PlanetMath: Galois group of a quartic polynomial

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Galois group of a quartic polynomial

(Topic)

Consider a general (monic) quartic polynomial over $\mathbb{Q}$

$\displaystyle f(x)=x^4 + ax^3+bx^2+cx+d$

and denote the Galois group of

The Galois group is isomorphic to a subgroup of (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 (embedded in ) - again, see the article on the Galois group of a cubic polynomial.

If it factors as two irreducible quadratics, then the splitting field of is the compositum of $\mathbb{Q}(\sqrt{D_1})$ and $\mathbb{Q}(\sqrt{D_2})$ , where and are the discriminants of the two quadratics. This is either a biquadratic extension and thus has Galois group isomorphic to , or else is a square, and $\mathbb{Q}(\sqrt{D_1},\sqrt{D_2})=\mathbb{Q}(\sqrt{D_1})$ and the Galois group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ .

This leaves us with the most interesting case, where is irreducible. In this case, the Galois group acts transitively on the roots of , so it must be isomorphic to a transitive subgroup of . The transitive subgroups of are

$\displaystyle S_4$
$\displaystyle A_4$
$\displaystyle D_8$	$\displaystyle \cong\{e,\ (1234),\ (13)(24),\ (1432),\ (12)(34),\ (14)(23),\ (13),\ (24)\}$ and its conjugates
$\displaystyle V_4$	$\displaystyle \cong\{e,\ (12)(34),\ (13)(24),\ (14)(23)\}$
$\displaystyle \mathbb{Z}/4\mathbb{Z}$	$\displaystyle \cong \{e,\ (1234),\ (13)(24),\ (1432)\}$ and its conjugates

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 is

$\displaystyle C(x) = x^3 - 2b x^2 + (b^2+ac-4d) x + (c^2+a^2d - abc)$

and has roots

$\displaystyle r_1=(\alpha_1+\alpha_2)(\alpha_3+\alpha_4)$
$\displaystyle r_2=(\alpha_1+\alpha_3)(\alpha_2+\alpha_4)$
$\displaystyle r_3=(\alpha_1+\alpha_4)(\alpha_2+\alpha_3)$

But then a short computation shows that the discriminant

is the same as the discriminant of

. Also, since $r_1,r_2,r_3\in\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ , it follows that the splitting field of

is a subfield of the splitting field of

and thus that the Galois group of

is a quotient of the Galois group of

. There are four cases:

If is irreducible, and is not a rational square, then does not fix and thus is not contained in . But in this case, where is not a square, the Galois group of is , which has order . The only subgroup of not contained in with order a multiple of (and thus capable of having a subgroup of index ) is itself, so in this case $G\cong S_4$ .
If is irreducible but is a rational square, then fixes , so $G\leq A_4$ . In addition, the Galois group of is , so divides the order of a transitive subgroup of , which means that $G\cong A_4$ itself.
If is reducible, suppose first that it splits completely in $\mathbb{Q}$ . Then each of $r_1,r_2,r_3\in \mathbb{Q}$ and thus each element of fixes each . Thus $G\cong V_4$ .
Finally, if splits into a linear factor and an irreducible quadratic, then one of the , say , is in $\mathbb{Q}$ . Then fixes $r_2=(\alpha_1+\alpha_3)(\alpha_2+\alpha_4)$ but not or . The only possibilities from among the transitive groups are then that $G\cong D_8$ or $G\cong \mathbb{Z}/4\mathbb{Z}$ . In this case, the discriminant of the quadratic is not a rational square, but it is a rational square times .
Now, $G\cap A_4$ fixes $\mathbb{Q}(\sqrt{D})$ , since fixes $\sqrt{D}$ up to sign and restricts our attention to even permutations. But $\lvert G:G\cap A_4 \rvert =2$ , so the fixed field of $G\cap A_4$ has dimension over $\mathbb{Q}$ and thus is exactly $\mathbb{Q}(\sqrt{D})$ . If $G\cong D_8$ , then $G\cap A_4\cong V_4$ , while if $G\cong \mathbb{Z}/4\mathbb{Z}$ , then $G\cap A_4\cong \mathbb{Z}/2\mathbb{Z}$ ; in the first case only, $G\cap A_4$ acts transitively on the roots of . Thus $G\cap A_4\cong V_4$ if and only if is irreducible over $\mathbb{Q}(\sqrt{D})$ .

So, in summary, for irreducible, we have the following:

Condition	Galois group
irreducible, not a rational square
irreducible, a rational square
splits completely
factors as linear times irreducible quadratic, irreducible over $\mathbb{Q}(\sqrt{D})$
factors as linear times irreducible quadratic, reducible over $\mathbb{Q}(\sqrt{D})$	$\mathbb{Z}/4\mathbb{Z}$