2.2. Splitting fields.

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Let f be a polynomial with coefficients in F. A field E containing F is said to split f if f

splits in E[X], i.e., if f(X) =

_

(X −αi) with αi ∈ E. If E is also generated by the αi, then

it is called a splitting field for f.

Note that if f(X) =

_

fi(X)mi , then a splitting field for

_

fi(X) is also a splitting field

for f (and conversely).

Example 2.3. (a) Let f(X) = aX2 + bX + c ∈ Q[√ X] be irreducible, and let α =

b2 − 4ac;the n the subfield Q[α] of C generated by α is a splitting field for f.

(b) Let f(X) = X3 + aX2 + bX + c ∈ Q[X] be irreducible, and let α1, α2, α3 be its roots

in C. Then Q[α1, α2, α3] = Q[α1, α2] is a splitting field for f(X). Note that [Q[α1] : Q] = 3

and that [Q[α1, α2] : Q[α1]] = 1 or 2, and so [Q[α1, α2] : Q] = 3 or 6. We’ll see later that

the degree is 3 if and only if the discriminant of f(X) is a square in F. For example, the

discriminant of X3 + bX + c is −4b3 − 27c2, and so the splitting field of X3 + 10X + 1 has

degree 6 over Q.

Proposition 2.4. Every polynomial has a splitting field.

Proof. Let f ∈ F[X]. Let g1 be an irreducible factor of f(X), and let F1 =

F[X]/(g1(X)) = F[α1], α1 = X + (g1). Then α1 is a root of f(X) in F1, and we define

f1(X) to be the quotient f(X)/(X − α1) (in F1[X]). Then f1 ∈ F1[X], and the same construction

gives us a field F2 = F1[α2] with α2 a root of f1. By continuing in this fashion, we

obtain a splitting field.

Remark 2.5. Let n = degf. In the proof, [F1 : F] ≤ n, [F2 : F1] ≤ n − 1, ..., and so

the degree of the splitting field over F is ≤ n!. Whether or not there exist polynomials of

degree n in F[X] whose splitting field has degree n! depends on F. For example, there don’t

for n > 1 if F = C or Fp, nor for n > 2 if F = R. However, later we shall see how to

write down large numbers (in fact infinitely many) polynomials of degree n in Q[X] whose

splitting fields have degree n!.

Example 2.6. (a) Let f = (Xp − 1)/(X − 1);a ny field generated by a root of f is a

splitting field (if ζ is one root, the remainder are ζ2, ζ3, . . . , ζp−1).

(b) Suppose F is of characteristic p, and let f = Xp − X − a;a ny field generated by a

root of f is a splitting field (if α is one root, the remainder are α + 1, ..., α + p − 1).

(c) If α is one root of Xn−a, then the remaining roots are all of the form ζα, where ζn = 1.

Therefore, if F contains all the nth roots of 1, i.e., if Xn − 1 splits in F[X], then F[α] is a

splitting field for Xn − a. Note that if p is the characteristic of F, then Xp − 1 = (X − 1)p,

and so F automatically contains all the pth roots of 1.

Proposition 2.7. Let f ∈ F[X], and let E be a splitting field for f, and let Ω ⊃ F be a

second field splitting f.

(a) There exists at least one F-homomorphism ϕ : E → Ω.

14 J.S. MILNE

(b) The number of F-homomorphisms E → Ω is ≤ [E : F], and = [E : F] if f has deg(f)

distinct roots in Ω.

(c) If Ω is also a splitting field for f, then each F-homomorphism E → Ω is an isomorphism.

In particular, any two splitting fields for f are F-isomorphic.

Proof. Write E = F[α1, ..., αm], m ≤ deg(f), with the αi the distinct roots of f(X).

The minimum polynomial of α1 is an irreducible polynomial f1 dividing f. As f (hence f1)

splits in Ω, Proposition 2.1 shows that there exists an F-homomorphism ϕ1 : F[α1] → Ω,

and the number of ϕ1’s is ≤ deg(f1) = [F[α1] : F], with equality holding when f (hence also

f1) has distinct roots in Ω.

Next, the minimum polynomial of α2 over F[α1] is an irreducible factor f2 of f(X)

in F[α1][X]. According to Proposition 2.2, each ϕ1 extends to a homomorphism ϕ2 :

F[α1, α2] → Ω, and the number of extensions is ≤ deg(f2) = [F[α1, α2] : F[α1]], with

equality holding when f (hence also f2) has distinct roots in Ω.

On combining these statements we conclude that there exists an F-homomorphism ϕ :

F[α1, α2] → Ω, and the number of such homomorphisms is ≤ [F[α1, α2] : F], with equality

holding when f has deg(f) distinct roots in Ω.

After repeating the argument m times, we obtain (a) and (b). For (c), note that, because

an F-homomorphism E → Ω is injective, we must have [E : F] ≤ [Ω : F]. If Ω is also a

splitting field, then we obtain the reverse inequality also. We therefore have equality, and so

any F-homomorphism E → Ω is an isomorphism.

Corollary 2.8. Let E and L be extension fields of F, with E finite over F; then there

exists an extension field Ω of L and an F-homomorphism E → Ω.

Proof. Write E = F[α1, . . . , αm], and let fi be the minimum polynomial of αi over F.

Let E_ be a splitting field of f =df

_

fi regarded as an element of E[X], and replace E with

the subfield of E_ generated by F and all the roots of f(X). Thus E is now the splitting

field of f(X) ∈ F[X]. Let Ω be a splitting field for f regarded as an element of L[X]. The

proposition shows that there is an F-homomorphism E → Ω.

Remark 2.9. After replacing E by its (isomorphic) image in Ω, we will have that E and

L are subfields of Ω. This will allow us to assume that E and L are subfields of a common

field.

Warning! If E and E_ are splitting fields of f(X) ∈ F[X], then we know there is an

F-isomorphism E → E_, but there will in general be no preferred such isomorphism. Error

and confusion can result if you simply identify the fields.

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