# Galois group of a quartic polynomial

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

 $f(x)=x^{4}+ax^{3}+bx^{2}+cx+d$

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 $\mathbb{Q}(\sqrt{D_{1}})$ 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

 $\displaystyle S_{4}$ $\displaystyle A_{4}$ $\displaystyle D_{8}$ $\displaystyle\cong\{e,\ (1234),\ (13)(24),\ (1432),\ (12)(34),\ (14)(23),\ (13% ),\ (24)\}\text{ and its conjugates}$ $\displaystyle V_{4}$ $\displaystyle\cong\{e,\ (12)(34),\ (13)(24),\ (14)(23)\}$ $\displaystyle\mathbb{Z}/4\mathbb{Z}$ $\displaystyle\cong\{e,\ (1234),\ (13)(24),\ (1432)\}\text{ and its conjugates}$

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

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

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 $D$ 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 $G\leq A_{4}$. 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 $\lvert G:G\cap A_{4}\rvert=2$, 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 $\mathbb{Q}(\sqrt{D})$ $\mathbb{Z}/4\mathbb{Z}$

## References

Title Galois group of a quartic polynomial GaloisGroupOfAQuarticPolynomial 2013-03-22 17:41:35 2013-03-22 17:41:35 rm50 (10146) rm50 (10146) 7 rm50 (10146) Topic msc 12D10