n) \ar@{-}[dl] \\ & & F & } \] Now, $L'/F'$ is Galois since $L/F$ is. But $K'$ is a Kummer extension of $F'$, so has cyclic Galois group and thus $L'/F'$ has cyclic Galois group as well (being a quotient of $\Gal(K'/F')$). Thus $L'$ is a Kummer extension of $F'$, so that $L'=F'(\sqrt[n]{\beta})$ for some $\beta\in L$. It follows that $L=F(\sqrt[n]{\beta})$. But since $L$ is Galois over $F$, it follows that $n\leq 2$ (since otherwise in order to be Galois, $L$ would have to contain the non-real $n^{\mathrm{th}}$ roots of unity).