    # Galois extensions and groups II

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/KG). 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 σ1, ···, σn: K → K be distinct field automorphisms. Suppose that σ1, ···, σn forms a group under composition. If F is a fixed field of σ1, ···, σn. 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 σ1,···, σn+1: K → L are mutually distinct field homomorphisms, then [K : KS] ≥ n+1] ≤ [K : KS] ≤ [Notice that F ⊆ KS ⊆ K, KS = {a ∈ K| σ1(a) = ··· = σn+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 G1, G2 two finite groups of automorphism with the same field F. Then, G1 = [ 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/KG).] Gal(K/F) = G2

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 ⊆ KGal(K/F) ⊆ K. In particular, F ⊆ KGal(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.

# Examples

• K = ℚ(i), F = ℚ, Is K/F Galois? G = {id, σ} where σ: K → K, σ(i) = -i. G is indeed a group.

Recall. Let K be any field, and let σ1, ···, σn: K → K be distinct field automorphisms. Suppose that σ1, ···, σn forms a group under composition. If F is a fixed field of 1, ···, σn. Then, [K: F] = n ⇒ [K:KG] = |G| = 2 ⇒ [ℚ ⊆ KG ⊆ ℚ(i) and [K : ℚ] = 2]KG = ℚ.

We have ℚ ⊆ KG ⊆ 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/KG)= [KG = ℚ] Gal(ℚ(i)/ ℚ) ⇒ KG = KGal(ℚ(i)/ ℚ) = ℚ and the extension ℚ(i)/ℚ is Galois.

• K = $ℚ(\sqrt{2})$, F = ℚ. There is only one automorphism, namely the identity: K → K because $\sqrt{2}$ must go to $\sqrt{2}, w\sqrt{2}, ~or~ w^2\sqrt{2}$ and only the first root is real (K ⊆ ℝ). So Gal(K/F) = {identity}, and KGal(K/F) = K ≠ F ⇒ [By definition] $ℚ(\sqrt{2})/ℚ$ is not Galois.

• K = $ℚ(\sqrt{2})$, F = ℚ. If α = $\sqrt{2}$, 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 |KGal(K/F)| = |Gal(K/F)| = 2 ⇒ since [$ℚ(\sqrt{2}):ℚ$]=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/ℤ2xℤ/ℤ2. Then, G = Gal(K/F) = [K/F is Galois if F = KG] Gal(K/KG) = 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 dimK = 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) $\sqrt{3}∉\mathbb{Q}$, x2 -3 is irreducible in ℚ, and degree(x2-3) = 2 ⇒ [$\mathbb{Q}(\sqrt{3}) : \mathbb{Q}$] = 2. B) $\sqrt{5}∉\mathbb{Q}(\sqrt{3})$, x2 -5 is irreducible in $\mathbb{Q}(\sqrt{3})$, and degree(x2-5) = 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

• A completely similar argument as a previous example shows that $K = \mathbb{Q}(\sqrt{2},\sqrt{3})$, $σ: \sqrt{2} → -\sqrt{2}, τ: \sqrt{3} → -\sqrt{3}, μ = στ: \sqrt{2} → -\sqrt{2}, \sqrt{3} → -\sqrt{3}, \sqrt{6} → \sqrt{6}$. Gal($\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$) = {id, σ, τ, μ} ≋ ℤ/ℤ2xℤ/ℤ2. Then, G = Gal(K/F) = [K/F is Galois if F = KG] Gal(K/KG) = Gal($\mathbb{Q}(\sqrt{2},\sqrt{3})/ℚ$) and the extension is obviously Galois.

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 $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2} +\sqrt{3})$

• K = $ℚ(\sqrt{2}, i)$, F = ℚ. $Gal(ℚ(\sqrt{2}, i)/\mathbb{Q})$ = {1, σ1, σ2, σ3} is indeed a group where σ3 = σ1σ2. |Gal(K/ℚ)| = 4.

Following the same reasoning, [K : KG] = |G| = 4. ℚ ⊆ KGal(K,ℚ) ⊆ K ⇒ [KGal(K, ℚ) : ℚ] = 1 ⇒ KGal(K, ℚ) = ℚ ⇒ K/Q is Galois.

To sum up. Let K be any field. [K : KS] ≥ |S|, [K : KG] = |G|, Gal(K, KG) = G. F ⊆ KGal(K/F) ⊆ K. K/F is Galois if F = KGal(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 = KGal(K/F) ⇒ [K : KGal(K/F)] = [K : F] = [Let K be any field, and let σ1, ···, σn: K → K be distinct field automorphisms. Suppose that σ1, ···, σn forms a group under composition. If F is a fixed field of σ1, ···, σn. Then, [K: F] = n] = |Gal(K/F)|∎

Let’s assume that |Gal(K/F)| = [K : F], let L = KGal(K/F). We claim that L = F.

[K:L] = [Let K be any field, and let σ1, ···, σn: K → K be distinct field automorphisms. Suppose that σ1, ···, σn forms a group under composition. If F is a fixed field of σ1, ···, σn. Then, [K: F] = n] |Gal(K/F)| = [By our hypothesis] [K : F] ≥ [K : L] because L = KGal(K/F) is an intermediate field ⇒ [L : F] = 1 ⇒ L = F ⇒ F = KGal(K/F)

# Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
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