ProofOfFundamentalTheoremOfGaloisTheory 2004-06-23 09:37:08 2004-06-29 18:46:10 Proof fundamental theorem of Galois 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} \newtheorem{lemma}{Lemma} The theorem is a consequence of the following lemmas, roughly corresponding to the various assertions in the theorem. We assume $L/F$ to be a finite-dimensional Galois extension of fields with Galois group \[ G=\Gal(L/F). \] The first two lemmas establish the correspondence between subgroups of $G$ and extension fields of $F$ contained in $L$. \begin{lemma} \label{lemma1} Let $K$ be an extension field of $F$ contained in $L$. Then $L$ is Galois over $K$, and $\Gal(L/K)$ is a subgroup of $G$. \end{lemma} \begin{proof} Note that $L/F$ is normal and separable because it is a Galois extension; it remains to prove that $L/K$ is also normal and separable. Since $L$ is normal and finite over $F$, it is the splitting field of a polynomial $f\in F[X]$ over $F$. Now $L$ is also the splitting field of $f$ over $K$ (because $F\subset K\subset L$), so $L/K$ is normal. To see that $L/K$ is also separable, suppose $\alpha\in L$, and let $f^\alpha_F\in F[X]$ be its minimal polynomial over $F$. Then the minimal polynomial $f^\alpha_K$ of $\alpha$ over $K$ divides $f^\alpha_F$, which has no double roots in its splitting field by the separability of $L/F$. Therefore $f^\alpha_K$ has no double roots in its splitting field for any $\alpha\in L$, so $L$ is separable over $K$. The assertion that $\Gal(L/K)$ is a subgroup of $G$ is clear from the fact that $K\supset F$. \end{proof} \begin{lemma} \label{lemma2} The function $\phi$ from the set of extension fields of $F$ contained in $L$ to the set of subgroups of $G$ defined by \[ \phi(K)=\Gal(L/K) \] is an inclusion-reversing bijection. The inverse is given by \[ \phi^{-1}(H)=L^H, \] where $L^H$ is the fixed field of $H$ in $L$. \end{lemma} \begin{proof} The definition of $\phi$ makes sense because of Lemma \ref{lemma1}. The \PMlinkescapetext{identities} \[ \phi^{-1}\circ\phi(K)=K\quad\hbox{and}\quad \phi\circ\phi^{-1}(H)=H \] for all subgroups $H\subset G$ and all fields $K$ with $F\subset K\subset L$ follow from the properties of the Galois group. The fixed field of $\Gal(L/K)$ is precisely $K$; on the other hand, since $L^H$ is the fixed field of $H$ in $L$, $H$ is the Galois group of $L/L^H$. For extensions $K$ and $K'$ of $F$ with $F\subset K\subset K'\subset L$, we have \[ \sigma\in\Gal(L/K')\iff\sigma\in\Gal(L/K), \] so $\phi(K)\supset\phi(K')$. This shows that $\phi$ is inclusion-reversing. \end{proof} The following lemmas show that normal subextensions of $L/F$ are Galois extensions and that their Galois groups are quotient groups of $G$. \begin{lemma} \label{lemma3} Let $H$ be a subgroup of $G$. Then the following are equivalent: \begin{enumerate} \item $L^H$ is normal over $F$. \item $\sigma\left(L^H\right)=L^H$ for all $\sigma\in G$. \item $\sigma H\sigma^{-1}=H$ for all $\sigma\in G$. \end{enumerate} In particular, $L^H$ is normal over $F$ if and only if $H$ is a normal subgroup of $G$. \end{lemma} \begin{proof} $1\Rightarrow2$: Since for all $\sigma\in G$ and $\alpha\in L^H$, $\sigma(\alpha)$ is a zero of the minimal polynomial of $\alpha$ over $F$, we have $\sigma(\alpha)\in L^H$ by the \PMlinkescapetext{normality} of $L^H/F$. $2\Rightarrow3$: For all $\sigma\in G,\tau\in H$ the equality \[ \sigma\tau\sigma^{-1}(x)=\sigma\sigma^{-1}(x)=x \] holds for all $x\in L^H$ (from the assumption it follows that $\sigma^{-1}(x)\in L^H$, which is fixed by $\tau$). This implies that \[ \sigma\tau\sigma^{-1}\in\Gal(L/L^H)=H \] for all $\sigma\in G,\tau\in H$. $3\Rightarrow1$: Let $\alpha\in L^H$, and let $f$ be the minimal polynomial of $\alpha$ over $F$. Since $L/F$ is normal, $f$ splits into linear factors in $L[X]$. Suppose $\alpha'\in L$ is another zero of $f$, and let $\sigma\in G$ be such that $\sigma(\alpha')=\alpha$ (such a $\sigma$ always exists). By assumption, for all $\tau\in H$ we have $\tau':=\sigma\tau\sigma^{-1}\in H$, so that \[ \tau(\alpha')=\sigma^{-1}\tau'\sigma(\alpha') =\sigma^{-1}\tau'(\alpha)=\sigma^{-1}(\alpha)=\alpha'. \] This shows that $\alpha'$ lies in $L^H$ as well, so $f$ splits in $L^H[X]$. We conclude that $L^H$ is normal over $F$. \end{proof} \begin{lemma} Let $H$ be a normal subgroup of $G$. Then $L^H$ is a Galois extension of $F$, and the homomorphism \begin{eqnarray*} r\colon G&\to&\Gal(L^H/F) \\ \sigma&\mapsto&\sigma\vert_{L^H} \end{eqnarray*} induces a natural identification \[ \Gal(L^H/F)\cong G/H. \] \end{lemma} \begin{proof} By Lemma \ref{lemma3}, $L^H$ is normal over $F$, and because a subextension of a separable extension is separable, $L^H/F$ is a Galois extension. The map $r$ is well-defined by the implication $1\Rightarrow2$ from Lemma \ref{lemma3}. It is surjective since every automorphism of $L^H$ that fixes $F$ can be extended to an automorphism of $L$ (if $L\ne L^H$, for example, we can choose an $\alpha\in L\setminus L^H$ such that $L=L^H(\alpha)$ using the primitive element theorem, and we can extend $\sigma\in\Gal(L^H/F)$ to $L$ by putting $\sigma(\alpha)=\alpha$). The kernel of $r$ is clearly equal to $H$, so the first isomorphism theorem gives the claimed identification. \end{proof}