4.4. Quartic polynomials.

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Let f(X) be a quartic polynomial, and assume that the roots of f are simple. In order to

determine Gf we shall exploit the fact that S4 has

V = {1, (12)(34), (13)(24), (14)(23)}

as a normal subgroup—it is normal because it contains all elements of type 2+2—see Groups

p34. Let E be the splitting field of f, and let f(X) =

_

(X − αi) in E. We identify the

30 J.S. MILNE

Galois group Gf of f with a subgroup of the symmetric group S4 = Sym({α1, α2, α3, α4}).

Consider the partially symmetric elements

α = α1α2 + α3α4

β = α1α3 + α2α4

γ = α1α4 + α2α3.

They are distinct elements of E because the αi are distinct, e.g.,

α − β = α1(α2 − α3) + α4(α3 − α2) = (α1 − α4)(α2 − α3).

The group Sym({α1, α2, α3, α4}) permutes {α, β, γ} transitively. The stabilizer of each of

α, β, γ must therefore be a subgroup of index 3 in S4, and hence has order 8. For example,

the stabilizer of β is < (1234), (13) >. Groups of order 8 in S4 are Sylow 2-subgroups. There

are three of them, all isomorphic to D4. By the Sylow theorems, V is contained in a Sylow 2-

subgroup, and, because they are conjugate and it is normal, it must be contained in all three.

It follows that V is the intersection of the three Sylow 2-subgroups. Each Sylow 2-subgroup

stabilizes exactly one of α, β, or γ, and therefore their intersection V is the subgroup of S4

fixing α, β, and γ.

Lemma 4.7. The field M = F[α, β, γ] corresponds to Gf ∩ V. Hence M is Galois over

F, with Galois group G/G ∩ V .

Proof. The first statement follows from the above discussion, and the second follows

from the Fundamental Theorem of Galois Theory.

Picture:

E 1

G ∩ V | |

M G∩ V

G/G ∩ V | |

F G

Let g(X) = (X − α)(X − β)(X − γ) ∈ M[X]—it is called the resolvant cubic of f. Any

permutation of the αi (a fortiori, any element of Gf ) merely permutes α, β, γ, and so fixes

g(X). Therefore (by the Fundamental Theorem) g(X) has coefficients in F. More explicitly,

we have:

Lemma 4.8. If f = X4+bX3+cX2+dX+e, then g = X3−cX2+(bd−4e)X−b2e+4ce−d2.

The discriminants of f and g are equal.

Proof. Compute everything in terms of the αi’s. (Cf. Hungerford, V.4.10.)

Now let f be an irreducible separable quartic. Then G = Gf is a transitive subgroup of

S4 whose order is divisible by 4. There are the following possibilities:

G (G ∩ V : 1) (G : V ∩ G)

S4 4 6

A4 4 3

V 4 1

D4 4 2

C4 2 2

FIELDS AND GALOIS THEORY 31

(G ∩ V : 1) =[E : M], (G : V ∩ G) = [M : F].

Note that G can’t, for example, be the group generated by (12) and (34) because this is

not transitive. The groups of type D4 are the Sylow 2-subgroups discussed above, and the

groups of type C4 are those generated by cycles of length 4.

We can compute (G : V ∩ G) from the resolvant cubic g, because G/V ∩ G = Gal(M/F),

and M is the splitting field of g. Once we know (G : V ∩G), we can deduce G except in the

case that it is 2. If [M : F] = 2, then G∩V = V or C2. Only the first group acts transitively

on the roots of f, and so (from 4.4) we see that (in this case) G = D4 or C4 according as f

is irreducible or not in M[X].

Example 4.9. Consider f(X) = X4 + 4X2 + 2 ∈ Q[X]. It is irreducible by Eisenstein’s

criterion, and its resolvant cubic is (X − 4)(X2 − 8);t hus M = Q[

√

2]. Note that f, when

regarded as a polynomial in X2, factors over M;he nce Gf = C4.

Example 4.10. Consider f(X) = X4 −10X2 +4 ∈ Q[X]. One can check directly (using

1.6) that it is irreducible, and its resolvant cubic is (X +10)(X +4)(X −4). Hence Gf = V .

Example 4.11. Consider f(X) = X4 − 2 ∈ Q[X]. It is irreducible by Eisenstein’s

criterion, and its resolvant cubic is g(X) = X3 + 8X. Hence M = Q[i

√

2]. One can check

that f is irreducible over M, and so its Galois group is D4.

Alternatively, analyze the equation as in (3.20).

Maple knows how to factor polynomials over Q and over Q[α] where α is a root of an

irreducible polynomial. To learn the syntax, type: ?Factor.

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