quartic polynomial with Galois group D8

The polynomialMathworldPlanetmathPlanetmathPlanetmath f(x)=x4-2x2-2 is Eisenstein at 2 and thus irreducible over . Solving f(x) as a quadratic in x2, we see that the roots of f(x) are

α1=1+3 α3=-1+3
α2=1-3 α4=-1-3

Note that the discriminant of f(x) is -4608=-2932, and that its resolvent cubicMathworldPlanetmath is


which factors over into a linear and an irreducible quadratic. Additionally, f(x) remains irreducible over (-4608)=(-2), since none of the roots of f(x) lie in this field and the discriminant of f(x), regarded as a quadratic in x2, does not lie in this field either, so f(x) cannot factor as a productMathworldPlanetmathPlanetmathPlanetmath of two quadratics. So according to the article on the Galois groupMathworldPlanetmath of a quartic polynomial, f(x) should indeed have Galois group isomorphicPlanetmathPlanetmathPlanetmath to D8. We show that this is the case by explicitly examining the structureMathworldPlanetmath of its splitting fieldMathworldPlanetmath.

Let K be the splitting field of f(x) over , and let G=Gal(K/).

Let K1=(α1)=(α3) and K2=(α2)=(α4). Clearly K contains both K1 and K2 and thus contains K1K2=(α1,α2). But obviously f(x) splits in K1K2, so that K=K1K2. We next determine the degree of K over .

Note that K1K2 since K1 is a real field while K2 is not. Thus K1K2K1,K2. Clearly [K1:]=[K2:]=4, so [K1K2:]2. But


so 3K1K2. Hence K1K2=(3); call this field F.

Since K1K2, we also have K=K1K2K1 and K=K1K2K2; thus K is a quadratic extension of each and [K:F]=4.

Putting these results together, we see that


so that G has order 8.

Now, neither K1 nor K2 is Galois over (since the Galois closure of either one is K), so that the subgroupMathworldPlanetmathPlanetmath of G fixing (say) K1 is a nonnormal subgroup of G. Thus G must be nonabelianPlanetmathPlanetmathPlanetmath, so must be isomorphic to either D8 or Q8 (the quaternions). But the subgroups of G corresponding to K1 and K2 are distinct subgroups of order 2 in G, and Q8 has only one subgroup of order 2. Thus GD8. (Alternatively, note that all subgroups of Q8 are normal, so GD8 since it has a nonnormal subgroup).

Title quartic polynomial with Galois group D8
Canonical name QuarticPolynomialWithGaloisGroupD8
Date of creation 2013-03-22 17:44:09
Last modified on 2013-03-22 17:44:09
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 6
Author rm50 (10146)
Entry type Example
Classification msc 12D10