Galois group of a quartic polynomial
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 and , 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 and the Galois group is isomorphic to .
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 (http://planetmath.org/GroupAction) subgroup of . The transitive subgroups of are
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
and has roots
But then a short computation shows that the discriminant of is the same as the discriminant of . Also, since , 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 .
If is irreducible but is a rational square, then fixes , so . In addition, the Galois group of is , so divides the order of a transitive subgroup of , which means that itself.
If is reducible, suppose first that it splits completely in . Then each of and thus each element of fixes each . Thus .
Finally, if splits into a linear factor and an irreducible quadratic, then one of the , say , is in . Then fixes but not or . The only possibilities from among the transitive groups are then that or . In this case, the discriminant of the quadratic is not a rational square, but it is a rational square times .
So, in summary, for irreducible, we have the following:
|irreducible, not a rational square|
|irreducible, a rational square|
|factors as linear times irreducible quadratic, irreducible over|
|factors as linear times irreducible quadratic, reducible over|
- 1 D.S. Dummit, R.M. Foote, Abstract Algebra, Wiley and Sons, 2004.