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Motivating the fundamental theorem of Galois Theory

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

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Motivation

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.

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 x3-2, 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, α2} is $ℚ(\sqrt{2},i)$, e is the identity automorphism.

![Image](/maths/images/galoistheory.jpeg ./maths/images/galoistheory4.jpeg ./maths/images/galoistheory2.jpeg ./maths/images/galoistheory3.jpeg) 

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 x2 + 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, $\sqrt[4]{2} → \sqrt[4]{2}$
σ2: i → i, $\sqrt[4]{2} → -\sqrt[4]{2}$
σ3: i → i, $\sqrt[4]{2} → -i\sqrt[4]{2}$
στ: i → -i, $\sqrt[4]{2} → i\sqrt[4]{2}$
σ2τ: i → -i, $\sqrt[4]{2} → -\sqrt[4]{2}$
σ3τ: i → -i, $\sqrt[4]{2} → -i\sqrt[4]{2}$

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.

![Image](/maths/images/galoistheory.jpeg ./maths/images/galoistheory4.jpeg ./maths/images/galoistheory2.jpeg ./maths/images/galoistheory3.jpeg) 

  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 ≋ $\mathbb{V}_4$

4 K | K ^ = 4 G , ( [ [ K K 3 : ^ , K G ^ : 5 G ) ] ] = = | 1 G | = K ^ 4 G = G a l o i s

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

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.

H 2 1 2 / \ { 2 H 2 1 | 2 | G } \ H / 2 3 2 Q ( i 2 ) / 2 2 \ K | Q | Q ( 2 2 \ ) 2 / Q 2 ( i 2 )

K/F Galois, K/L1, K/L2, K/L3 are Galois, L1/F, L2/F, and L3/F are Galois, too.

Futhermore, αβ2 = βα, β2α = αβ, αβ ≠ βα (not Abelian), β3 = α2 = 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 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, $\sqrt[3]{2}$)/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, $\sqrt[3]{2}$)/ℚ) ≈ S3.
  3. The minimal polynomial of $\sqrt[3]{2}$ is x3 - 2 = 0, and the minimal polynomial of w is x2 + x +1, and K = ℚ(w, $\sqrt[3]{2}$) 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.

Image 

r s / s { 1 | H | G } r / K s s = | K | F ^ F H = p ^ F r p

[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 $\mathbb{ℚ}(\sqrt[4]{2})/\mathbb{Q}$ 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 → ap), so $K/\mathbb{F_p}$ is a cyclic extension.

G a l o i s K F F p G a l o i s

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

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.
  5. Fields and Galois Theory. Morandi. P., Springer.
  6. Fields and Galois Theory. By Evan Dummit, 2020.
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