Just because we can’t find a solution, it doesn’t mean there isn’t one, Andrew Wiles
Proposition. Let K be any field and let G be a finite group of automorphisms of K. Suppose F is the fixed field of G. Then G = Gal(K/F) = Gal(K/K^{G}).
Proof.
Let G be a finite group of automorphism of K, F be the fixed field of G.
Recall. Gal(K/F) = {σ ∈ Aut(K)| σ(a) = a ∀a ∈ F}. Let K be any field, and let σ_{1}, ···, σ_{n}: K → K be distinct field automorphisms. Suppose that σ_{1}, ···, σ_{n} forms a group under composition. If F is a fixed field of σ_{1}, ···, σ_{n}. Then, [K: F] = n ⇒ [K:F] = |G| = n.
Clearly G ⊆ Gal(K/F), this is because any σ ∈ G fixes every element of F (F is the fixed field of G), so σ is an F-automorphism and G is the group of all F-automorphism ∴ σ ∈ Gal(K/F)
We claim Gal(K/F) ⊆ G, suppose σ ∈ Gal(K/F), and for the sake of contradiction, let’s suppose σ ∉ G. Let S = G ∪ {σ}.
n + 1 = |S| [If σ_{1},···, σ_{n+1}: K → L are mutually distinct field homomorphisms, then [K : K^{S}] ≥ n+1] ≤ [K : K^{S}] ≤ [Notice that F ⊆ K^{S} ⊆ K, K^{S} = {a ∈ K| σ_{1}(a) = ··· = σ_{n+1}(a)} ⊆ K] [K : F] = n ⊥
Corollary. Let K be a field. There cannot be two different finite groups of automorphism of K with the same fixed field.
Proof.
Let’s suppose G_{1}, G_{2} two finite groups of automorphism with the same field F. Then, G_{1} = [ Let K be any field and let G be a finite group of automorphisms of K. Suppose F is the fixed field of G. Then G = Gal(K/F) = Gal(K/K^{G}).] Gal(K/F) = G_{2} ∎
Definition. Let K/F be a finite extension of fields. Then, K/F is called a Galois extension if the fixed field by the automorphism (Galois) group Aut(K/F), the group of all F-automorphism of K, is precisely the base field F.
Notice that F ⊆ K^{Gal(K/F)} ⊆ K. In particular, F ⊆ K^{Gal(K/F)}, F is always contained in the fixed field of Gal(K/F) (∀ σ ∈ Gal(K/F), σ fixes F, ∀a ∈ F, σ(a) = a ⇒ ∀a ∈ F, a ∈ K ^{Gal(K/F)}. The extension K/F is Galois if K ^{Gal(K/F)} = F.
Recall. Let K be any field, and let σ_{1}, ···, σ_{n}: K → K be distinct field automorphisms. Suppose that σ_{1}, ···, σ_{n} forms a group under composition. If F is a fixed field of _{1}, ···, σ_{n}. Then, [K: F] = n ⇒ [K:K^{G}] = |G| = 2 ⇒ [ℚ ⊆ K^{G} ⊆ ℚ(i) and [K : ℚ] = 2]K^{G} = ℚ.
We have ℚ ⊆ K^{G} ⊆ ℚ(i) and by the previous theorem (Let K be any field and let G be a finite group of automorphisms of K. Suppose F is the fixed field of G. Then G = Gal(K/F) = Gal(K/K^{G})= [K^{G} = ℚ] Gal(ℚ(i)/ ℚ) ⇒ K^{G} = K^{Gal(ℚ(i)/ ℚ)} = ℚ and the extension ℚ(i)/ℚ is Galois.
K = $ℚ(\sqrt[3]{2})$, F = ℚ. There is only one automorphism, namely the identity: K → K because $\sqrt[3]{2}$ must go to $\sqrt[3]{2}, w\sqrt[3]{2}, ~or~ w^2\sqrt[3]{2}$ and only the first root is real (K ⊆ ℝ). So Gal(K/F) = {identity}, and K^{Gal(K/F)} = K ≠ F ⇒ [By definition] $ℚ(\sqrt[3]{2})/ℚ$ is not Galois.
K = $ℚ(\sqrt[4]{2})$, F = ℚ. If α = $\sqrt[4]{2}$, then the fourth roots of 2 are ±α and ±iα, but there are only two automorphisms because α must go to ±α since ±iα are not real (K ⊆ ℝ). So Gal(K/F) = {identity, σ}, where σ: α → -α and |K^{Gal(K/F)}| = |Gal(K/F)| = 2 ⇒ since [$ℚ(\sqrt[4]{2}):ℚ$]=4, F ≠ ℚ, K/ℚ is not Galois.
$K = \mathbb{Q}(\sqrt{3},\sqrt{5})$, $σ: \sqrt{3} → -\sqrt{3}, τ: \sqrt{5} → -\sqrt{5}, μ = στ: \sqrt{3} → -\sqrt{3}, \sqrt{5} → -\sqrt{5}, \sqrt{15} → \sqrt{15}$. Gal($\mathbb{Q}(\sqrt{3},\sqrt{5})/\mathbb{Q}$) = {id, σ, τ, μ} ≋ ℤ/ℤ_{2}xℤ/ℤ_{2}. Then, G = Gal(K/F) = [K/F is Galois if F = K^{G}] Gal(K/K^{G}) = Gal($\mathbb{Q}(\sqrt{3},\sqrt{5})/ℚ$) and the extension is obviously Galois.
[K: ℚ] = 4, and K as a vector space over ℚ has a basis {$1, \sqrt{3}, \sqrt{5}, \sqrt{15}$} and dim_{ℚ}K = 4, so every element a ∈ K can be written uniquely as $a + b\sqrt{3} + c\sqrt{5} + d\sqrt{15}$. σ is defined uniquely as $a + b\sqrt{3} + c\sqrt{5} + d\sqrt{15} → a - b\sqrt{3} + c\sqrt{5} - d\sqrt{15}$, τ: $a + b\sqrt{3} + c\sqrt{5} + d\sqrt{15} → a + b\sqrt{3} - c\sqrt{5} - d\sqrt{15}$, μ: $a + b\sqrt{3} + c\sqrt{5} + d\sqrt{15} → a - b\sqrt{3} - c\sqrt{5} + d\sqrt{15}$
Why [K : ℚ] = 4? We build the tower, $\mathbb{Q} ⊆ \mathbb{Q}(\sqrt{3}) ⊆ \mathbb{Q}(\sqrt{3}, \sqrt{5})$. A) $\sqrt{3}∉\mathbb{Q}$, x^{2} -3 is irreducible in ℚ, and degree(x^{2}-3) = 2 ⇒ [$\mathbb{Q}(\sqrt{3}) : \mathbb{Q}$] = 2. B) $\sqrt{5}∉\mathbb{Q}(\sqrt{3})$, x^{2} -5 is irreducible in $\mathbb{Q}(\sqrt{3})$, and degree(x^{2}-5) = 2 ⇒ [$\mathbb{Q}(\sqrt{3}, \sqrt{5}):\mathbb{Q}(\sqrt{3})$] = 2 ⇒ C) [K : ℚ] = [$\mathbb{Q}(\sqrt{3}, \sqrt{5}):\mathbb{Q}(\sqrt{3})$][$\mathbb{Q}(\sqrt{3}) : \mathbb{Q}$] = 2·2 = 4
Futhermore, K has 3 subfields L such that [K : L] = 2 (⇒ [L : ℚ] = 2), namely $ℚ(\sqrt{2}),ℚ(\sqrt{3}),ℚ(\sqrt{6})$, and $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2} +\sqrt{3})$
Following the same reasoning, [K : K^{G}] = |G| = 4. ℚ ⊆ K^{Gal(K,ℚ)} ⊆ K ⇒ [K^{Gal(K, ℚ)} : ℚ] = 1 ⇒ K^{Gal(K, ℚ)} = ℚ ⇒ K/Q is Galois.
To sum up. Let K be any field. [K : K^{S}] ≥ |S|, [K : K^{G}] = |G|, Gal(K, K^{G}) = G. F ⊆ K^{Gal(K/F)} ⊆ K. K/F is Galois if F = K^{Gal(K/F)}
Theorem. A finite extension K/F is Galois iff the order of the Galois group equals the degree of the extension, i.e., |Gal(K/F)| = [K : F]
Proof:
Let’s suppose K/F is Galois ⇒ F = K^{Gal(K/F)} ⇒ [K : K^{Gal(K/F)}] = [K : F] = [Let K be any field, and let σ_{1}, ···, σ_{n}: K → K be distinct field automorphisms. Suppose that σ_{1}, ···, σ_{n} forms a group under composition. If F is a fixed field of σ_{1}, ···, σ_{n}. Then, [K: F] = n] = |Gal(K/F)|∎
Let’s assume that |Gal(K/F)| = [K : F], let L = K^{Gal(K/F)}. We claim that L = F.
[K:L] = [Let K be any field, and let σ_{1}, ···, σ_{n}: K → K be distinct field automorphisms. Suppose that σ_{1}, ···, σ_{n} forms a group under composition. If F is a fixed field of σ_{1}, ···, σ_{n}. Then, [K: F] = n] |Gal(K/F)| = [By our hypothesis] [K : F] ≥ [K : L] because L = K^{Gal(K/F)} is an intermediate field ⇒ [L : F] = 1 ⇒ L = F ⇒ F = K^{Gal(K/F)} ∎