3.3. Separable, normal, and Galois extensions.

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 An algebraic extension E/F is said

to be separable if the minimum polynomial of every element of E is separable, i.e., doesn’t

have multiple roots (in a splitting field);equ ivalently, if every irreducible polynomial in F[X]

having a root in E is separable. Thus E/F is inseparable if and only if

(a) F is nonperfect, and in particular has characteristic p _= 0, and

(b) there is an element α of E whose minimal polynomial is of the form g(Xp), g ∈ F[X].

For example, E = Fp(T ) is an inseparable extension of Fp(T p).

An algebraic extension E/F is normal if the minimum polynomial of every element of E

splits in E;equ ivalently, if every irreducible polynomial f ∈ F[X] having a root in E splits

in E.

Thus if f ∈ F[X] is irreducible of degree m and has a root in E, then

E/F separable =⇒ roots of f distinct

E/F normal =⇒ f splits in E



=⇒ f has m distinct roots in E.

Therefore, E/F is normal and separable if and only if, for each α ∈ E, the minimum

polynomial of α has [F[α] : F] distinct roots in E.

22 J.S. MILNE

Example 3.11. (a) The field Q[ 3

√

2], where 3

√

2 is the real cube root of 2, is separable

but not normal over Q (X3 − 2 doesn’t split in Q[α]).

(b) The field Fp(T ) is normal but not separable over Fp(T p)—it is the splitting field of the

inseparable polynomial Xp − T p.

Theorem 3.12. Let E be an extension field of F. The following statements are equivalent:

(a) E is the splitting field of a separable polynomial f ∈ F[X];

(b) F = EG for some finite group of automorphisms of E;

(c) E is normal and separable, and of finite degree, over F.

Moreover, if E is as in (a), then F = EAut(E/F); if G and F are as in (b) then G =

Aut(E/F).

Proof. (a) =⇒ (b). Let G = Aut(E/F), and let F_ = EG ⊃ F. Then E is also the

splitting field of f ∈ F_[X], and f is still separable when regarded as a polynomial over F_.

Hence Proposition 3.8 shows that

[E : F

_

] = #Aut(E/F

_

)

[E : F] = #Aut(E/F).

Since Aut(E/F_) = Aut(E/F) = G, we conclude that F = F_, and so F = EAut(E/F).

(b) =⇒ (c). By Artin’s lemma, we know that [E : F] ≤ (G : 1);i n particular, it is finite.

Let α ∈ E and let f be the minimum polynomial of α;w e have to prove that f splits into

distinct factors in E. Let {α1 = α, ..., αm} be the orbit of α under G, and let

g(X) =

_

(X − αi) = Xm + a1Xm−1 + ・ ・ ・ + am.

Any σ ∈ G merely permutes the αi. Since the ai are symmetric polynomials in the αi, we

find that σai = ai for all i, and so g(X) ∈ F[X]. It is monic, and g(α) = 0, and so f(X)|g(X)

(see p7). But also g(X)|f(X), because each αi is a root of f(X) (if αi = σα, then applying

σ to the equation f(α) = 0 gives f(αi) = 0). We conclude that f(X) = g(X), and so f(X)

splits into distinct factors in E.

(c) =⇒ (a). Because E has finite degree over F, it is generated over F by a finite number

of elements, say, E = F[α1, ..., αm], αi ∈ E, αi algebraic over F. Let fi be the minimum

polynomial of αi over F. Because E is normal over F, each fi splits in E, and so E is the

splitting field of f =

_

fi. Because E is separable over F, f is separable.

Finally, we have to show that if G is a finite group acting on a field E, then G =

Aut(E/EG). We know that:

• [E : EG] ≤ (G : 1) (Artin),

• G ⊂ Aut(E/EG), and,

• E is the splitting field of a separable polynomial in EG[X] (because b =⇒ a), and so

(by 3.8) the order of Aut(E/EG) is [E : EG].

Now the inequalities

[E : EG] ≤ ____________(G : 1) ≤ (Aut(E/EG) : 1) = [E : EG]

must be equalities, and so G = Aut(E/EG).

FIELDS AND GALOIS THEORY 23

An extension of fields E ⊃ F satisfying the equivalent conditions of the proposition is

called a Galois extension, and Aut(E/F) is called the Galois group Gal(E/F) of E over F.

Note that we have shown that F = EGal(E/F).

Remark 3.13. Let E be Galois over F with Galois group G, and let α ∈ E. The elements

α1 = α, α2, ..., αm of the orbit of α are called the conjugates of α. In the course of the proof

of the the above theorem we showed that the minimum polynomial of α is

_

(X − αi).

Corollary 3.14. Every finite separable extension E of F is contained in a finite Galois

extension.

Proof. Let E = F[α1, ..., αm]. Let fi = minimum polynomial of αi over F, and take E_

to be the splitting field of

_

fi over F.

Corollary 3.15. Let E ⊃ M ⊃ F; if E is Galois over F, then it is Galois over M.

Proof. We know E is the splitting field of some f ∈ F[X];it is also the splitting field of

f regarded as an element of M[X].

Remark 3.16. When we drop the assumption that E is separable over F, we can still

say something. Let E be a finite extension of F. An element α ∈ E is said to be separable

over F if its minimum polynomial over F is separable. The elements of E separable over

F form a subfield E_ of E that is separable over F;wr ite [E : F]sep = [E_ : F] (separable

degree of E over F). If Ω is an algebraically closed field containing F, then there are exactly

[E : F]sep F-homomorphisms E → Ω. When E ⊃ M ⊃ F (finite extensions),

[E : F]sep = [E : M]sep[M : F]sep.

In particular,

E is separable over F ⇐⇒ E is separable over M and M is separable over F.

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