Theorem. Let K/F be a finite extension. The following statements are equivalent:
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 → K^{H}, its inverse is defined by L → Gal(K/L). H_{1} ⊇ H_{2} ⇒ K^{H1} ⊆ K^{H2} and L_{1} ⊆ L_{2} ⇒ Gal(K/L_{1}) ⊇ Gal(K/L_{2}). Futhermore, it satisfies the following equality |H| = [K : K^{H}] and [G : H] = [K^{H} : 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.
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.
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.
Roots of x^{4}+4? x^{4} + 4 = (x^{2}+2x+2)(x^{2}-2x+2)
The roots of (x^{2}+2x+2) is $\frac{-2±\sqrt{4-8}}{2}=-1±i$, and the roots of (x^{2}-2x+2) is $\frac{2±\sqrt{4-8}}{2}=1±i$, so the roots of x^{4} + 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).
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 x^{3} -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).
There are only 4 homomorphisms:
[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 : 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.
Notice that: σ_{1}^{2} = σ_{2}^{2} = σ_{3}^{2} = identity. There are only two groups of order 4 up to isomorphism, namely, C_{4}, the cyclic group of order 4, and K_{4}, the Klein group. The Klein group is a group in which each element is self-inverse, Gal(K/F) ≋ K_{4} ≋ ℤ/2ℤ x ℤ/2ℤ.
[K : ℚ] = 4 = |Gal(K/Q)|, it is normal because it is a splitting field of (x^{2}-2)(x^{2}+1) and it is generated by the roots of this polynomial.
Reclaim: $ζ_5 = cos\frac{2π}{5}+isin\frac{2π}{5}$
Let’s prove that $\sqrt{5} ∈ K$. 1 + ξ_{5} + ξ_{5}^{2} + ξ_{5}^{3} + ξ_{5}^{4} = 0. Let S = ξ_{5} + ξ_{5}^{4} ⇒ S^{2} = ξ_{5}^{2} + 2ξ_{5}^{5} +ξ_{5}^{3} = 2 + ξ_{5}^{2} + ξ_{5}^{3} = [1 + ξ_{5} + ξ_{5}^{2} + ξ_{5}^{3} + ξ_{5}^{4} = 0] 1 - (ξ_{5} + ξ_{5}^{4}) = 1 - S.
Therefore, S^{2} = 1 -S ⇒ S^{2} + S = 1 ⇒ (2S + 1)^{2} = 4(S^{2} + S) + 1 = [S^{2} + S = 1] 5. Therefore, 5 is a square in K, so K contains $\sqrt{5}$
K/ℚ is Galois because:
Futhermore, notice that σ_{2}(ξ_{5}) = ξ_{5}^{2}, σ_{2}^{2}(ξ_{5}) = ξ_{5}^{4} (i.e. σ_{2}^{2} = σ_{4}), σ_{2}^{3}(ξ_{5}) = σ_{2}(ξ_{5}^{4}) = ξ_{5}^{8} = ξ_{5}^{3} (i.e. σ_{2}^{3} = σ_{3}), and σ_{2}^{4}(ξ_{5}) = σ_{2}(ξ_{5}^{3}) = ξ_{5}^{6} = ξ_{5} (i.e. σ_{2}^{4} = σ_{1}). Therefore, σ_{2} 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 {σ_{1}, σ_{4}} ≋ ℤ/4ℤ, σ_{1} = id, σ_{4}(ξ_{5}) = ξ_{5}^{4}, σ_{4}^{2}(ξ_{5}) = σ_{4}(ξ_{5}^{4}) = ξ_{5}^{16} = ξ_{5} (i.e. σ_{4}^{2} = σ_{1 })
x → x+1, f(x+1) = (x+1)^{4} + 1 = x^{4} + 4x^{3} + 6x^{2} +4x +2 ⇒ [By Eisenstein criterion. If ∃p prime (=2): p|a_{i} 0≤ i ≤ n, p does not divide a_{n}(=1), p^{2}(4) ɫ a_{0}(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
ζ_{8} is the primitive eighth root of unity ⇒ ζ_{8}^{4} ≠ 1 ⇒ [x^{8} -1 = (x^{4} -1)(x^{4}+1) = (x^{4} -1)f(x)] f(ζ_{8}) = 0, so the irreducible polynomial of ζ_{8} over ℚ is x^{4}+1, then [K=ℚ(ζ_{8}) : 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(ζ_{8}).
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:
|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 x^{3} -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 x^{4} -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 $\mathbb{Q}(\sqrt[4]{5}, i)/Q$.