# Recall

Theorem. Let K/F be a finite extension. The following statements are equivalent:

1. K/F is Galois.
2. F is the fixed field of Aut(K/F), i.e., F = KGal(K/F)
3. 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]
4. K/F is a normal, finite, and separable extension.
5. K is the splitting field of a separable polynomial f ∈ F[x] over F.
6. Let Char(F)=0 or F be finite (or more generally F is perfect). Then, a finite extension K/F is Galois if and only if K/F is normal.

Fundamental Theorem of Galois Theorem. Let K/F be a finite separable extension, i.e., a Galois extension. Let G = Gal(K/F) be the Galois group. Then, the following statements holds, (1) There is an inclusion-reversing bijective map or correspondence from subgroups of G and intermediate fields of K/F, given by H → KH, its inverse is defined by L → Gal(K/L). H1 ⊇ H2 ⇒ KH1 ⊆ KH2 and L1 ⊆ L2 ⇒ Gal(K/L1) ⊇ Gal(K/L2). Futhermore, it satisfies the following equality |H| = [K : KH] and [G : H] = [KH : F]

(2) Suppose an intermediate field L (F ⊆ L ⊆ K) corresponds to the subgroup H under the Galois correspondence. K/L is always normal (hence Galois). L is Galois over F if and only if H = Gal(K/L) is a normal subgroup of G, H = Gal(K/L) ◁ G In this case, the Galois group of L/F is isomorphic to the quotient group G/H, i.e., Gal(L/F) ≋ G/Gal(K/L)

Mathematics consists of proving the most obvious thing in the least obvious way, George Pólya.

Let’s find the Galois groups and determine if the extensions are Galois or normal. In some examples, we will determine the intermediate fields.

• 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.

A field extension of degree 2 is a normal extension. Let K be any extension of a basis field of characteristic 0 (e.g., ℚ) of degree 2. Then, Gal(K/Q) = ℤ/2ℤ, K/Q is both normal and Galois. Therefore, ℚ(i)/ℚ is both normal and Galois.

• Let K be the splitting field of x4 + 4 over ℚ, F = ℚ ⇒ K/ℚ normal because it is the splitting field of a polynomial and separable (char(ℚ)=0) ⇒ K/ℚ is Galois

Roots of x4+4? x4 + 4 = (x2+2x+2)(x2-2x+2)

The roots of (x2+2x+2) is $\frac{-2±\sqrt{4-8}}{2}=-1±i$, and the roots of (x2-2x+2) is $\frac{2±\sqrt{4-8}}{2}=1±i$, so the roots of x4 + 4 = {1+i, 1-i, -1+i, -1-i} ∈ K ⇒ [1+i+-1+i ∈ K, 2i ∈K] i ∈K ⇒ K = ℚ(i), where {1+i, 1-i, -1+i, -1-i} ∈ K = ℚ(i) and [K : ℚ(i)] = 2, Gal(K/ℚ)≋ℤ/2ℤ, and the only intermediate fields are K and ℚ (the trivial ones).

• 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 KGal(K/F) = K ≠ F ⇒ [By definition] $ℚ(\sqrt[3]{2})/ℚ$ is not Galois, or equivalently, [A finite extension K/F is Galois iff |Gal(K/F)| = [K : F]] [K : ℚ] = 3, but |Gal(K/ℚ)| = 1.

It is not normal either, [If f ∈ F[x] is an irreducible polynomial with at least one root in K, then f splits (that is, it splits as a product of linear factors) completely in K[x]] because x3 -2 ∈ ℚ[x] is irreducible, it has a root in K, namely $\sqrt[3]{2}$, but does not split completely (it does not have the non-real cubic roots of 2).

• 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. [K : ℚ] = 4.

There are only 4 homomorphisms:

1. identity, $i → i, \sqrt{2}→\sqrt{2}$
2. σ1, $i → -i, \sqrt{2}→\sqrt{2}$
3. σ2, $i → i, \sqrt{2}→-\sqrt{2}$
4. σ3, $i → -i, \sqrt{2}→-\sqrt{2}$

[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]] It is a Galois group because |Gal(K/F)| = [K : F] = 4. Futhermore, [K : KG] = |G| = 4. ℚ ⊆ KGal(K,ℚ) ⊆ K ⇒ [KGal(K, ℚ) : ℚ] = 1 ⇒ KGal(K, ℚ) = ℚ ⇒ K/Q is Galois.

Notice that: σ12 = σ22 = σ32 = identity. There are only two groups of order 4 up to isomorphism, namely, C4, the cyclic group of order 4, and K4, the Klein group. The Klein group is a group in which each element is self-inverse, Gal(K/F) ≋ K4 ≋ ℤ/2ℤ x ℤ/2ℤ.

[K : ℚ] = 4 = |Gal(K/Q)|, it is normal because it is a splitting field of (x2-2)(x2+1) and it is generated by the roots of this polynomial.

• K = ℚ(ζ5), F = ℚ where ζ5 is the primitive fifth root of unity. Let’s find the irreducible polynomial of ξ5 over ℚ. ξ55 - 1 = 0, so it satisfies x5 -1 = (x -1)(x4 + x3 + x2 + x + 1), and since ξ5 ≠ 1, ξ is the root of an irreducible (cyclotomic) polynomial of degree 4, namely x4 + x3 + x2 + x + 1 ⇒ [K : ℚ] = 4
The splitting field of x5 -1 over ℚ is K = ℚ(ξ) because x5 - 1 = (x -1)(x4 + x3 + x2 + x + 1), and the roots of x5 - 1 are 1, ξ5, ξ52, ξ53, and ξ54.

Reclaim: $ζ_5 = cos\frac{2π}{5}+isin\frac{2π}{5}$

Let’s prove that $\sqrt{5} ∈ K$. 1 + ξ5 + ξ52 + ξ53 + ξ54 = 0. Let S = ξ5 + ξ54 ⇒ S2 = ξ52 + 2ξ5553 = 2 + ξ52 + ξ53 = [1 + ξ5 + ξ52 + ξ53 + ξ54 = 0] 1 - (ξ5 + ξ54) = 1 - S.

Therefore, S2 = 1 -S ⇒ S2 + S = 1 ⇒ (2S + 1)2 = 4(S2 + S) + 1 = [S2 + S = 1] 5. Therefore, 5 is a square in K, so K contains $\sqrt{5}$

K/ℚ is Galois because:

1. First argument: K is the splitting field of x4 + x3 + x2 + x + 1 (K/ℚ is normal), char(ℚ) = 0 (in characteristic zero all irreducible polynomials are separable, K/ℚ is separable).
2. Second argument: σi send ξ5 to ξ5i. Then, σ1, ···, σ4 are all distinct automorphisms of K. |Gal(K/ℚ)| = 4 = [K : ℚ].

Futhermore, notice that σ25) = ξ52, σ225) = ξ54 (i.e. σ22 = σ4), σ235) = σ254) = ξ58 = ξ53 (i.e. σ23 = σ3), and σ245) = σ253) = ξ56 = ξ5 (i.e. σ24 = σ1). Therefore, σ2 is of order 4, and [There are only two groups of order 4 up to isomorphism: ℤ/4ℤ (cyclic) or Gal(K/Q) ≋ ℤ/2ℤ x ℤ/2ℤ] Gal(K/ℚ) ≋ ℤ/4ℤ.

Besides, K/ℚ has exactly one intermediate field L such that [K : L] = 2 because Gal(K/Q) ≋ ℤ/4ℤ has exactly one subgroup of order 2, that is L = ℚ($\sqrt{5}$). The subgroup associated is {σ1, σ4} ≋ ℤ/4ℤ, σ1 = id, σ45) = ξ54, σ425) = σ454) = ξ516 = ξ5 (i.e. σ42 = σ1 )

• K = ℚ(ζ8), F = ℚ where ζ8 is the primitive eighth root of unity. Let’s find the irreducible polynomial of ζ8 over ℚ. It satisfies x8 -1 = (x4 -1)(x4+1). Claim: f(x) = x4 + 1 is irreducible.

x → x+1, f(x+1) = (x+1)4 + 1 = x4 + 4x3 + 6x2 +4x +2 ⇒ [By Eisenstein criterion. If ∃p prime (=2): p|ai 0≤ i ≤ n, p does not divide an(=1), p2(4) ɫ a0(2) ⇒ p is irreducible over ℚ] f(x+1) is irreducible over ℚ.

p(x) is irreducible ↭ p(x+a) is irreducible. Suppose p(x) is reducible, p(x) = u(x)v(x)··· ⇒ p(x+a) = u(x+a)v(x+a)···, and therefore p(x) is reducible

ζ8 is the primitive eighth root of unity ⇒ ζ84 ≠ 1 ⇒ [x8 -1 = (x4 -1)(x4+1) = (x4 -1)f(x)] f(ζ8) = 0, so the irreducible polynomial of ζ8 over ℚ is x4+1, then [K=ℚ(ζ8) : ℚ] = 4

Reclaim: $ζ_8 = cos\frac{2π}{8}+isin\frac{2π}{8}=cos\frac{π}{4}+isin\frac{π}{4}$ The primitive 8th roots of unity are: $±\frac{1}{\sqrt{2}}±\frac{i}{\sqrt{2}}$

Let’s prove that K = $\mathbb{Q(i, \sqrt{2})}$

$(ζ_8^2)^{4} = ζ_8^8 = 1 ⇒ ζ_8^2= i~ or~ ζ_8^2=-i$ ⇒ i ∈ K. Besides, $\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}∈K, \frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}∈K ⇒\frac{2}{\sqrt{2}}=\sqrt{2}∈ K$. Therefore, $\mathbb{Q(i, \sqrt{2})}$ ⊆ K = ℚ(ζ8).

Therefore, [$\mathbb{K} = ℚ(ζ_8) : \mathbb{Q(i, \sqrt{2})}$ ] = 1 ⇒ $ℚ(ζ_8) = \mathbb{Q(i, \sqrt{2})}$

K/ℚ is Galois because:

1. First argument: K is the splitting field of x4 + 1 (K/ℚ is normal), char(ℚ) = 0 (in characteristic zero all irreducible polynomials are separable, K/ℚ is separable).
2. Second argument: |Gal(K/ℚ)| = [See the exercise before i→±i, $\sqrt{2}→±\sqrt{2}$] [K : ℚ] = 4

|Gal(K/ℚ)| = 4 ⇒ [There are only two groups of order 4 up to isomorphism] Gal(K/Q) ≋ ℤ/4ℤ (cyclic) or Gal(K/Q) ≋ ℤ/2ℤ x ℤ/2ℤ

If [K : ℚ] = 4 ⇒ Gal(K/Q) ≋ ℤ/4ℤ, then K/ℚ has exactly one intermediate field L such that [K : L] = 2 because Gal(K/Q) ≋ ℤ/4ℤ has exactly one subgroup of order 2, but K/ℚ has more than one such intermediate field, namely $ℚ(\sqrt{2})~ (⊆ ℝ) and~ ℚ(i)$ (⊆ ℂ). Therefore, Gal(K/Q) ≋ ℤ/2ℤ x ℤ/2ℤ. ℤ/2ℤ x ℤ/2ℤ has three subgroups of order 2, and five total subgroups (the previous one plus the trivial subgroup and the group itself), and five intermediate fields.

• is $\mathbb{Q}(\sqrt[3]{5}, \sqrt{7})/Q$ a normal extension? No, the polynomial x3 -5 is irreducible over ℚ (Eisenstein’s criterion) and has a real root in K = $\mathbb{Q}(\sqrt[3]{5}, \sqrt{7})$, namely $\sqrt[3]{5}$, but does not split completely in K because the other roots of f are non-real and K ⊆ ℝ.

• is K = $\mathbb{Q}(\sqrt[4]{5})/Q$ a normal extension? It is not a normal extension, because the polynomial x4 -5 has a couple of roots in K, namely $±\sqrt[4]{5}$, but not all of them (it does not include $\sqrt[4]{5}i$). A normal extension that contains this field extension is $\mathbb{Q}(\sqrt[4]{5}, i)/Q$.