4.1. When is Gf ⊂ An?

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Consider a polynomial

f(X) = Xn + a1Xn−1 + ・ ・ ・ + an

and let f(X) =

_n

i=1(X − αi) in some splitting field. Set

Δ(f) =

_

1≤i<j≤n

(αi − αj), D(f) = Δ(f)2 =

_

1≤i<j≤n

(αi − αj)2.

Note that D(f) _= 0 if f has a only simple roots, i.e., if f is separable with no multiple

factors. Identify Gf with a subgroup of Sym({α1, . . . , αn}) (as in §3.6).

Proposition 4.1. Assume f is separable, and let σ ∈ Gf .

(a) σΔ(f) = sign(σ)Δ(f), where sign(σ) is the signature of σ.

(b) σD(f) = D(f).

Proof. The first equation follows immediately from the definition of the signature of σ

(see Groups, p31), and the second equation is obtained by squaring the first.

Corollary 4.2. Let f(X) ∈ F[X] be of degree n and have only simple roots. Let Ff be

a splitting field for f, so that Gf = Gal(Ff/F ).

(a) The discriminant D(f) ∈ F.

(b) The subfield of Ff corresponding to An ∩ Gf is F[Δ(f)]. Hence

Gf ⊂ An ⇐⇒ Δ(f) ∈ F ⇐⇒ D(f) is a square in F.

Proof. (a)We know that D(f) is an element of Ff fixed by Gf =df Gal(Ff/F ). Therefore

it lies in F (by the Fundamental Theorem of Galois Theory).

(b) Because f has simple roots, Δ(f) _= 0, and so the formula σΔ(f) = sign(σ)Δ(f) shows

that σ fixes Δ(f) ⇐⇒ σ ∈ An. Therefore Gf ∩ An is the subgroup of Gf corresponding to

F[Δ(f)], and so Gf ∩ An = Gf ⇐⇒ F[Δ(f)] = F.

The discriminant of f can be expressed as a universal polynomial in the coefficients of

f—we shall prove this later. For example:

D(aX2 + bX + c) = b2 − 4ac

D(X3 + bX + c) = −4b3 − 27c2.

By completing the cube, one can put any cubic polynomial in this form.

The formulas for the discriminant rapidly become very complicated, for example, that for

X5 + aX4 + bX3 + cX2 + dX + e has about 60 terms. Fortunately, Maple knows them: the

syntax is “discrim(f,X);” where f is a polynomial in the variable X.

Remark 4.3. Suppose F ⊂ R. Then D(f) will not be a square if it is negative. It is

known that the sign of D(f) is (−1)s where 2s is the number of nonreal roots of f in C.

Thus if s is odd, then Gf is not contained in An. This can be proved more directly by noting

that complex conjugation will act on the roots as the product of s transpositions (cf. the

proof of Proposition 4.13). Of course the converse is not true: when s is even, Gf is not

necessarily contained in An.

FIELDS AND GALOIS THEORY 29

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