Galois group of a quartic polynomial

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

$$f(x)=x^4+a{x}^3+b{x}^2+cx+d$$

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 $\mathbb{Q}(\sqrt{D_1})$ and $\mathbb{Q}(\sqrt{D_2})$, 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 $D_1{D}_2$ 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 $f(x)$ is irreducible. In this case, the Galois group acts transitively on the roots of $f(x)$, so it must be isomorphic to a transitive (http://planetmath.org/GroupAction) subgroup of $S_4$. The transitive subgroups of $S_4$ are

$S_4$

$A_4$

$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)\}$

$\mathbb{Z}/4\mathbb{Z}$$\cong \{e,(1234),(13)(24),(1432)\}\text{ 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 $f(x)$ is

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

and has roots

$$r_1=({\alpha }_1+{\alpha }_2)({\alpha }_3+{\alpha }_4)$$

$$r_2=({\alpha }_1+{\alpha }_3)({\alpha }_2+{\alpha }_4)$$

$$r_3=({\alpha }_1+{\alpha }_4)({\alpha }_2+{\alpha }_3)$$

But then a short computation shows that the discriminant $D$ of $C(x)$ is the same as the discriminant of $f(x)$. Also, since $r_1,r_2,r_3\in \mathbb{Q}({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)$, 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:

• 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 $G\cong S_4$.

• If $C(x)$ is irreducible but $D$ is a rational square, then $G$ fixes $D$, so $G\le A_4$. 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.

• If $C(x)$ 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 $G$ fixes each $r_i$. Thus $G\cong V_4$.

• Finally, if $C(x)$ splits into a linear factor and an irreducible quadratic, then one of the $r_i$, say $r_2$, is in $\mathbb{Q}$. 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 $G\cong \mathbb{Z}/4\mathbb{Z}$. In this case, the discriminant of the quadratic is not a rational square, but it is a rational square times $D$.

  Now, $G\cap A_4$ fixes $\mathbb{Q}(\sqrt{D})$, since $G$ fixes $\sqrt{D}$ up to sign and $A_4$ restricts our attention to even permutations. But $|G:G\cap A_4|=2$, so the fixed field of $G\cap A_4$ has dimension $2$ 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 $f(x)$. Thus $G\cap A_4\cong V_4$ if and only if $f(x)$ is irreducible over $\mathbb{Q}(\sqrt{D})$.

So, in summary, for $f(x)$ irreducible, we have the following:

Condition	Galois group
$C(x)$ 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 $\mathbb{Q}(\sqrt{D})$	$D_8$
$C(x)$ factors as linear times irreducible quadratic, $f(x)$ reducible over $\mathbb{Q}(\sqrt{D})$	$\mathbb{Z}/4\mathbb{Z}$