Normal Extension

We saw that given an irreducible polynomial in F[x] and one of its root α, is not generally true that the splitting field of f(x) is F(α), that is, the field containing one root of the irreducible polynomial may not contain all of them. For example ∛2 is a root of the irreducible polynomial x³ − 2 over ℚ, but ℚ(∛2) is not the splitting field for x³ − 2, because it does not possess the complex roots of the polynomial are not in it.

8.7.1 Definition. Let K be an extension of a field F. Two element α and β ∈ K algebraic over F, are said to be conjugate over F whenever they have the same minimal polynomial over F.  □

Example 8.7.2. ℝ is not a normal extension of the field ℚ, for x³ − 2 ∈ ℚ[x] is irreducible over ℚ and has a root ∛2, but it does not split into linear factors in ℝ, since it has complex roots. ■

For example ∛2, ω∛2 an ω²∛2, where ω is a 3th primitive root of unity, are all conjugate elements, since roots of a same irreducible polynomial which is as well the minimal polynomial (recall that every irreducible polynomial is a minimal polynomial for each of its roots).

8.7.3. Proposition. Let K be an extension of a field F with [K:F] = 2. Then K is normal over F.

Proof. In the first place K is algebraic over F. Suppose that a ∈ K is a root of the minimal polynomial p ∈ F[x]. Then deg(p) = [F(a) : F] ≤ [K: F] =2, which means deg(p) = 1 or 2. In the first case a is the only root of p. Suppose that deg(p) = 2 with say p = x² + mx + c and m,c ∈ F; if a' is another root of p, then aa' = b ∈ F, so that a' ∈ K. Therefore K is normal over F.  □

8.7.4 Proposition. Let K be an extension of a field F. Then the following proposition are equivalent:

1. The extension K over F is closed with respect to conjugate elements, that is F ⊆ K ⊆ L, a ∈ K and b (∈ L) is conjugate to a ∈ F, then b is in K (In words, if K contains an element, it contains all its conjugate elements).

2. if f(x) is an irreducible polynomial in F[x], and if α is a root of f(x), contained in K, then f(x) splits over K, that is K contains all roots of f(x).

Proof. (a) ⇒ (b) Let α a root of the irreducible polynomial f(x) ∈ F[x]. Let L the splitting field for f over K and β a root of f ∈ L. It results then F ⊆ K ⊆ L and hence being K closed with respect to conjugate elements β ∈ K. Thus f splits over K.

(b) ⇒ (a) Let F ⊆ K ⊆ L and let a ∈ K and b ∈ L, with b conjugate to a over F. Then a and b have the same minimal polynomial f over F, thus b ∈ K as well.  □

Recalling that a normal subgroup of a group G is such that if it contains an element it contains all conjugate elements to it. We give the following definition.

8.7.5 Definition. Let K be an extension of F. We say that K is a normal extension of F if, whenever f is an irreducible polynomial in F[x] which has one root in K, then all the roots of f lie in K (that is, f splits into linear factors in K[x]). Or alternatively, the extension K over F is normal if it is closed with respect to conjugate elements. □

For example, ℂ is a normal extension of ℝ, or of ℚ. However, if α is the real cube root of 2, then K = ℚ(α) is not a normal extension of ℚ. For the polynomial x³ − 2 is irreducible over ℚ, by Eisenstein’s criterion; it has a root α ∈ K; but it does not contain the other two roots of x³ − 2, since they are non-real, whereas K is contained in the real numbers. In fact, we have the factorisation x³ − 2 = (x − α)(x² + αx + α²) in K[x], where the second factor is irreducible in K[x].

That there is a close connection between normal extensions and splitting fields of polynomials is demosntrated by the following fundamental results.

8.7.6 Theorem. Let K be a finite extension of a field F. Then K is normal over F if and only if K is the splitting field of some polynomial in F[x].

Proof. Let K be a finite and normal extension of F. Then it results (for the sole fact of being a finite extension)

for some algebraic α_i over F. Let p_i(x) ∈ F[x] the minimal polynomial of α_i. Let

Every p_i is irreducible over F, possess a zero α_i in K, hence by normality of K over F it splits over K. Thus f as well splits over K.
Let now K be a splitting field for F of some polynomial f(x) ∈ F[x]. By definition of splitting field the extension K over F is finite. We must prove it is a normal extension i.e. that every irreducible polynomial g(x) in F[x], with a root α ∈ K, splits over K (Proposition 8.7.2). Let L ⊇ K the splitting field of the polynomial f(x)g(x), which contains the splitting field K of f(x). Let β ∈ L, another root of g(x). We must show that β ∈ K. Since α ∈ K, it results:

It suffices to show that also [K(β) : K] = [K(α) : K] (= 1). We've that L ⊇ K(α) and L ⊇ K(β) such that

And also L ⊇ F(α) and L ⊇ F(β)

Since α and β have the same minimal polynomial g, it results

further the following isomorphism holds (see Corollary 8.2.15)

Since K is the splitting field for f over F, K(α) is the splitting field of f over F(α). Analogously K(β) is the splitting field of f over F(β). There exists then an extension of φ to an isomorphism ψ : K(α) ⟶ K (β) such that ψ_|F(α) ≡ φ. Further [K(α): F(α)] = [K(β) : F(β)]. Using repeatedly the compound extension degree theorem (see Theorem 8.1.6)

by dividing by [K:F], the desired result follows:

Exercises

1. Determine whether the following exensions are normal.

   ℚ(√11i) over ℚ;    ℚ(√5,3^1/5) over ℚ(3^1/5);    ℚ(3^1/5) over ℚ.

2. Determine the conjugate element over ℚ of

   (−1 + √3i)/2    α primitive 8-th root of unity.

Solutions

1. We employ 8.7.6 Theorem to verify they are splitting field

   • x² + 11 in ℚ[x]. Clearly, if “b = √11i” is a root of that polynomial, then −b is the other root. Thus the field ℚ(√11i) is normal over ℚ.

   • The minimal polynomial of √5 is x² − 5 in ℚ[x], and ℚ(√5) contains −√5 as well. The minimal polynomial of 3^1/5 is x − 3^1/5; so ℚ(√5,3^1/5) is a normal extension over ℚ(3^1/5).

   • ℚ(3^1/5) over ℚ is not normal, since it has only one real root 3^1/5 whereas the four other conjugates 3^1/5 are non-real.

2. The minimal polynomial is x² + x + 1. The conjugate elements are the two roots of the polynomial, i.e. it is (−1 ± √3i)/2.

3. If ω is a primitive nth root of unity, its minimal polynomial is the nth cyclotomic polynomial, hence the conjugate elements of ω are the nth roots of unity, i.e. the ω^k with 1 ≤ n − 1, (n,k) = 1.  ■