A Worked out Galois Group for the Classroom

Abstract Let f=X6−3X2−1∈Q[X] and let Lf be the splitting field of f over Q. We show by hand that the Galois group Gal(Lf/Q) of the Galois extension Lf/Q is isomorphic to the alternating group A4. Moreover, we show that the six roots of f correspond to the six edges of a tetrahedron and that the four roots of the polynomial X4+18X2−72X+81 correspond to the four faces of a tetrahedron, which allows us to determine all eight proper intermediate fields of the extension Lf/Q.


INTRODUCTION.
Teaching Galois Theory, one often has the problem that the Galois group of a field extension of Q is either quite simple or too difficult to be computed by hand.An example of a Galois group which is isomorphic to the dihedral group of order 8 can be found in Stewart [3,Ch. 13].Introducing this example, Stewart writes that this Galois group has an "archetypal quality, since a simpler example would be too small to illustrate the theory adequately, and anything more complicated would be unwieldy" [3, p. 155].Moreover, it is usually rather tedious to compute the Galois group along with the intermediate fields and their relations.
The aim of this note is to provide a worked out field extension over Q whose Galois group is isomorphic to the alternating group A 4 (i.e., to the symmetry group of the tetrahedron), and to compute by hand all intermediate fields and their relations.If we do not require that the ground field is Q, a canonical way to obtain a field extension L/K with Gal(L/K) ∼ = A 4 for some fields L K Q, is to start with a polynomial f ∈ Q[X] of degree 4 such that the Galois group of the field extension L/Q -where L is the splitting field of f over Q -is isomorphic to the symmetry group S 4 .Then, since A 4 S 4 , by the Galois correspondence we find a quadratic extension K of Q such that Gal(L/K) ∼ = A 4 (see also Osofsky [2,p. 222]).However, since the ground field K of the field extension L/K is already a field extension of Q, it is quite exhausting to compute Gal(L/K) and the intermediate fields of L/K by hand.Before we present our example in the next section, we set up the terminology (according to [1,3]), where we assume that the reader is familiar with the basic facts of Galois Theory with respect to field extensions over are fields, then the group of all automorphisms of L which fix M point-wise is the Galois group of the field extension L/M , denoted Gal(L/M ).Let f ∈ Q[X] be a polynomial, L f its splitting field over Q, and M an intermediate subfield, so Q ⊆ M ⊆ L f .Let g ∈ M [X] and let K g ⊆ L f be its splitting field over M .Then K g /M is a Galois extension.We will only consider Galois extensions of this type.Now we can state the main theorem of Galois Theory.THE GALOIS CORRESPONDENCE.Let L/Q be an arbitrary Galois extension.Then the following holds: • To each subgroup H Gal(L/Q) there exists an intermediate field L H , such that • Let M 1 and M 2 be intermediate fields of some field extension L/Q, and let the conjugate class of M contains only M ), then the field extension M/Q is Galois and We start with the polynomial f = X 6 − 3X 2 − 1 and consider its splitting field L f over Q.The goal is to show that Gal(L f /Q) ∼ = A 4 , where A 4 is the alternating group of degree 4, which is isomorphic to the symmetry group of the tetrahedron.
In order to compute the roots of f , we replace X 2 by ξ and first compute the roots of the irreducible polynomial g = ξ 3 − 3ξ − 1.To see that g is irreducible, consider the polynomial By the Eisenstein-Schönemann Criterion (with p = 3), we see that g is irreducible over Q, and so is g.
Hence, the roots of g are given by Notice that −ξ 2 = e πi (e 7πi/9 + e −7πi/9 ) = e 16πi/9 + e 2πi/9 = e −2πi/9 + e 2πi/9 = α 2 + ᾱ2 , and similarly we have −ξ 3 = α 4 + ᾱ4 .Thus, we have For convenience in later arguments, we rewrite the three roots of g as follows: Then by construction we obtain the six pairwise distinct roots of f as ± √ ξ k for 1 ≤ k ≤ 3.In particular, we define This shows that Now, let us show that f is irreducible over Q.For this, assume on the contrary that f = p • q for some non-constant polynomials p, q Then, in the former case this would imply ξ 1 ∈ Q, and in the latter case this would imply Thus, in both cases we arrive at a contradiction.If deg(p) = 3 and p is of the form . Thus, there are no non-constant polynomials p, q ∈ Q[X] such that f = p • q, which shows that f is irreducible over Q.In particular, since f ∈ Q[X] is a monic, irreducible polynomial of degree 6 with the six roots ζ 1 , . . ., ζ 6 , we have , where L f and L g are the splitting fields of f and g, respectively.Then, since deg(g) = 3 and L g = Q(ξ 1 ), we have |G g | = 3 and therefore G g ∼ = C 3 , where C n denotes the cyclic group of order n.Furthermore, since the field extension for all 1 ≤ m ≤ 6.To see this, consider, for example, If we adjoin to the field L g a root ζ m (for 1 ≤ m ≤ 6), then we obtain the intermediate field Now, there are three possible intermediate fields of the form , and Q(ζ 3 ).To see that these three intermediate fields are pairwise distinct, notice first that, since , then, since (ζ 2 ) = 0, we can write In particular we have that Gal(L f /Q) is not cyclic.

SUBGROUPS AND INTERMEDIATE FIELDS.
Figure 1 illustrates all subgroups of A 4 .For some of these subgroups of A 4 , we already found the corresponding intermediate fields.In particular, we found that the field that corresponds to and Q(ζ 3 ).Notice that these three fields are pairwise conjugate.To see this, let σ ∈ Gal(L f /Q(ζ 1 )) and let, for example, Mathematical Assoc. of America American Mathematical Monthly 121:1 August 9, 2023 10:13 a.m.GaloisRevisionMonthlyStyle.tex page 6 A 4 In order to find the four intermediate fields we proceed as follows.First, we identify ζ 1 , . . ., ζ 6 with the numbers 1, . . ., 6 and the elements of the group A 4 with a subgroup of S 6 (i.e., the symmetry group of {1, . . ., 6}).Furthermore, let, again in cycle notation, Let ϑ 1 , ϑ 2 , ϑ 3 , ϑ 4 , be defined as follows: It is not hard to verify that for each . Furthermore, we can verify that for we have which shows that the four intermediate fields M 1 , . . ., M 4 are pairwise conjugate.For example, let τ : 3) and we have which shows that π ∈ Gal(L f /M 2 ).Moreover, we get that which shows that π is a cyclic permutation of ϑ 1 , ϑ 4 , and ϑ 3 .
Figure 2 illustrates all intermediate fields of the field extension Mathematical Assoc. of America American Mathematical Monthly 121:1 August 9, 2023 10:13 a.m.GaloisRevisionMonthlyStyle.tex page 8 and for 1 ≤ i ≤ 3, ) is isomorphic to either {ι}, C 3 , or A 4 .We have seen above that there is a π ∈ Gal(L h /Q) which is a cyclic permutation of ϑ 1 , ϑ 3 , ϑ 4 , and similarly, we find a π ∈ Gal(L h /Q) which is a cyclic permutation of ϑ 2 , ϑ 3 , ϑ 4 .Hence, Gal(L h /Q) must be isomorphic to A 4 .In particular, the fields L f and L h are isomorphic.
Let us consider again the tetrahedron T with the six edges ζ 1 , . . ., ζ 6 , where the pairs of edges ζ i , ζ i+3 (for 1 ≤ i ≤ 3) are opposite edges of T .We already know that the group Gal(L f /Q) is isomorphic to the symmetry group of the tetrahedron acting on its six edges.We show now that Gal(L h /Q) is isomorphic to the symmetry group of the tetrahedron acting on its four faces.For this, we identify the four faces of the tetrahedron with the four roots ϑ 1 , . . ., ϑ 4 of h as illustrated in Figure 3.  GaloisRevisionMonthlyStyle.tex page 9 In order to see that the elements of the symmetry group of the tetrahedron correspond simultaneously to the elements of Gal(L f /Q) and Gal(L h /Q), respectively, we consider two elements of the symmetry group of the tetrahedron.
First, let ρ 1 be the rotation by the angle π about the axis joining the midpoints of the edges ζ 1 and ζ 4 .Then ρ 1 acts on the edges and the faces of the tetrahedron as follows: Notice that the intermediate field which corresponds to ρ 1 is Q(ζ 1 ).Second, let ρ 2 be the rotation by the angle 2π/3 about the axis joining the center of the face ϑ 1 with the opposite vertex.Then ρ 2 acts on the edges and the faces of the tetrahedron as follows: Notice that the intermediate field which corresponds to ρ 2 is Q(ϑ 1 ).

Conclusion.
What we have achieved is a visualization of a Galois group in terms of the edges and faces of a tetrahedron.In particular, we found two polynomials f and h of degree six and four, respectively, such that the roots of f correspond to the six edges and the roots of h correspond to the to the four faces (or vertices) of the tetrahedron.Moreover, since we were able to carry out all the calculations by hand, we obtained a complete understanding of the field extension L f /Q, and in addition, we have an illustrative example of a Galois extension that shows the power and beauty of Galois Theory.

Figure 2 .
Figure 2. Diagram of intermediate fields.For two fields K and M , an arrow K −→ M or K − M indicates that K is a subfield of M , and K − M indicates that the field extension is Galois.
For two groups H and G, an arrow H −→ G or H − G indicates that H is a subgroup or a normal subgroup of G; and ι denotes the identity automorphism of L f .