    # 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.

The saddest aspect of life right now is that science gathers knowledge faster than society gathers wisdom, Isaac Asimov # Motivation

• Consider the extension $ℚ(\sqrt{2})$ of ℚ. $ℚ(\sqrt{2})$ = {a + b$\sqrt{2}$|a, b ∈ ℚ} and since any automorphism of a field containing ℚ must act as the identity on ℚ ⇒ An automorphism Φ of $ℚ(\sqrt{2})$ is completely determined by Φ($\sqrt{2}$ ).

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) = x2 -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.

• Let study the extension $ℚ(\sqrt{2})$ of ℚ. An automorphism Φ of $ℚ(\sqrt{2})$ is completely determined by $Φ(\sqrt{2})$, but $ℚ(\sqrt{2})$ ⊆ ℝ.

2 = [2 ∈ ℚ, Φ acts as the identity on ℚ] Φ(2) = $Φ (\sqrt{2}\sqrt{2}\sqrt{2})$ = [Φ is an automorphism] $(Φ(\sqrt{2}))^{3}⇒Φ(\sqrt{2})=\sqrt{2}$ (Another way of thinking about this is that $\sqrt{2}$, a root of x3-2, must go to $\sqrt{2}, \sqrt{2}w$ or $\sqrt{2}w^2$, but only the first root is real and $ℚ(\sqrt{2})$ ⊆ ℝ). Therefore, Φ is the identity automorphism, and Gal($ℚ(\sqrt{2})/ℚ$) has only one element, namely the identity, and $ℚ(\sqrt{2})/ℚ$ is not Galois ([K : ℚ] = 3, but |Gal(K/ℚ)| = 1).

• Let study the extension $ℚ(\sqrt{2}, i)$ of ℚ(i). An automorphism Φ of $ℚ(\sqrt{2}, i)$ fixing ℚ(i) is completely determined by $Φ(\sqrt{2})$

2 = Φ(2) = $Φ((\sqrt{2})^4) ⇒ Φ(\sqrt{2})$ is a fourth root of 2, then there are four automorphisms of $ℚ(\sqrt{2}, i)$ fixing ℚ(i). Let α automorphism: α(i) = i, $α(\sqrt{2})=i(\sqrt{2}) ⇒ α ∈ Gal(ℚ(\sqrt{2},i)/ℚ(i))$ and [ $α^2(\sqrt{2})=α(i\sqrt{2})=-\sqrt{2}, α^3(\sqrt{2})=α(-\sqrt{2})=-i\sqrt{2}, α^4=α(-i\sqrt{2})=\sqrt{2}$] α has order 4 ⇒ $Gal(ℚ(\sqrt{2},i)/ℚ(i))$ is a cyclic group of order 4, Figure 1.

$α^2(\sqrt{2})=α^2(\sqrt{2}\sqrt{2})=α^2(\sqrt{2})α^2(\sqrt{2})=-\sqrt{2}\sqrt{2}=\sqrt{2}$ ⇒ The fixed field of {e, α2} is $ℚ(\sqrt{2},i)$, e is the identity automorphism. • Compute the splitting field K of the polynomial f(x) = x4 -2 ∈ ℚ[x].

f(x) = $x^4 - 2 = (x^2-\sqrt{2})(x^2+\sqrt{2}) = (x - \sqrt{2})(x + \sqrt{2})(x - i\sqrt{2})(x + i\sqrt{2})$ ⇒ The splitting field of f is K = $\mathbb{Q}(i, \sqrt{2})$. 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 : $ℚ(\sqrt{2})$][$ℚ(\sqrt{2})$ : ℚ] = 2 · 4 = 8.

[ K : $ℚ(\sqrt{2})$] = 2 because the minimum polynomial of i over $ℚ(\sqrt{2})$ is x2 + 1. [$ℚ(\sqrt{2})$ : ℚ] = 4 because f(x) is the minimum polynomial of $\sqrt{2}$ over ℚ.

Let’s calculate the elements of Gal(K/ℚ):

id: i → i, $\sqrt{2} → \sqrt{2}$
σ: i → i, $\sqrt{2} → i\sqrt{2}$
τ: i → -i, $\sqrt{2} → \sqrt{2}$
σ2: i → i, $\sqrt{2} → -\sqrt{2}$
σ3: i → i, $\sqrt{2} → -i\sqrt{2}$
στ: i → -i, $\sqrt{2} → i\sqrt{2}$
σ2τ: i → -i, $\sqrt{2} → -\sqrt{2}$
σ3τ: i → -i, $\sqrt{2} → -i\sqrt{2}$

Gal(K/ℚ) ≋ D8, and the subgroups are: a) the trivial ones: G and {id}; b) Proper subgroups of order 4: {id, σ, σ2, σ3}, {1, σ2, τ, σ2τ}, and {1, σ2, στ, σ3τ}; c) Proper subgroups of order 2: {1, σ2}, {1, τ}, {1, στ}, {1, σ2τ}, {1, σ3τ}

The intermediate subfields of K/ℚ corresponding to {id, σ, σ2, σ3}, {1, σ2, τ, σ2τ}, and {1, σ2, στ, σ3τ} are ℚ(i), $\mathbb{Q}(\sqrt{2})$, and $\mathbb{Q}(i\sqrt{2})$ respectively.

• Let consider the extension $ℚ(\sqrt{3}, \sqrt{5})$ of ℚ. $ℚ(\sqrt{3}, \sqrt{5}) =$ {$a+b\sqrt{3}+c\sqrt{5}+d\sqrt{3}\sqrt{5}|~ a, b, c, d ∈ ℚ$} and any automorphism of $ℚ(\sqrt{3}, \sqrt{5})$ is determined by Φ($\sqrt{3}$) and Φ($\sqrt{5}$), and by the same reasoning as before there are four automorphisms and Gal($ℚ(\sqrt{3}, \sqrt{5})$/ℚ) ≋ ℤ2 ⊕ ℤ2, Figure 2. Then, G = Gal(K/F) and K is the splitting field of the separable polynomial (x2-3)(x2-5) over ℚ, so it is Galois (Computing Galois groups).

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 $ℚ(\sqrt{3}\sqrt{5})=ℚ(\sqrt{15})$. 1. 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

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$[(a + b\sqrt{3})+(c + d\sqrt{3})\sqrt{5})] = (a + b\sqrt{3})-(c + d\sqrt{3})\sqrt{5})$

3. 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} ≋ ℤ2 ⊕ ℤ2 ≋ $\mathbb{V}_4$

Why [K : ℚ] = 4? We build the tower, $\mathbb{Q} ⊆ \mathbb{Q}(\sqrt{3}) ⊆ \mathbb{Q}(\sqrt{3}, \sqrt{5})$. A) $\sqrt{3}∉\mathbb{Q}$, 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

• K = $\mathbb{Q}(\sqrt{2}, i), F = ℚ$. We know that (Computing Galois Groups) K/F is Galois (|Gal(K/F)| = [K : F] = 4), and Gal(K/F) = {1, σ1, σ2, σ3} ≋ ℤ/2ℤ x ℤ/2ℤ, where σ1: $i → i, \sqrt{2}→ -\sqrt{2}$, σ2: $i → -i, \sqrt{2}→\sqrt{2}$, and σ3: $i → -i, \sqrt{2}→-\sqrt{2}$

There are five subgroups of G, namely {1}, G itself, H1 = {1, σ1}, H2 = {1, σ2}, and H3 = {1, σ3}

Let’s find the fixed fields of these subgroups. K{1} = K, KG = ℚ, L1 = KH1 = ℚ(i), L2 = KH2 = $\mathbb{Q}(\sqrt{2})$, L3 = KH3 = $\mathbb{Q}(i\sqrt{2})$. 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/L1, K/L2, K/L3 are Galois, L1/F, L2/F, and L3/F are Galois, too.

• Let w = $e^{\frac{i2π}{3}}=\frac{-1}{2}+i\frac{\sqrt{3}}{2}$ and consider the extension ℚ(w, $\sqrt{2}$) of ℚ. w3 = 1, w2 = $e^{\frac{i4π}{3}}=e^{\frac{-i2π}{3}} = \frac{-1}{2}-i\frac{\sqrt{3}}{2} = -(1+w)$ ⇒ w2 + w + 1 = 0. An automorphism Φ of ℚ(w, $\sqrt{2}$) is completely determined by Φ(w) -2 choices: w and w2- and Φ($\sqrt{2}$) -the minimal polynomial of $\sqrt{2}$ is x3-2, so there are three choices: $\sqrt{2}, \sqrt{2}w, \sqrt{2}w^2$-. Therefore, there are six possibilities (see figure below).

Futhermore, αβ2 = βα, β2α = αβ, αβ ≠ βα (not Abelian), β3 = α2 = id, αβα-1 = β-1. Besides, |Gal(ℚ(w, $\sqrt{2}$) : ℚ)| = |Gal(ℚ(w, $\sqrt{2}): ℚ(\sqrt{2})$|·|$ℚ(\sqrt{2}):ℚ$| = [w is a complex number, w ∉ $ℚ(\sqrt{2})$, the irreducible polynomial is w2+w+1] 2·3 = 6.

1. βα (not Abelian), β3 = α2 = id, αβα-1 = β-1 -it gives a presentation of S3- |Gal(ℚ(w, $\sqrt{2}$)/ℚ)| = 6 ⇒ G ≋ D6 (the dihedral group of degree 3) ≋ S3 (the symmetric group).
2. We can get the same result by recalling the following theorem. Let G be a group, |G| = 2p, p prime, p > 2 ⇒ G ≈ ℤ2p or Dp, but we know that is not Abelian, αβ ≠ βα, so Gal(ℚ(w, $\sqrt{2}$)/ℚ) ≈ S3.
3. The minimal polynomial of $\sqrt{2}$ is x3 - 2 = 0, and the minimal polynomial of w is x2 + x +1, and K = ℚ(w, $\sqrt{2}$) is the splitting field of the separable polynomial (x3 - 2)(x2 + x +1) over ℚ. Futhermore, K/ ℚ is Galois because the degree of a finite Galois extension is the size of the Galois group, i.e., |Gal(K/F)| = [K : F] = 6.
4. The subgroups of G are {id}, G, H’= {1, β, β2}, H1 = {1, α}, H2 = {1, αβ}, H3 = {1, αβ2}.
5. H’ is normal in G ↭ Q(w) is a Galois extension of ℚ. However, H1, H2, and H3 are not normal in G ↭ KH1, KH2, and KH3 are not Galois extensions of ℚ. • Recall. Let p be a prime, every finite field extension $\mathbb{K}/\mathbb{F_p}$ is exactly K = $\mathbb{F}_{p^r}$, K is a Galois extension of $\mathbb{F_p}$ and is the splitting field of the polynomial xq -x (where q=pr), Gal(K/F) ≋ ℤ/rℤ (the Galois group has a generator σ: a → ap). r = [$\mathbb{F_p^r} : \mathbb{F_p}$] Subgroups of G are cyclic groups of order (say s) dividing r (s | r)

[KH : F] = r/s ⇒ |KH| = pr/s ⇒ [Structure theorem of finite fields] A field of order pk where k=r/s exists, KH = $\mathbb{F_{p^k}},~ and~ \mathbb{F_{p^k}}⊆\mathbb{F_{p^r}}$ 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 $\mathbb{ℚ}(\sqrt{2})/\mathbb{Q}$ is neither Galois, nor cyclic, and $\mathbb{ℚ}(w,\sqrt{2}/\mathbb{Q})$ 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 → ap), 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

# Bibliography 