4.2. When is Gf transitive?

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Proposition 4.4. Let f(X) ∈ F[X] have only simple roots. Then f(X) is irreducible if

and only if Gf permutes the roots of f transitively.

Proof. =⇒ : If α and β are two roots of f(X) in a splitting field Ff for f, then they

both have f(X) as their minimum polynomial, and so there is a natural F-isomorphism

F[α] → F[β], namely,

F[α] ≈ F[X]/(f(X)) ≈ F[β], α↔ X ↔ β.

Write Ff = F[α1, α2, ...] with α1 = α and α2, α3, . . . the other roots of f(X). Then the

F-isomorphism F[α] → F[β] extends (step by step) to a homomorphism F[α1, α2, ...]→ Ff

(see 2.7), which must be an isomorphism.

⇐= : Let g(X) ∈ F[X] be an irreducible factor of f, and let α be one of its roots. If β is

a second root of f, then (by assumption) β = σα for some σ ∈ Gf. Now the equation

0 = σg(α)

g(X)∈F[X]

= g(σα)

shows that β is also a root of g, and we see that we must have f(X) = g(X).

Note that when f(X) is irreducible of degree n, then n|(Gf : 1) because [F[α] : F] = n

and [F[α] : F]|[Ff : F] = (Gf : 1). Thus Gf is a transitive subgroup of Sn whose order is

divisible by n.

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