# 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

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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