ProofOfTheCompositumOfAGaloisExtensionAndAnotherExtensionIsGalois 2009-01-05 17:55:26 2009-01-06 20:31:41 Proof the compositum of a Galois extension and another extension is Galois 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} \newtheorem{thm}{Theorem} \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \newtheorem{ax}{Axiom} \begin{proof} The diagram of the situation of the theorem is: \[\xymatrix @R1pc@C1pc{ & & & \ar@{-}[llld]\ar@{-}[rdd]EF \\ \ar@{-}[rdd]E \\ & & & & \ar@{-}[llld]F \\ & \ar@{-}[d]E\cap F \\ & K } \] To see that $EF/F$ is Galois, note that since $E/K$ is Galois, $E$ is a splitting field of a set of polynomials over $K$; clearly $EF$ is a splitting field of the same set of polynomials over $F$. Also, if $f\in K[x]$ is separable over $K$, then also $f$ is separable over $F$. Thus $EF$ is normal and separable over $F$, so is Galois. $E$ is obviously Galois over $E\cap F$ since $E\cap F\supset K$. Let $r$ be the restriction map \[ r: H = \Gal(EF/F) \to \Gal(E/K) : \sigma \mapsto \sigma |_E\] $r$ is clearly a group homomorphism, and since $E$ is normal over $K$, $r$ is well-defined. Claim $r$ is injective. For suppose $\sigma\in\Gal(EF/F)$ and $\sigma|_E$ is the identity. Then $\sigma$ is fixed on $F$ (since it is in $\Gal(EF/F)$ and on $E$ (since its restriction to $E$ is the identity), so is fixed on $EF$ and thus is itself the identity. Now, the image of $r$ is a subgroup of $\Gal(E/K)$ with fixed field $L$, and thus the image of $r$ is $\Gal(E/L)$. Claim $E\cap F=L$. $\subset$ is obvious: any element $x\in E\cap F$ is fixed by $\sigma|_E$ for each $\sigma\in H$ since $\sigma$ fixes $F$. Thus $E\cap F\subset L$. To see the reverse inclusion, choose $x\in L$; then $x$ is fixed by each $r(\sigma)$ for $\sigma\in H$. But $x\in L\subset E$, so that (as an element of $E$), $x$ is fixed by each $\sigma\in H$. Thus $x\in F$ so that $x\in E\cap F$. Thus $L = E\cap F$, and $r$ is then an isomorphism $\Gal(EF/F)\cong \Gal(E/E\cap F)$. \end{proof} \begin{thebibliography}{10} \bibitem{bib:df} Morandi,~P., \emph{Field and Galois Theory}, Springer, 1996. \end{thebibliography}