Galois Extensions; The Fundamental Theorem of Galois Theory

Let K an extension of the field F with characteristic zero.

and

We define the followin operations

and

By Proposition 9.2.17, we have

and

Thus these are mappings, inverting inclusion relations. Further

We have studied cases in which the inclusion in 9.3.1 is proper, e.g. if K = ℚ(∛5), F = ℚ, T = ℚ,

Regarding the second inclusion 9.3.2, the next theorem shows that for finite extension the equality sign always holds.

9.3.1 Theorem. Let K a finite extension of a field F and let H, be a subgroup of G(K,F). Let K_H the field fixed by H, then

1. [K: K_H] = |H|;

2. H = G(K,K_H).

Proof. In proposition , we seen the following properties

since K is a finite extension, by Theorem 9.2.11, we know that

Thus

So it suffices to prove point a) to have also point b), because if |H| = [K : K_H], then H, subgroup of G(K,K_H) with the same order of G(K,K_H), must coincide wih G(K,K_H). We prove that

Since K is a finite extension of F, (hence as well of K_H), there exists an element a ∈ K such that K = K_H(a). If [K : K_H] = m, the minimal polynomial p(x) ∈ K_H[x] of a will have m as degree. Let σ₁ = id, σ₂, ..., σ_h, the h elements of H. We construct the h elementary symmetric functions in the h elements a = σ₁(a), σ₂(a), ..., σ_h(a)

An arbitrary element σ_i ∈ H fixes such functions, as it easily verified. Thus the elements α_i are in K_H. The polynomial

which has clearly a as root, and coefficients in K_H since

which as we saw, has coefficients in K_H. Since the miminal polynomial of a has degree m then h ≥ m, that is |H| ≥ [K: H]. Thus |H| = [K:K_H].  □

The functions Φ and Ψ are known as Galois correspondence. We want to study now the circumstances under which the two mappings are one the inverse of the other. To this end we give another characterization of a normal extension.

9.3.2 Theorem. Let K a finite extension of F. The following statements are equivalents

1. K is a normal extension of F;

2. K_G(K,F) = F  (9.3.3).

Proof. (i) ⇒ (ii) Let K normal over F. We prove 9.3.3 by induction on n = [K:F] (strong induction). For [K:F] = 1, we have K = F and K_G(K,F) = K_id = K = F, thus 9.3.3, holds. We suppose 9.3.3 true, for every normal extension K' of F', with [K', F'] < [K : F] and we prove it for the extension K over F. As normal extension over F, we known by theorem 8.7.6 that, K, is also a splitting field of a polynomial f(x) ∈ F[x]. We can suppose that f(x) has a irreducible polynomial p(x) ∈ F[x] of degree greater than 1, otherwise we would have K = F, and 9.3.3 would be consequently true (deg(p(x)) = 1 just means p(x) = mx − a, with m,a ∈ F). Let α₁, α₂, ..., α_s the s distinct roots of p(x) in K. K in addition to be the splitting field for f as polynomial in F[x], is also a splitting field of f as polynomial with coefficients in F(α₁). We have

and by the inductive hyphothesis we can conclude that

To prove 9.3.3 we must show that, if k ∈ K is left fixed by every automorphisms of G(K,F), then k belongs to F, that is it does not exist outside F, an element that is left fixed by every element of G(K,F). Such an element k (fixed by every element G(K,F)) is certainly left fixed by every element of G(K,F(α₁)). By 9.3.4, k must be in F(α) and can be written in the following form

Owing to splitting fields' theorems, for every i = 1, ..., s there exists an automorphism σ_i of K, such that

with α_i, i = 1, ..., s, all the distinct roots of p(x). Every σ_i fixes k, such that k ∈ K_G(K,F) and also every a_i. So applying σ_i to 9.3.4 yields

But then the polynomial

is a polynomial of degree s − 1 that, by 9.3.5 has s roots. This implies that it must be the null polynomial, thus k −a₀ = 0, i.e. k = a₀ ∈ F.
We now prove the other direction (ii) ⟶ (i). By the hypotheis it holds that K_G(K,F) = F. We must prove that K is a splitting field of a polynomial with coefficients in F. Let a ∈ K such that K = F(a) and let

Consider the polynomial

the coefficients of this polynomial (as we've already shown in theorem 9.3.1) are the elementary symmetric fuctions α_i in the roots σ₁(a), σ₂(a),..., σ_n(a), with every σ_i ∈ K_G(K,F) = F. Thus f is a polynomial of F[x], and since K contains all roots of f there cannot be a smaller field containing all the roots of f and F, because such a field should contain the root a and F, thus coincide with K = F(a). We conclude that K is the splitting field of f(x) ∈ F[x], and thus normal extension of F.  □

Definition 9.3.3. A Galois extensions K of F is a finite and normal extension of K.  □

We're now able to prove two important theorems involved the in the proof of Galois correspondence theorem.

Theorem 9.3.4. Let K a Galois extension of F. Then

Proof. From theorem 9.3.1, we know that for whatsoever subgroup H of G(K,F) we have

Then for H = G(K,F), since K_G(K,F) = F for a Galois extension, we have

The next proposition shows that an alternative definition of a Galois extension is that of a finite extension K of F such that, if φ ∈ I(K,F), then φ(K) ⊆ K.

Proposition 9.3.5. Let K a Galois extension F and let T an intermediate field, F ⊆ T ⊆ K. The following statements are equivalent

1. T is a normal extension of F;

2. σ(T) ⊆ T, ∀σ ∈ G(K,F).

Proof. (i) ⇒ (ii) Let σ ∈ G(K,F), t ∈ T and s = σ(t). t is algebraic over F (because it belongs to a finite extension see Corollary 8.2.14); let p(x) ∈ F[x] it minimal polynomial, σ(t) is root of the same polynomial (9.2.4 Proposition). But then t and s are conjugates over F. Since T is normal over F, s must belong to T.

(ii) ⇒ (i). Let α ∈ T and β ∈ K conjugate over F. We must prove that β ∈ T. α and β, as conjugate over F, are roots of the same irreducible polynomial with coefficients in F, hence there exists a F-isomorphism τ such that

Then τ can be extended by an autormorpshim σ: K ⟶ K such that σ ∈ G(K,F). Thus, by the hypothesis (ii), σ(T) ⊆ T from which β = σ(α) ∈ T.  □

Definition 9.3.6. Let f(x) a polynomial with coefficients in a field F. We define the Galois group of a polynomial f(x), the Galois group of the extension K over F, with K splitting field of f(x).  □

In some cases there is no algebraic difference between conjugate elements over a field as they satisfy the same algebraic equation. Some examples of algebraic equations with reational coefficients are

(√2)² = 2  the same equation is true for −√2⟶   (−√2)² = 2
(√2)³ = 2 √2  ⟶   (−√2)³ = 2 −√2
$\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}=\sqrt{2}$  ⟶   $\frac{1}{-\sqrt{2}}+\frac{-\sqrt{2}}{2}=-\sqrt{2}$

Another example is the following involving the complex roots of unity. Consider ℚ(ζ) with ζ fifth root of unity.

Consider the algebraic equation with rational coefficients

The automorpshim

will satisfy as well the relation above, as

The same is true for the automorphisms

And the function composition of these automorpshisms leads to other automorphisms preserving the algebraic relations: φ₂ ∘ φ₃ = φ₁. This means that the set of automorphsims preserving algebraic relations is a group, with group operation, the composition of functions and identity element φ_id which leaves the elements unchanged. So

We've endend up with G(ℚ(ζ),ℚ) being cyclic of order 4.

But there are automorphisms that do not preserve the relations, for example just swapping ζ⁴ ⟶ ζ³, or just ζ ⟶ ζ², and the equation above does not hold anymore. We have the following mapping between the set of all the subgroups of G(ℚ(ζ),ℚ) the set of all the subfields T of ℚ(ζ) containing ℚ

The basic question of Galois theory is, "given the conjugates elements over a field how many ways are there to swap them preserving algebraic relations?"

9.3.7 Galois Correspondence Theorem. Let f(x) a polynomial with coefficients in a field F. Let K the splitting field of F and G(K,F) its Galois group. Let ℱ the set of all its subfields T of K containing F and let 𝒢 the set of all the subgroups of G(K,F), then

is a bijection between ℱ and 𝒢, whose inverse is

that is

In addition the following properties hold:

1. [K:T] = |G(K,T)|,    [T:F] = |G(K,F)|/|G(K,T)|

2. T is a normal extension of F if and only if G(K,T) is a normal subgroup of G(K,F);

3. If T is a normal extension of F, then the group G(T,F) is such that

   G(T,F) ≃ G(K,F)/G(K,T)

Proof. We saw in theorem 9.3.1 that H = G(K,K_H). Regarding K_G(T,H) = T, it derives from theorem 9.3.2, since K normal extension of F, implies that K is as well normal extension of T. Both relations imply that Ψ and Φ are one the inverse of the other.

1. K is normal extension of T, hence [K:T] = |G(K,T)| based on theorem 9.3.4. Further

   |K:F| = [K:T] [T:F],    [K:F] = |G(K,F)|,    [K:T] = |G(K,T)|

   from which

   $$[T:F]=\frac{|G(K,F)|}{|G(K,T)|}$$

2. In the proposition 9.3.5 we already seen that T is a normal extension of F if and only if for every σ ∈ G(K,F), σ(T) ⊆ T. But then, if we take τ ∈ G(K,T), it results

   τ(σ(t)) = σ(t)   ∀t ∈ T, ∀σ ∈ G(K,F)

   This relation is equivalente to

   σ⁻¹ τσ(t) = t

   that is to

   σ⁻¹G(K,T)σ ⊆ G(K,T)   ∀σ ∈ G(K,F)

   which is the condition for G(K,T) is normal in G(K,F).
   Conversely, if G(K,T) ⊲̲ G(K,F), then T is a normal extension of F, indeed from

   σ⁻¹ τ σ(t) = t,   ∀t ∈ T,  ∀τ ∈ G(K,T),  ∀σ ∈ G(K,F)

   follows that

   τ (σ(t)) = σ(t),   ∀t ∈ T,  ∀σ ∈ G(K,F),  ∀τ ∈ G(K,T)

   Then, since σ(t) is fixed by τ and that K_G(T,H) = T, as K is a normal extension, it follows that σ(t) ∈ T.

3. Since σ(T) ⊆ T for every σ ∈ G(K,F), then every σ ∈ G(K,F) induces an automorphism σ' ∈ G(T,F), defined by

   σ'(t) = σ(t)    ∀t ∈ T

   Clearly σ'(a) = a ∀a ∈ F, hence σ' ∈ G(T,F). The mapping

   φ : G(K,F) ⟶ G(T,F)
   σ ⟼ σ'

   is a homomorphism of G(K,F) onto G(T,F) whose kernel is

   Ker φ := {σ ∈ G(K,F) | σ' = e_G(T,F)} = {σ ∈ G(K,F) | σ'(t) = t ∀t ∈ T } = G(T,F)

   By the fundamental homomorphism theorem

   Im φ ≃ G(K,F) /G(T,F)

   hence

   |Im φ| = |G(K,F)| /|G(T,F)| = [T: F] = |G(T,F|

   thus Im φ = G(K,T), which proves the claim.  □

9.3.8 Example. We illustrate Galois correspondence theorem for the polynomial f(x) = x³ − 2 ∈ ℚ[x]. It results x³ − 2 = (x − ∛2) (x − ω∛2)(x − ω²∛2), with ω primitive cubic root of the unit. The splitting field K of x³ − 2 is K = ℚ(∛2, ω). An element σ ∈ G(K,ℚ) is found by knowing σ(∛2) and σ(ω). The minimal polynomial over ℚ, of ∛2 is x³ − 2, and the minimal polynomial of ω is x² + x + 1. There exist in total six automorphisms

It results G(K, ℚ) ≃ S₃. The non-trivial subgroups of G(K, ℚ) are H₁ = ⟨σ₂⟩, H₂ = ⟨σ₅⟩, H₄ = ⟨σ₃⟩. |H₁| = |H₂| = |H₃| = 2, |H₄| = 3. The fixed fields are the following

9.3.9 Proposition. Let n ∈ ℕ. The Galois group of the polynomial xⁿ − 1 is isomorphic to the multiplicative group U(ℤ_n) of the invertible elements of ℤ_n.

Proof. The splitting field of xⁿ − 1 over ℚ is ℚ(ζ), where ζ is a nth primitive root of unity. As seen in example 9.2.17, U(ℤ_n ≃ G(K, ℚ).  □

9.3.10 Proposition. Let K a Galois extension of F, H₁ and H₂ subgroups of G(K,F) and T₁, T₂ subfields of K containing F such that G(K,T₁) = H₁ and G(K,T₂) = H₂. Then

with ⟨H₁, H₂⟩ subgroup generated by H₁ and H₂, that is the subgroup made by all finite products of elements in H₁ and H₂ and T₁(T₂) the subfield made by T₁ and T₂ i.e. the subfield generated over T₁ by T₂ (which coincides with the subfield generated over T₂ by T₁).

Proof. G(K,T₁(T₂)) consists of automorphisms of K which fix the elements T₁(T₂), thus they fix the elements of both T₁ and T₂. Then G(K,T₁(T₂)) ⊆ H₁ ∩ H₂. Conversely if σ ∈ H₁ ∩ H₂, fixes both the elements of T₁ and T₂ and consequently those of T₁(T₂), thus H₁ ∩ H₂ ⊆ G(K,T₁(T₂)), hence the equality holds.

Let now σ ∈ ⟨H₁, H₂⟩. It is clear that σ fixes the elements of T₁ ∩ T₂, thus ⟨H₁, H₂⟩ ⊆ G(K,T₁(T₂)). Conversely, let k ∈ K, such that k ∉ T₁ ∩ T₂: suppose that k ∉ T₁. Since T₁ = K_H1, there exists at least one element σ ∈ H₁ such that σ(k) ≠ k, hence k is not left fixed by H₁ ∩ H₂. Thus K_{⟨H1, H2⟩}, ⊆ T₁ ∩ T₂ and K_{⟨H1, H2⟩} = T₁ ∩ T₂. By Galois correspondence theorem, ⟨H₁, H₂⟩ = G(K,T₁(T₂)).  □

9.3.11 Proposition. If φ ∈ I(K,F) is such that φ(K) ⊆ K, then φ(K) = K, that is φ is an automorphism. Thus a Galois extension is such that I(K,F) = G(K,F).

Proof. [K:F] = n finite means K = F(b) for some b in K that satisfies a min poly p(x) in F[x]. Let {b = b₁, b₂, ..., b_k} be all the roots in K of p(x). If φ: K ⟶ ℂ is an F-monomorphism sending K to some subset of K, we know that a F-homomorphism preseves the minimal polynomial relation: Say p(x) = x^m + f_(m−1) x^(m−1) + ... + f₁ x + f₀ is the min poly of b. So b^m + ... + f₁ b + f₀ = 0, so also φ(that LHS) = φ(b)^m + ... + f₁ φ(b) + f₀ = 0, so φ(b) is one of the conjugates of b in K, meaning it's one of those b_i up there. The complete set of conjugates of b in K just means the complete set of roots of p(x) containined in K. We need φ(K) ⊆ K to conclude that φ(b) is one of the things in that set. Without knowing φ(K) ⊆ K, then φ(b) could be some *other* root of p(x), not in that set. That could happen if K is not the splitting field of p(x) over F. Thus a F-homomorphism has to send each root b_i up there to some b_j in there, injectively. But an injective map of a finite set to itself is also surjective, so φ(b_j) = b for some b_j. So the image includes b and all of F, thus it's all of F(b) = K.  □

the requirement of the previous porposition is K at least algebraic but not necessarily finite: [Q(pi):Q] = [Q(pi^2):Q] = aleph_0, but the map sending pi to pi^2 (and identity on Q) sends Q(pi) isomorphically onto its proper subfield Q(pi^2).