Galois group of a cubic polynomial

Definition 1.

If f(x)k[x], the Galois groupMathworldPlanetmath of f(x) is the Galois group Gal(K/k) of a splitting fieldMathworldPlanetmath K of f(x).

Theorem 1.

For f(x)k[x], the Galois group of f(x) permutes the set of roots of f(x). Therefore, if the roots of f(x) are α1,,αnK, the Galois group of f(x) is isomorphicPlanetmathPlanetmathPlanetmathPlanetmath to a subgroupMathworldPlanetmathPlanetmath of Sn.

Proof. K=k(α1,,αn), so any automorphismPlanetmathPlanetmathPlanetmathPlanetmath σ of K fixing k is determined by the image of each αi. But σ must take each αi to some αj (where possibly i=j), since σ is a homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of K and thus f(σ(αi))=σ(f(αi))=0. Thus σ permutes the roots of f(x) and is determined by the resulting permutationMathworldPlanetmath.

We now restrict our attention to the case k=. If f(x)[x] is a cubic, its Galois group is a subgroup of S3. We can then use the knowledge of the group structureMathworldPlanetmath of S3 to anticipate the possible Galois groups of a cubic polynomial. There are six subgroups of S3, and the three subgroups of order 2 are conjugatePlanetmathPlanetmathPlanetmath. This leaves four essentially different subgroups of S3: the trivial group, the group (1,2) that consists of a single transpositionMathworldPlanetmath, the group A3=(1,2,3), and the full group S3. All four of these groups can in fact appear as the Galois group of a cubic.

Let K be a splitting field of f(x) over .

If f(x) splits completely in , then K= and so the Galois group of f(x) is trivial. So any cubic (in fact, a polynomialMathworldPlanetmathPlanetmathPlanetmath of any degree) that factors completely into linear factors in will have trivial Galois group.

If f(x) factors into a linear and an irreduciblePlanetmathPlanetmath quadratic term, then K=(D), where D is the discriminantPlanetmathPlanetmathPlanetmath of the quadratic. Hence [K:]=2 and the order of Gal(K/) is 2, so Gal(K/)1,2/2; the nontrivial element of the Galois group takes each root of the quadratic to its complex conjugate (i.e. it maps D-D). Thus any cubic that has exactly one rational root will have Galois group isomorphic to 1,2/2.

The Galois groups A3 and S3 arise when considering irreducible cubics. Let f(x) be irreducible with roots r1,r2,r3. Since f is irreducible, the roots are distinct. Thus Gal(K/) has at least 3 elements, since the image of r1 may be any of the three roots. Since Gal(K/)S3, it follows that Gal(K/)A3 or Gal(K/)S3 and thus by the fundamental theorem of Galois theoryMathworldPlanetmath that [K:]=3 or 6.

Now, the discriminant of f is


This is a symmetric polynomialMathworldPlanetmath in the ri. The coefficients of f(x) are the elementary symmetric polynomials in the ri: if f(x)=x3+ax2+bx+c, then


Thus D can be written as a polynomial in the coefficients of f, so D. D0 since f(x) is irreducible and therefore has distinct roots; also clearly DK and thus (r1,D)K. If f(x)=x3+ax2+bx+c, then its discriminant D is 18abc+a2b2-4b3-4a3c-27c2 (see the article on the discriminant for a longer discussion).

If D, it follows that D has degree 2 over , so that [(r1,D):]=6. Hence K=(r1,D), so we can derive the splitting field for f by adjoining any root of f and the square root of the discriminant. This can happen for either positive or negative D, clearly. Note in particular that if D<0, then D is imaginaryPlanetmathPlanetmath and thus K is not a real field, so that f has one real and two imaginary roots. So any cubic with only one real root has Galois group S3.

If D, then any element of Gal(K/) must fix D. But a transposition of two roots does not fix D - for example, the map




Then Gal(K/) does not include transpositions and so it must in this case be isomorphic to A3. Thus [K:]=3, so K=(r1)=(r1,D) since D. This proves:

Theorem 2.

Let f(x)Q[x] be an irreducible cubic and K its splitting field. Then if α is any root of f,


where D is the discriminant of f. Thus if D is rational, [K:Q]=3 and the Galois group is isomorphic to A3, otherwise [K:Q]=6 and the Galois group is isomorphic to S3.

Thus if f is an irreducible cubic with Galois group A3, it must have three real roots. However, the converseMathworldPlanetmath does not hold (see Example 4 below).

of looking at the above analysis is that for a “general” polynomial of degree n, the Galois group is Sn. If the Galois group of some polynomial is not Sn, there must be algebraicMathworldPlanetmath relationsMathworldPlanetmathPlanetmath among the roots that restrict the available set of permutations. In the case of a cubic whose discriminant is a rational square, this relation is that D, which is a polynomial in the roots, must be preserved.

  1. Example 1

    f(x)=x3-6x2+11x-6. By the rational root test, this polynomial has the three rational roots 1,2,3, so it factors as f(x)=(x-1)(x-2)(x-3) over . Its Galois group is therefore trivial.

  2. Example 2

    f(x)=x3-x2+x-1. Again by the rational root test, this polynomial factors as (x-1)(x2+1), so its Galois group has two elements, and a splitting field K for f is derived by adjoining the square root of the discriminant of the quadratic: K=(-1). The nontrivial element of the Galois group maps -1--1.

  3. Example 3

    f(x)=x3-2. This polynomial has discriminant -108=-362. This is not a rational square, so the Galois group of f over is S3, and the splitting field for f is (23,-108)=(23,-3). This is in agreement with what we already know, namely that the cube roots of 2 are


    where ω=-1+-32 is a primitive cube root of unity.

  4. Example 4

    f(x)=x3-4x+2. This is irreducible since it is Eisenstein at 2 (or by the rational root test), and its discriminant is 202, which is not a rational square. Thus the Galois group for this polynomial is also S3; note, however, that f has three real roots (since f(0)>0 but f(1)<0).

  5. Example 5

    f(x)=x3-3x+1. This is also irreducible by the rational root test. Its discriminant is 81, which is a rational square, so the Galois group for this polynomial is A3. Explicitly, the roots of f(x) are


    and we see that


    Let’s consider an automorphism of K sending r1 to r3, i.e. sending cos(2π/9)cos(14π/9). Given the relations above, it is clear that this mapping uniquely determines the image of r3 as well, since


    and thus we see how the relation imposed by the discriminant actually manifests itself in terms of restrictionsPlanetmathPlanetmathPlanetmathPlanetmath on the permutation groupMathworldPlanetmath.

Title Galois group of a cubic polynomial
Canonical name GaloisGroupOfACubicPolynomial
Date of creation 2013-03-22 17:40:33
Last modified on 2013-03-22 17:40:33
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 10
Author rm50 (10146)
Entry type Topic
Classification msc 12F10
Classification msc 12D10
Related topic CardanosFormulae