F-automorphisms of an extension

We continue to work as we did in the last section with fields of characteristic zero.

9.2.1 Definition. Let K an extension of a field F. A monomorphism of K in ℂ which fixes the elements of F is said F-monomorphism of K over F.  □

Example 9.2.2. Let F = ℚ, K = ℚ(α), where α is the third real root of x³ − 3, we have the following F-monomorphism φ from K to ℂ which sends α to ωα (with ω primitive third root of unity) and every a ∈ F to itself.
We indicate with I(K,F) the set of all F-monomorphisms. ■

9.2.3 Proposition. If F = ℚ, every monomorphism of K is a ℚ-monomorphism.

Proof. Let φ a monomorphism of K. Then it results

Let K a finite extension of F. We shall find all the F-monomorphism of K. Since K is a finite extension we have K = F(α) for some α ∈ K (see Proposition 8.8.2). If [K:F] = n, every element k ∈ K can be written in the form

Let φ a F-monomorphism of K. Then

where φ is completely determined once we know φ(α). Further φ(α) is conjugated to α over F (see definition 8.7.1. Indeed we have the following proposition

9.2.4 Proposition. Let K = F(α) a finite extension of F and let φ a F-monomorphism of K. Then φ(α) is a conjugate of α over F, that is φ(α) and α have the same minimal polynomial over F.

Proof. Let p(x) = a₀ + a₁ x + a₂ x² + ... + anxⁿ, (a_i ∈ F) the minimal polynomial of α. We shall prove that φ(α) is also a root of p(x). We have the following equalities:

Thus the possible transformed element of a root α of an irreducible polynomial with coefficients in F can be found among the roots of the polynomial itself. For example consider the extension K = ℚ(√3) of ℚ. If f(x) = x² − 3, then the transformed √3 through of K can only be either √3 or −√3. As

As a consequence of the previous results we have the following theorem.

9.2.5 Theorem. Let K a finite extension of F, with degree n. Then there exist at most n F-monomorphism of K. In other words:

Before proving that there exist exactly n F-monomorphism, we introduce the following lemma.

9.2.6 Lemma. Let F a field and let F(α) an extension of degree n. Let θ a monomorphism from F to ℂ, and let p(x) = a₀ + a₁x + a₂x² + ... + a_nxⁿ (a_i ∈ F) the minimal polynomial of α. Let p_θ the polynomial with coefficients in ℂ, p_θ = θ(a₀) + θ(a₁)x + ... + θ(a_n)xⁿ with β a root of p_θ, there exists a monomorphism, extension of θ to F(α) which sends α to β.

Proof. The proof of this lemma is analogous to that of theorem Theorem 8.2.13, but we repeat it for sake of completeness. Every element s ∈ F(α) can be written univocally as

We define φ: F(α) ⟶ ℂ as

It is obvious that φ is an extension of θ which sends α to β, preserving addition. We show that it preserves the product as well. Let t = b₀ + b₁ α + ...+ b_nαⁿ⁻¹ another element of F(α). We consider the following polynomial associate respectively to s and t:

Dividing f ⋅ g by p

the value in α of this relation, knowing that p(α) = 0, is

For every polynomial v(x) = ∑^m_i=0 c_ixⁱ, we indicate with v_θ(x) the polynomial ∑^m_i=0 θ(c_i)xⁱ, we have the following chains of equalities

Thus φ preserves the product as well: it is a homomorphism. Further, since φ(1) = 1, φ is not the null homomorphism, hence being a non-null homomorphism between fields, it is a monomorphism.  □

We are now able to prove that there are exactly n F-monomorphisms.

9.2.7 Theorem. Let K an extension of F, of degree n, and let α ∈ K such that K = F(α). Let α₁ = α, α₂, ..., α_n the conjugates elements of α over F. Then

1. α, α₂, ..., α_n are all distinct;

2. for every j = 1, ..,n there exits a unique F-monomorphism φ_j of K in ℂ such that φ_j(α) = α_j;

3. φ_j of the previous points cover all the possible F-monomorphisms from K to ℂ.

Proof. a) α₁, α₂, ..., α_n are distinct since zeros of an irreducible polynomial with characteristic zero. (see proposition 8.3.6).

b) If in lemma 9.2.5 we consider as θ the mapping that fixes all the elements of F, then p_θ = p, the lemma says there exists a F-monomorphism φ_j which sends α to α_j given by

c) We found n distinct F-monomorphisms because the α_i are all distinct and thus there cannot be any other, owing to theorem 9.2.5.  □

Among the F-monomorphism, there exists a class of particular importance.

9.2.8 Definition. Let K an extension of a field F. A F-automorphism σ of K is a F-monomorphism of K over K itself. We indicate with G(K,F) the set of all F-automorphism of K.  □

Not every F-monomorphism is a F-automorphism, e.g. in Example 9.2.2 the image through φ of α is not contained in ℚ(α). (α is real, while φ(α) = ωα is complex and not real).

The important fact regarding the F-automorphisms is that with respect to the product of functions the set G(K,F) is a group.

9.2.9 Proposition. The set G(K,F) of all F-automorphisms of an extension K of F is a subgroup of the group of all automorphisms of K.

Proof. By definition of F-automorphism we have

Let σ, τ ∈ G(K,F). Then their product σ ∘ τ is an automorphism which fixes F, because

And σ⁻¹ is an automorphism and

Hence G(K,F) is a subgroup of the group of all automorphisms of K. □

9.2.10 Definition. The group G(K,F) of all utomorphisms of K which fixes the elements of F is known as Galois group of the extension K over F.  □

9.2.11 Theorem. Let K a finite extension of a field F and let n its degree. Then

Proof. It is an immediate consequence of theorem 9.2.5, observing that G(K,F) ⊆ I(K,F).  □

9.2.12 Definition. Let K a field and G a group of automorphisms of K. We define

which is known as field fixed by G.  □

The following proposition show that K_G is indeed a field.

9.2.13 Proposition. Let K a field and G a group of automorphisms of K. The set K_G = {k ∈ K | σ(k) = k ∀σ ∈ G} is a subfield of K.

Proof. Let k₁,k₂ ∈ K_G. Then for every σ ∈ G.

Let now K an extension of a field F. If G(K,F) is the Galois groups of the extension K over F, it results as immediate consequence of the definition of G(K,F), that

9.2.14 Example. Let K = ℂ, field of the complex numbers and let F = ℝ.
If σ ∈ G(ℂ, ℝ), since ℂ = ℝ(i). A field automorphism, is such that σ(x+y) = σ(x) + σ(y), and σ(xy) = σ(x) σ(y), so

σ(−1) = σ(i²) = σ(i)σ(i). Since σ(−1) = −1, the square of σ(i), must be −1, that is σ(i) must be a root of x² + 1, i.e. ±i. Now that we know that σ(i) = ± i, it's clear that can have only two possible mappings

and

are respectively the identity automorphism and the automorpism which maps an element to its complex conjugate. Thus

We evaluate now the field fixed by G(ℂ, ℝ), that is

Since that the identity fixes all the elements in ℂ, it suffices to consider the element k ∈ ℂ that are fixed by σ₂. Thus

In this case, K_G(F,K), which as we know must contain F, is equal to F.  □

9.2.15 Example. Let K = ℚ(∛5) and F = ℚ. By Eisenstein and Gauss x³ − 5 is irreducible in ℚ[x], so [ℚ(∛5): ℚ] = 3 and a basis is {1, ∛5, ∛5²}. So we can express K as

The action of σ on a element of K is such that

So to find all possible automorphisms of ℚ(∛5) (which fix ℚ), by proposition 9.2.4, we only need to find the images σ(∛5). Since the minimal polynomial of α over ℚ is x³ − 5, σ(α) must be another root of the same polynomial. Apart from α the other two roots are complex, hence not in ℚ(∛5) ⊂ ℝ. It follows that the only possibility is σ(α) = α, that is the unique automoprhism of G(ℚ(∛5), ℚ) is the identity automorphism.

It is obvious that the field fixed by G(ℚ(∛5), ℚ) is all G(ℚ(∛5)), hence in this case

Notice that in this case, respect to the previous example we have

9.2.16 Example. Let K = ℚ(∛5), ω), where ω is a primitive third root of unity. Let's calculate ℚ(∛5), ω) as vector space over ℚ. We start by ℚ(∛5) (already calculated in the previous exercise) and then adjoin ω. ω is root of the irreducible polynomial x² + x + 1 over ℚ(∛5) (as over ℚ, see exercise 7.d), so

Since a basis of ℚ(∛5,ω) over ℚ(∛5) is given by {1, ω} and a basis of ℚ(∛5) over ℚ is {1, ∛5, ∛5²} we have that a basis of ℚ(∛5,ω) over ℚ is given by

The roots of x³ − 5, are ∛5, ∛5ω, ∛5ω².
ω is the root of x³ − 1 which is reducible as (x − 1) (x² + x + 1), only the last is irreducible over ℚ with roots −1/2 ± √3i/2. There are in total six automoprhisms

Notice that in this case

it results G(K, ℚ) ≃ S₃.  ■

9.2.17 Example. Let ζ a primitive nth root of unity. We calculate ℚ(ζ). We know ζ satisfies the nth cyclotomic polynomial Φ_n(x) which is irreducible over ℚ, thus the minimal polynomial of ζ over ℚ. It follows that

Every element of G(ℚ(ζ), ℚ), maps ζ to another root of Φ_n(x), i.e. an element of the form ζ^r with 1 ≤ r ≤ n and (n,r) = 1. Conversely if r is an integer such that 1 ≤ r ≤ n and (n,r) = 1, the mapping σ_r which maps ζ to ζ^r is an automorphism of ℚ(ζ). This fact suggests the possibility to define the following mapping between the group U(ℤ_n) of the invertible elements of ℤ_n and G(ℚ(ζ), ℚ):

This is a bijective mapping. We prove it is a group homomorphism. If ī, j̄ ∈ U(ℤ_n), since we are dealing with a multiplicative group we have that ij ≡ k (mod n). G(ℚ(ζ), ℚ) is a group under composition of function so we show that σ_i σ_j = σ_k, indeed:

We have to confirm that ζ^ij = ζ^k. Now since ij ≡ k (mod n) means that there exists m such that ij = k + mn, therefore,

which proves that

In the examples above, it results

In some cases the equality sign holds and as we shall prove soon it happens when K is a normal extension.

Through definition 9.2.10 it is possible to associate to every extension K of a field F a group, its Galois group G(K,F). We shall now establish a relatin between the set of intermediate fields of an extension K of F and the set of subgroups of G(K,F).
Let K an extension of F and let G(K,F) its Galois group. For every intermediate field, T of K containing F we define

For every subgroup H of G(K,F) we define

The following properties hold true.

9.2.18 Proposition. Let K an extension of F and let G(K,F) its Galois group. For every subgroup H of G(K,F) and every intermediate subfield T such that F ⊆ T ⊆ K, we have

1. G(K,K) = id;

2. if T₁, T₂ are fields such that F ⊆ T₁ ⊆ T₂ ⊆ K, then

G(K,T₂) ⊆ G(K,T₁)

3. G(K,T) is a subgroup of G(K,F);

4. K_id = K;

5. If H₁, H₂ are subgroups of G(K,F) such that H₁ ⊆ H₂ ⊆ G(K,F), then

   K_H2 ⊆ K_H1;

6. K_H is an intermediate field: F ⊆ K_H ⊆ K;

7. H ⊆ G(K,K_H);

8. T ⊆ K_G(K,T).

Proof.

1. G(K,K) = {σ ∈ G(K,F) | σ(k) = k ∀k ∈ K}, and this is exactly what σ = id means.

2. Supose that T₁ ⊆ T₂ and let σ ∈ G(K,T₂). Then σ fixes every element of T₂ and so certainly fixes every element of T₁. Hence σ ∈ G(K,T₁)

3. G(K,T) is made by the automorphisms that fix the elements of T, which contains F, but all the automorphism that fix T must also fix F.

4. K_id = {k ∈ K | σ(k) = k ∀σ ∈ {id}} = K.

5. The elements k ∈ K that are fixed by the automorpshims of H₂, must be fixed as well by the automorphisms of H₁, because H₁ is a subgroup of H₂. Thus K_H2 ⊆ K_H1

6. By definition of K_H is clear that K_H ⊆ K. The other inclusion F ⊆ K_H derives from the fact that H is a subgroup of G(K,F) which by definition the group of automorphisms that fix the elements of F.

7. H is a subgroup of G(K,F)

8. G(K,K_H) is made of the automorphisms that fix all the elements in K_H, and we proved in f) that it contains F. G(K,K_H) must contain all the automorpshisms of H, because this latter is a subgroup of G(K,F) which fixes only the element of F.

9. The automorphisms in G(K,T) fix all the elements of T, and K_G(K,T) contains all the elements of K which are fixed by the elements of G(K,T), thus it contains T. □

9.2.19 Proposition. In the field ℝ of all real numbers, the identity mapping of ℝ is the only automorphism.

Before proving it we introduce two lemmas.

9.2.20 Lemma. Let ϕ: ℝ → ℝ be an automorphism of the field of real numbers ℝ, then for any positive real number x , we have ϕ(x) > 0

Proof. ϕ(x²) = ϕ(x) ϕ(x) ≥ 0. Note that since ϕ(0) = 0 and ϕ is bijective, ϕ(x) ≠ 0 for any x ≠ 0. Thus, it follows that ϕ(x) > 0 for each positive real number x. □

9.2.21 Lemma. For any x,y ∈ ℝ such that x > y, we have ϕ(x) > ϕ(y). Thus, in any automorphism of the field of real numbers the order of the elements is preserved.

Proof. Since x > y, we have x−y > 0 and it follows from the previous lemma that

Hence, ϕ(x) > ϕ(y). □

Proof of 9.2.19. Let σ an automorphsim of ℝ, and let a,b ∈ ℝ such that a < b. Then by the previous lemma σ(a) < σ(b). Suppose that there exists a ∈ ℝ such that σ(a) > a. Let q a rational number, such that σ(a) > q > a, we have σ(q) = q. This leads to a contradiction since q > a but σ(a) > σ(q) = q.  □

Exercise 9.2.22. Determine the Galois group G(ℚ(∛2), ℚ) and its fixed field.

All the images of ∛2 are to be searched among the roots of x³ − 2. Since ℚ(∛2) has only one root, the only automorphism is the identity, G(ℚ(∛2), ℚ) = {id}, whereas I(ℚ(∛2), ℚ) has three elements. It is obvious that its fixed field is the entire ℚ(∛2).  ■