GaloisGroupOfAQuarticPolynomial 2007-12-16 20:39:49 2008-01-24 22:48:53 Topic Galois group % 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{\Order}[1]{\lvert #1 \rvert} \newcommand{\Rats}{\mathbb{Q}} \newcommand{\Ints}{\mathbb{Z}} \newcommand{\subgroup}{\leq} \PMlinkescapeword{addition} \PMlinkescapeword{class} Consider a general (monic) quartic polynomial over $\Rats$ \[f(x)=x^4 + ax^3+bx^2+cx+d\] and denote the Galois group of $f(x)$ by $G$. The Galois group $G$ is isomorphic to a subgroup of $S_4$ (see the article on the Galois group of a cubic polynomial for a discussion of this question). If the quartic splits into a linear factor and an irreducible cubic, then its Galois group is simply the Galois group of the cubic portion and thus is isomorphic to a subgroup of $S_3$ (embedded in $S_4$) - again, see the article on the Galois group of a cubic polynomial. If it factors as two irreducible quadratics, then the splitting field of $f(x)$ is the compositum of $\Rats(\sqrt{D_1})$ and $\Rats(\sqrt{D_2})$, where $D_1$ and $D_2$ are the discriminants of the two quadratics. This is either a biquadratic extension and thus has Galois group isomorphic to $V_4$, or else $D_1D_2$ is a square, and $\Rats(\sqrt{D_1},\sqrt{D_2})=\Rats(\sqrt{D_1})$ and the Galois group is isomorphic to $\Ints/2\Ints$. This leaves us with the most interesting case, where $f(x)$ is irreducible. In this case, the Galois group acts transitively on the roots of $f(x)$, so it must be isomorphic to a \PMlinkname{transitive}{GroupAction} subgroup of $S_4$. The transitive subgroups of $S_4$ are \begin{align*} S_4&\\ A_4&\\ D_8&\cong\{e,\ (1234),\ (13)(24),\ (1432),\ (12)(34),\ (14)(23),\ (13),\ (24)\}\text{ and its conjugates}\\ V_4&\cong\{e,\ (12)(34),\ (13)(24),\ (14)(23)\}\\ \Ints/4\Ints &\cong \{e,\ (1234),\ (13)(24),\ (1432)\}\text{ and its conjugates} \end{align*} We will see that each of these transitive subgroups actually appears as the Galois group of some class of irreducible quartics. The resolvent cubic of $f(x)$ is \[C(x) = x^3 - 2b x^2 + (b^2+ac-4d) x + (c^2+a^2d - abc)\] and has roots \begin{gather*} r_1=(\alpha_1+\alpha_2)(\alpha_3+\alpha_4)\\ r_2=(\alpha_1+\alpha_3)(\alpha_2+\alpha_4)\\ r_3=(\alpha_1+\alpha_4)(\alpha_2+\alpha_3) \end{gather*} But then a short computation shows that the discriminant $D$ of $C(x)$ is the same as the discriminant of $f(x)$. Also, since $r_1,r_2,r_3\in\Rats(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$, it follows that the splitting field of $C(x)$ is a subfield of the splitting field of $f(x)$ and thus that the Galois group of $C(x)$ is a quotient of the Galois group of $f(x)$. There are four cases: \begin{itemize} \item If $C(x)$ is irreducible, and $D$ is not a rational square, then $G$ does not fix $D$ and thus is not contained in $A_4$. But in this case, where $D$ is not a square, the Galois group of $C(x)$ is $S_3$, which has order $6$. The only subgroup of $S_4$ not contained in $A_4$ with order a multiple of $6$ (and thus capable of having a subgroup of index $6$) is $S_4$ itself, so in this case $G\cong S_4$. \item If $C(x)$ is irreducible but $D$ is a rational square, then $G$ fixes $D$, so $G\subgroup A_4$. In addition, the Galois group of $C(x)$ is $A_3$, so $3$ divides the order of a transitive subgroup of $A_4$, which means that $G\cong A_4$ itself. \item If $C(x)$ is reducible, suppose first that it splits completely in $\Rats$. Then each of $r_1,r_2,r_3\in \Rats$ and thus each element of $G$ fixes each $r_i$. Thus $G\cong V_4$. \item Finally, if $C(x)$ splits into a linear factor and an irreducible quadratic, then one of the $r_i$, say $r_2$, is in $\Rats$. Then $G$ fixes $r_2=(\alpha_1+\alpha_3)(\alpha_2+\alpha_4)$ but not $r_1$ or $r_3$. The only possibilities from among the transitive groups are then that $G\cong D_8$ or $G\cong \Ints/4\Ints$. In this case, the discriminant of the quadratic is not a rational square, but it is a rational square times $D$. Now, $G\cap A_4$ fixes $\Rats(\sqrt{D})$, since $G$ fixes $\sqrt{D}$ up to sign and $A_4$ restricts our attention to even permutations. But $\Order{G:G\cap A_4}=2$, so the fixed field of $G\cap A_4$ has dimension $2$ over $\Rats$ and thus is exactly $\Rats(\sqrt{D})$. If $G\cong D_8$, then $G\cap A_4\cong V_4$, while if $G\cong \Ints/4\Ints$, then $G\cap A_4\cong \Ints/2\Ints$; in the first case only, $G\cap A_4$ acts transitively on the roots of $f(x)$. Thus $G\cap A_4\cong V_4$ if and only if $f(x)$ is irreducible over $\Rats(\sqrt{D})$. \end{itemize} So, in summary, for $f(x)$ irreducible, we have the following: \begin{center} \begin{tabular}{lc} Condition & Galois group\\ \hline $C(x)$ irreducible, $D$ not a rational square & $S_4$\\ $C(x)$ irreducible, $D$ a rational square & $A_4$\\ $C(x)$ splits completely & $V_4$\\ $C(x)$ factors as linear times irreducible quadratic, $f(x)$ irreducible over $\Rats(\sqrt{D})$ & $D_8$\\ $C(x)$ factors as linear times irreducible quadratic, $f(x)$ reducible over $\Rats(\sqrt{D})$ & $\Ints/4\Ints$ \end{tabular} \end{center} \begin{thebibliography}{10} \bibitem{dummit} D.S.~Dummit, R.M.~Foote, \emph{Abstract Algebra}, Wiley and Sons, 2004. \end{thebibliography}