    # Recall

Since the mathematicians have invaded the theory of relativity I do not understand it myself any more, Albert Einstein

They shouldn’t be allowed to teach math so early in the morning, Kendare Blake, Anna Dressed in Blood.

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) Exercise. Let K/ℚ be a normal extension such that |Gal(K/ℚ)| = 8 and σ2 = id ∀σ ∈Gal(K/ℚ), σ ≠ id. Find the number of intermediate fields of the extensions K/ℚ.

Solution.

Note that K/ℚ is Galois. Since char(ℚ) = 0, K/Q is separable and it is normal by assumption ⇒ K/Q is Galois, [K : ℚ] = |Gal(K/ℚ)| = 8

For convenient sake, let G = Gal(K/ℚ). By assumption, σ2 = id ∀σ ∈Gal(K/ℚ), σ ≠ id so ord(σ) = 2 ∀σ ∈Gal(K/ℚ), σ(id) = id.

Besides, G is Abelian, σ, τ ∈ G, στ = (στ)-1 = [Shoe and sock theorem] τ-1σ-1 = τσ

G is Abelian and |Gal(K/ℚ)| = 8 ⇒ G ≋ ℤ/2ℤ x ℤ/2ℤ x ℤ/2ℤ (it is left to the reader to demonstrate completely this statement)

Let L be a non-trivial intermediate field:

Recall: If L = KH, then [K : L] = [K : KH] = |H|

Case 1, [K : L] = 2, L = KH where H ≤ G has order 2. # number of such L = # subgroups of G of order 2 # elements of order 2 in G = 7 because every element except the identity has order 2, say L’1, L’2, ···, L’7

Case 2, [K : L] = 4, L = KH where H ≤ G has order 4. # number of such L = # subgroups of G of order 4 (they need to be isomorphic to ℤ/2ℤ x ℤ/2ℤ)

G = { (0, 0, 0) -0-, (1, 0, 0) -x1-, (0, 1, 0) -x2-, (0, 0, 1) -x3-, (1, 1, 0) -x1+x2-, (1, 0, 1) -x1+x3-, (0, 1, 1) -x2+x3-, (1, 1, 1) -x1+x2+x3-}

Let’s compute them: {0, x1, x2, x1+ x2}, {0, x1, x3, x1+ x3}, {0, x2, x3, x2+ x3}, {0, x1, x2 + x3, x1+ x2 + x3}, {0, x2, x1 + x3, x1+ x2 + x3}, {0, x3, x1 + x2, x1+ x2 + x3}, {0, x1+x2, x2 + x3, x1 + x3}, there are seven subgroups and intermediate fields, say L1, ···, L7

Therefore, there are exactly 2 (trivial ones) + 7 + 7 = 16 subgroups.

To name a few, L1 = K{0, x1}, L2 = K{0, x2}, L3 = K{0, x3}, L4 = K{0, x1+x2}···, L’1 = K{0, x1, x2, x1+x2}, etc. It is left as an exercise that each L’i is contained in exactly three Ljs, e..g, L’1 is contained in L1, L2, and L4 or {0, x1}, {0, x2}, {0, x1+x2} ≤ {0, x1, x2, x1+x2}

Example (Particular case of the exercise previously done): $\mathbb{ℚ}(\sqrt{2},\sqrt{3},\sqrt{5})/ℚ$

Let K = $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$, F = ℚ. Computing Gal(K, ℚ) is relatively easy, it is Gal(K, ℚ) = ⟨σ2, σ3, σ5⟩ with σk the automorphism K → K that interchanges $\sqrt{k}→-\sqrt{k}$ (k = 2, 3, and 5) and does not move the rest of the elements.

Gal(K, ℚ) = ⟨σ2, σ3, σ5⟩ ≋ ℤ/2ℤ x ℤ/2ℤ x ℤ/2ℤ. You can think (a, b, c) ∈ ℤ/2ℤ x ℤ/2ℤ x ℤ/2ℤ as representing $\sqrt{2}→(-1)^a,\sqrt{3}→(-1)^b, \sqrt{5}→(-1)^c$.

K/ℚ is a Galois extension (normal and separable) since K is the splitting field of the polynomial (x2 -2)(x2 -3)(x2 -5) and ℚ is the base field (char(ℚ) = 0). Another way of seeing this is [K : ℚ] = 8 [K/ℚ is Galois] = |Gal(K/ℚ)|, and yet another is as follows!

[K : ℚ] = 8.

$ℚ(\sqrt{2},\sqrt{3})=ℚ(\sqrt{2})(\sqrt{3})$ an extension over $ℚ(\sqrt{2})$ and satisfying a degree 2 polynomial, namely x2-3 ∈ ℚ[x] ⊆ $ℚ(\sqrt{2})[x]$, we know that this extension is at the most 2. To prove equality to 2 (·), we need to demonstrate that $\sqrt{3}∉ℚ(\sqrt{2})$ or $\sqrt{3}~ and~ \sqrt{2}$ are linearly independent over ℚ.

For the sake of contradiction, let’s assume that $\sqrt{3}∈ℚ(\sqrt{2}), \sqrt{3} =a + b\sqrt{2}$ where a, b ∈ ℚ

1. a = 0 ⇒ $\sqrt{3} = b\sqrt{2} ⇒ 3 = 2b^2 ⇒ b^2=\frac{2}{3}$ but this impossible because 2/3 is irrational.
2. b = 0 ⇒ $\sqrt{3} = a ∈ ℚ$ ⊥
3. ab ≠ 0 ⇒ $3 = a^2+2b^2+2ab\sqrt{2} ⇒ \sqrt{2}=\frac{3-a^2-2b^2}{2ab}∈ ℚ$ ⊥

Using the same reasoning as before K = $ℚ(\sqrt{2},\sqrt{3},\sqrt{5})=ℚ(\sqrt{2},\sqrt{3})(\sqrt{5})$, [K : $ℚ(\sqrt{2},\sqrt{3})$] = 2 by showing that $\sqrt{5}∉ℚ(\sqrt{2},\sqrt{3})$ and satisfy the irreducible polynomial x2 -5.

For the sake of contradiction, suppose $\sqrt{5}∈ℚ(\sqrt{2},\sqrt{3})$ ⇒

By the main theorem of Galois theory we know that there are only four intermediate fields of $ℚ(\sqrt{2},\sqrt{3})/ℚ$, namely:

1. $ℚ(\sqrt{2},\sqrt{3})$. This is not possible because $[ℚ(\sqrt{2},\sqrt{3}):ℚ]=4≠2=[(ℚ\sqrt{5}):ℚ]$
2. ℚ ⊥ 2 = $[(ℚ\sqrt{5}):ℚ]$
3. $ℚ(\sqrt{2})$ or $ℚ(\sqrt{3})$. Using the same argument as before, it can be demonstrated that $\sqrt{5}∉ℚ(\sqrt{2}), \sqrt{5}∉ℚ(\sqrt{3})$

Therefore, [K : ℚ] = 2·2·2 = 8.

What is Gal(K, ℚ)? What are all the possibilities automorphism of K that fixes ℚ?

$\sqrt{2}→\sqrt{2},-\sqrt{2},~\sqrt{3}→\sqrt{3},-\sqrt{3}, \sqrt{5}→\sqrt{5},-\sqrt{5}.$ where $\sqrt{2}, \sqrt{3},~ and~ \sqrt{5}$ are linear independent over ℚ, and their images can be chosen independently of each other.

1. 1(id) = $\sqrt{2}→\sqrt{2}, \sqrt{3}→\sqrt{3}, \sqrt{5}→\sqrt{5}$
2. σ1 = $\sqrt{2}→-\sqrt{2}, \sqrt{3}→\sqrt{3}, \sqrt{5}→\sqrt{5}$
3. σ2 = $\sqrt{2}→\sqrt{2}, \sqrt{3}→-\sqrt{3}, \sqrt{5}→\sqrt{5}$
4. σ3 = $\sqrt{2}→\sqrt{2}, \sqrt{3}→\sqrt{3}, \sqrt{5}→-\sqrt{5}$
5. σ4 = σ1σ2 = $\sqrt{2}→-\sqrt{2}, \sqrt{3}→-\sqrt{3}, \sqrt{5}→\sqrt{5}$
6. σ5 = σ1σ3 = $\sqrt{2}→-\sqrt{2}, \sqrt{3}→\sqrt{3}, \sqrt{5}→-\sqrt{5}$
7. σ6 = σ2σ3 = $\sqrt{2}→\sqrt{2}, \sqrt{3}→-\sqrt{3}, \sqrt{5}→-\sqrt{5}$
8. σ7 = σ1σ2σ3 = $\sqrt{2}→-\sqrt{2}, \sqrt{3}→-\sqrt{3}, \sqrt{5}→-\sqrt{5}$

i)2 = 1 ∀i, (σiσj) = 1, (σ1σ2σ3)2 = 1.

Gal(K, ℚ) = {1, σ1, σ2,···, σ7} ≋ ℤ/2ℤ x ℤ/2ℤ x ℤ/2ℤ, so ∃ sixteen intermediate fields, 7 of order 4, 7 of order 2, and so on.

Li: fixes fields of order 2 subgroup, e.g., L1 = K{1, σ1} = [Claim] $\mathbb{Q}(\sqrt{3},\sqrt{5})$

Let’s prove it: [K : K{1, σ1}] = |{1, σ1}| = 2. We know that $\sqrt{3},\sqrt{5}$ ∈ K{1, σ1} because σ1 fixes both of them, and therefore $ℚ(\sqrt{3},\sqrt{5})⊆$K{1, σ1}

Therefore, [K{1, σ1}: $ℚ(\sqrt{3},\sqrt{5})$] = 1 ⇒ K{1, σ1} = $ℚ(\sqrt{3},\sqrt{5})$

Similarly:

• L2 = K{1, σ2} = $\mathbb{Q}(\sqrt{2},\sqrt{5})$
• L3 = K{1, σ3} = $\mathbb{Q}(\sqrt{3},\sqrt{5})$
• L4 = K{1, σ1σ2} = [$σ_1σ_2(\sqrt{2}\sqrt{3})=(-\sqrt{2})(-\sqrt{3})=(\sqrt{2}\sqrt{3})$, so σ1σ2 fixes $\sqrt{5}$, and $\sqrt{2}\sqrt{3}=\sqrt{6}$] $\mathbb{Q}(\sqrt{5},\sqrt{6})$
• L5 = K{1, σ1σ3} = [$σ_1σ_2(\sqrt{2}\sqrt{5})=(-\sqrt{2})(-\sqrt{5})=(\sqrt{2}\sqrt{5})$, so σ1σ2 fixes $\sqrt{3}$, and $\sqrt{2}\sqrt{5}=\sqrt{10}$] $\mathbb{Q}(\sqrt{3},\sqrt{10})$
• L6 = K{1, σ2σ3} = $\mathbb{Q}(\sqrt{2},\sqrt{15})$
• L7 = K{1, σ1σ2σ3} = $\mathbb{Q}(\sqrt{6},\sqrt{15}) = \mathbb{Q}(\sqrt{6},\sqrt{10}) = \mathbb{Q}(\sqrt{10},\sqrt{15})$

L7 fixes all products of σi, but $\sqrt{6}\sqrt{15}=\sqrt{90}=3\sqrt{10}, \sqrt{10}$ is already L7. Similar argument, $\sqrt{6}\sqrt{10}=\sqrt{60}=2\sqrt{15}, \sqrt{10}\sqrt{15}=\sqrt{150}=5\sqrt{6}$

Consider $ℚ(\sqrt{2}+\sqrt{3})$? Clearly $ℚ(\sqrt{2}+\sqrt{3})⊆ℚ(\sqrt{2},\sqrt{3})$

Suppose that the extension is degree 2 ⇒ there is a minimal polynomial of degree 2 such that $\sqrt{2}+\sqrt{3}$ is a root, say x2 +bx + c ⇒ $5 + c + 2\sqrt{6}+b(\sqrt{3}+\sqrt{2})=0$ which is impossible as $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ is a basis of $ℚ(\sqrt{2},\sqrt{3})$ and the equation implies the elements in the basis are linearly dependent ⊥

Futhermore, the extension is not degree 1 because $\sqrt{2} +\sqrt{3}$ is irrational ⇒ $[ℚ(\sqrt{2}+\sqrt{3}):ℚ]=4⇒[ℚ(\sqrt{2},\sqrt{3}):ℚ(\sqrt{2}+\sqrt{3})] = 1 ⇒ ℚ(\sqrt{2},\sqrt{3})=ℚ(\sqrt{2}+\sqrt{3})$, it is a primitive or simple extension, it is generated by a simple extension. Similarly, $L_2 = ℚ(\sqrt{2}+\sqrt{5}), L_3 = ℚ(\sqrt{3}+\sqrt{5})$.

Recall that L’i are fixed fields of order four subgroups of Gal(K/ℚ).

L’1 = {1, σ1, σ2, σ1σ2} = $\mathbb{Q}(\sqrt{5})$ Justification:

Obviously, $\sqrt{5}∈\mathbb{Q}(\sqrt{5})$ because 1, σ1, σ2 and σ1σ2 fix it.

Gal(K/L’1) = |L’1| = |{1, σ1, σ2, σ1σ2}| = 4 ⇒ [K : ℚ] = 8 = [K : KL'1] · [KL'1 : ℚ] = 4 · [KL'1 : ℚ] ⇒ [KL'1 : ℚ] = 2 ⇒ [KL'1 : $ℚ(\sqrt{5})$] = 1 ⇒ KL'1 = $ℚ(\sqrt{5})$

Similarly $L’_2 = \mathbb{Q}(\sqrt{2}), L’_3 = \mathbb{Q}(\sqrt{3}), L’_4 = \mathbb{Q}(\sqrt{10}), L’_5 = \mathbb{Q}(\sqrt{6}), L’_6 = \mathbb{Q}(\sqrt{15}), L’_7 = \mathbb{Q}(\sqrt{30})$

Finally, K = $ℚ(\sqrt{2}+\sqrt{3}+\sqrt{5})$ (a primitive extensive) Consider that $σ_i(\sqrt{2}+\sqrt{3}+\sqrt{5})≠\sqrt{2}+\sqrt{3}+\sqrt{5}$ ∀i = 1, ··· 7, e.g., $σ_1(\sqrt{2}+\sqrt{3}+\sqrt{5}) = -\sqrt{2}+\sqrt{3}+\sqrt{5}$. You can verify it one by one or take into consideration that $\sqrt{2},\sqrt{3},\sqrt{5}$ are linear independent over ℚ.

Suppose for the sake of contradiction K ≠ $ℚ(\sqrt{2}+\sqrt{3}+\sqrt{5})$ ⇒ ∃H ≤ G, H ≠ {1}, $ℚ(\sqrt{2}+\sqrt{3}+\sqrt{5}) = K^H$, then ∃σi∈ H for some i, $σ_i(\sqrt{2}+\sqrt{3}+\sqrt{5}) = \sqrt{2}+\sqrt{3}+\sqrt{5}$ ⊥

# Bibliography 