Theorem. Let K/F be a finite extension. The following statements are equivalent:
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2 = [2 ∈ ℚ, Φ acts as the identity on ℚ] Φ(2) = $Φ (\sqrt{2}\sqrt{2})$= [Φ is an automorphism] $(Φ(\sqrt{2}))^{2}⇒Φ(\sqrt{2})=±\sqrt{2}$ ⇒ Gal($ℚ(\sqrt{2})/ℚ$) has only two elements, namely the identity and the mapping Φ, $Φ(a + b(\sqrt{2}))=a - b(\sqrt{2})$. Therefore, Gal($ℚ(\sqrt{2})/ℚ$) = {id, Φ} ≋ ℤ/2ℤ, a cyclic group of order 2.
f(x) = x^{2} -2 is separable over ℚ because the roots of the polynomial f(x) are $±\sqrt{2}$. Since all the roots of f(x) are distinct, f(x) is separable. Futhermore, since the roots of f(x) are $±\sqrt{2}$, the splitting field of f(x), which is separable, is $ℚ(\sqrt{2})$, therefore the extension is Galois.
2 = [2 ∈ ℚ, Φ acts as the identity on ℚ] Φ(2) = $Φ (\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2})$ = [Φ is an automorphism] $(Φ(\sqrt[3]{2}))^{3}⇒Φ(\sqrt[3]{2})=\sqrt[3]{2}$ (Another way of thinking about this is that $\sqrt[3]{2}$, a root of x^{3}-2, must go to $\sqrt[3]{2}, \sqrt[3]{2}w$ or $\sqrt[3]{2}w^2$, but only the first root is real and $ℚ(\sqrt[3]{2})$ ⊆ ℝ). Therefore, Φ is the identity automorphism, and Gal($ℚ(\sqrt{2})/ℚ$) has only one element, namely the identity, and $ℚ(\sqrt[3]{2})/ℚ$ is not Galois ([K : ℚ] = 3, but |Gal(K/ℚ)| = 1).
2 = Φ(2) = $Φ((\sqrt[4]{2})^4) ⇒ Φ(\sqrt[4]{2})$ is a fourth root of 2, then there are four automorphisms of $ℚ(\sqrt[4]{2}, i)$ fixing ℚ(i). Let α automorphism: α(i) = i, $α(\sqrt[4]{2})=i(\sqrt[4]{2}) ⇒ α ∈ Gal(ℚ(\sqrt[4]{2},i)/ℚ(i))$ and [ $α^2(\sqrt[4]{2})=α(i\sqrt[4]{2})=-\sqrt[4]{2}, α^3(\sqrt[4]{2})=α(-\sqrt[4]{2})=-i\sqrt[4]{2}, α^4=α(-i\sqrt[4]{2})=\sqrt[4]{2}$] α has order 4 ⇒ $Gal(ℚ(\sqrt[4]{2},i)/ℚ(i))$ is a cyclic group of order 4, Figure 1.
$α^2(\sqrt{2})=α^2(\sqrt[4]{2}\sqrt[4]{2})=α^2(\sqrt[4]{2})α^2(\sqrt[4]{2})=-\sqrt[4]{2}\sqrt[4]{2}=\sqrt{2}$ ⇒ The fixed field of {e, α^{2}} is $ℚ(\sqrt{2},i)$, e is the identity automorphism.
f(x) = $x^4 - 2 = (x^2-\sqrt{2})(x^2+\sqrt{2}) = (x - \sqrt[4]{2})(x + \sqrt[4]{2})(x - i\sqrt[4]{2})(x + i\sqrt[4]{2})$ ⇒ The splitting field of f is K = $\mathbb{Q}(i, \sqrt[4]{2})$. K/F is both normal and separable, and therefore Galois because F = ℚ (char(ℚ) = 0).
[K : ℚ] = [ K : ℚ(i) ] [ ℚ(i) : ℚ] = 4 · 2 = 8 = [K : $ℚ(\sqrt[4]{2})$][$ℚ(\sqrt[4]{2})$ : ℚ] = 2 · 4 = 8.
[ K : $ℚ(\sqrt[4]{2})$] = 2 because the minimum polynomial of i over $ℚ(\sqrt[4]{2})$ is x^{2} + 1. [$ℚ(\sqrt[4]{2})$ : ℚ] = 4 because f(x) is the minimum polynomial of $\sqrt[4]{2}$ over ℚ.
Let’s calculate the elements of Gal(K/ℚ):
id: i → i, $\sqrt[4]{2} → \sqrt[4]{2}$
σ: i → i, $\sqrt[4]{2} → i\sqrt[4]{2}$
τ: i → -i, $\sqrt[4]{2} → \sqrt[4]{2}$
σ^{2}: i → i, $\sqrt[4]{2} → -\sqrt[4]{2}$
σ^{3}: i → i, $\sqrt[4]{2} → -i\sqrt[4]{2}$
στ: i → -i, $\sqrt[4]{2} → i\sqrt[4]{2}$
σ^{2}τ: i → -i, $\sqrt[4]{2} → -\sqrt[4]{2}$
σ^{3}τ: i → -i, $\sqrt[4]{2} → -i\sqrt[4]{2}$
Gal(K/ℚ) ≋ D_{8}, and the subgroups are: a) the trivial ones: G and {id}; b) Proper subgroups of order 4: {id, σ, σ^{2}, σ^{3}}, {1, σ^{2}, τ, σ^{2}τ}, and {1, σ^{2}, στ, σ^{3}τ}; c) Proper subgroups of order 2: {1, σ^{2}}, {1, τ}, {1, στ}, {1, σ^{2}τ}, {1, σ^{3}τ}
The intermediate subfields of K/ℚ corresponding to {id, σ, σ^{2}, σ^{3}}, {1, σ^{2}, τ, σ^{2}τ}, and {1, σ^{2}, στ, σ^{3}τ} are ℚ(i), $\mathbb{Q}(\sqrt{2})$, and $\mathbb{Q}(i\sqrt{2})$ respectively.
The fixed field of {e, α} is $ℚ(\sqrt{5})$. The fixed field of {e, β} is $ℚ(\sqrt{3}).$ The fixed field of {e, αβ} is $ℚ(\sqrt{3}\sqrt{5})=ℚ(\sqrt{15})$.
σ ∈ $Gal(ℚ(\sqrt{3})/ℚ)$, σ defined as $σ(a +b\sqrt{3})=a -b\sqrt{3}$ and extended in $Gal(ℚ(\sqrt{3}, \sqrt{5})/ℚ)$ by saying that σ is the unique linear extension of the original map σ into $Gal(ℚ(\sqrt{3}, \sqrt{5})/ℚ)$ which fixes $ℚ(\sqrt{5})$: σ$[(a + b\sqrt{5})+(c + d\sqrt{5})\sqrt{3})] = (a + b\sqrt{5})-(c + d\sqrt{5})\sqrt{3}) = a +b\sqrt{5} -c\sqrt{3} -d\sqrt{15}$. Notice that [$Gal(ℚ(\sqrt{3}, \sqrt{5}):Gal(ℚ(\sqrt{5})$] = 2
Mutatis Mutandis, τ ∈ $Gal(ℚ(\sqrt{5})/ℚ)$, τ defined as $τ(a +b\sqrt{5})=a -b\sqrt{5}$ and extended in $Gal(ℚ(\sqrt{3}, \sqrt{5})/ℚ)$ by saying that τ is the unique linear extension of the original map τ into $Gal(ℚ(\sqrt{3}, \sqrt{5})/ℚ)$ which fixes $ℚ(\sqrt{3})$: σ$[(a + b\sqrt{3})+(c + d\sqrt{3})\sqrt{5})] = (a + b\sqrt{3})-(c + d\sqrt{3})\sqrt{5})$
So we have id, σ, τ ∈ $Gal(ℚ(\sqrt{3}, \sqrt{5})/ℚ)$, their composition στ ∈ $Gal(ℚ(\sqrt{3}, \sqrt{5})/ℚ)$ (∀a ∈ ℚ: στ(a) = σ(a) = a, so it fixes all rational numbers). Futhermore, $Gal(ℚ(\sqrt{3}, \sqrt{5})/ℚ)$ = {1, σ, τ, στ} ≋ ℤ_{2} ⊕ ℤ_{2} ≋ $\mathbb{V}_4$
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
There are five subgroups of G, namely {1}, G itself, H_{1} = {1, σ_{1}}, H_{2} = {1, σ_{2}}, and H_{3} = {1, σ_{3}}
Let’s find the fixed fields of these subgroups. K^{{1}} = K, K^{G} = ℚ, L_{1} = K^{H1} = ℚ(i), L_{2} = K^{H2} = $\mathbb{Q}(\sqrt{2})$, L_{3} = K^{H3} = $\mathbb{Q}(i\sqrt{2})$. The diagram below shows both all the intermediate fields of the extension K/F and the subgroups of the Galois group, the main theorem of Galois demonstrates that there is a correspondence between the intermediate fields of the extension K/F and the Galois subgroups.
K/F Galois, K/L_{1}, K/L_{2}, K/L_{3} are Galois, L_{1}/F, L_{2}/F, and L_{3}/F are Galois, too.
Futhermore, αβ^{2} = βα, β^{2}α = αβ, αβ ≠ βα (not Abelian), β^{3} = α^{2} = id, αβα^{-1} = β^{-1}. Besides, |Gal(ℚ(w, $\sqrt[3]{2}$) : ℚ)| = |Gal(ℚ(w, $\sqrt[3]{2}): ℚ(\sqrt[3]{2})$|·|$ℚ(\sqrt[3]{2}):ℚ$| = [w is a complex number, w ∉ $ℚ(\sqrt[3]{2})$, the irreducible polynomial is w^{2}+w+1] 2·3 = 6.
[K^{H} : F] = r/s ⇒ |K^{H}| = p^{r/s} ⇒ [Structure theorem of finite fields] A field of order p^{k} where k=r/s exists, K^{H} = $\mathbb{F_{p^k}},~ and~ \mathbb{F_{p^k}}⊆\mathbb{F_{p^r}}$ because k | r.
Any subfield of K must be of the form $\mathbb{F_{p^n}}$ where n | r ↭ it corresponds to a subgroup of G (cyclic) ≋ ℤ/rℤ. Futhermore, since G is Abelian ⇒ [Every subgroup of an Abelian group is normal] every subgroup is normal, and [Fundamental Theorem of Galois Theorem] all extensions of finite fields are Galois.
Definition. A finite extension K/F is called cyclic if the extension is Galois (K/F is Galois) and its Galois Gal(K/F) is cyclic,e.g., $\mathbb{ℚ}(\sqrt{2})/\mathbb{Q}$ is cyclic, but $\mathbb{ℚ}(\sqrt[4]{2})/\mathbb{Q}$ is neither Galois, nor cyclic, and $\mathbb{ℚ}(w,\sqrt[3]{2}/\mathbb{Q})$ is Galois, but not cyclic.
Theorem. All extensions of finite fields are cyclic.
Proof.
Let Char(F) = Char(K) = p. We know, $K/\mathbb{F_p}$ is Galois, $Gal(K/\mathbb{F_p})$ is a cyclic group, G ≋ ℤ/rℤ (the Galois group has a generator σ: a → a^{p}), so $K/\mathbb{F_p}$ is a cyclic extension.
K/F is Galois and Gal(K/F) ≤ Gal(K/$\mathbb{F_p}$), Gal(K/$\mathbb{F_p}$) is cyclic ⇒ [A subgroup of a cyclic group is cyclic, too] Gal(K/F) is cyclic ∎