Galois group of a quartic polynomial


Consider a general (monic) quartic polynomial over

f(x)=x4+ax3+bx2+cx+d

and denote the Galois groupMathworldPlanetmath of f(x) by G.

The Galois group G is isomorphicPlanetmathPlanetmathPlanetmathPlanetmath to a subgroupMathworldPlanetmathPlanetmath of S4 (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 S3 (embedded in S4) - again, see the article on the Galois group of a cubic polynomial.

If it factors as two irreducible quadratics, then the splitting fieldMathworldPlanetmath of f(x) is the compositum of (D1) and (D2), where D1 and D2 are the discriminants of the two quadratics. This is either a biquadratic extension and thus has Galois group isomorphic to V4, or else D1D2 is a square, and (D1,D2)=(D1) and the Galois group is isomorphic to /2.

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 transitiveMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath (http://planetmath.org/GroupAction) subgroup of S4. The transitive subgroups of S4 are

S4
A4
D8 {e,(1234),(13)(24),(1432),(12)(34),(14)(23),(13),(24)} and its conjugates
V4 {e,(12)(34),(13)(24),(14)(23)}
/4 {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 cubicMathworldPlanetmath of f(x) is

C(x)=x3-2bx2+(b2+ac-4d)x+(c2+a2d-abc)

and has roots

r1=(α1+α2)(α3+α4)
r2=(α1+α3)(α2+α4)
r3=(α1+α4)(α2+α3)

But then a short computation shows that the discriminant D of C(x) is the same as the discriminant of f(x). Also, since r1,r2,r3(α1,α2,α3,α4), it follows that the splitting field of C(x) is a subfieldMathworldPlanetmath of the splitting field of f(x) and thus that the Galois group of C(x) is a quotientPlanetmathPlanetmath 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 A4. But in this case, where D is not a square, the Galois group of C(x) is S3, which has order 6. The only subgroup of S4 not contained in A4 with order a multiple of 6 (and thus capable of having a subgroup of index 6) is S4 itself, so in this case GS4.

  • If C(x) is irreducible but D is a rational square, then G fixes D, so GA4. In additionPlanetmathPlanetmath, the Galois group of C(x) is A3, so 3 divides the order of a transitive subgroup of A4, which means that GA4 itself.

  • If C(x) is reducible, suppose first that it splits completely in . Then each of r1,r2,r3 and thus each element of G fixes each ri. Thus GV4.

  • Finally, if C(x) splits into a linear factor and an irreducible quadratic, then one of the ri, say r2, is in . Then G fixes r2=(α1+α3)(α2+α4) but not r1 or r3. The only possibilities from among the transitive groups are then that GD8 or G/4. In this case, the discriminant of the quadratic is not a rational square, but it is a rational square times D.

    Now, GA4 fixes (D), since G fixes D up to sign and A4 restricts our attention to even permutationsMathworldPlanetmath. But |G:GA4|=2, so the fixed field of GA4 has dimension 2 over and thus is exactly (D). If GD8, then GA4V4, while if G/4, then GA4/2; in the first case only, GA4 acts transitively on the roots of f(x). Thus GA4V4 if and only if f(x) is irreducible over (D).

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

Condition Galois group
C(x) irreducible, D not a rational square S4
C(x) irreducible, D a rational square A4
C(x) splits completely V4
C(x) factors as linear times irreducible quadratic, f(x) irreducible over (D) D8
C(x) factors as linear times irreducible quadratic, f(x) reducible over (D) /4

References

Title Galois group of a quartic polynomial
Canonical name GaloisGroupOfAQuarticPolynomial
Date of creation 2013-03-22 17:41:35
Last modified on 2013-03-22 17:41:35
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 7
Author rm50 (10146)
Entry type Topic
Classification msc 12D10