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:

- K/F is Galois.
- F is the fixed field of Aut(K/F), i.e., F = K
^{Gal(K/F)} - 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]
- K/F is a normal, finite, and separable extension.
- K is the splitting field of a separable polynomial f ∈ F[x] over F.
- 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 → K^{H}, its inverse is defined by L → Gal(K/L). H_{1} ⊇ H_{2} ⇒ K^{H1} ⊆ K^{H2} and L_{1} ⊆ L_{2} ⇒ Gal(K/L_{1}) ⊇ Gal(K/L_{2}). Futhermore, it satisfies the following equality |H| = [K : K^{H}] and [G : H] = [K^{H} : 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 σ^{2} = 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, σ^{2} = 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^{-1}σ^{-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 = K^{H}, then [K : L] = [K : K^{H}] = |H|

Case 1, [K : L] = 2, L = K^{H} 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 = K^{H} 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) -x_{1}-, (0, 1, 0) -x_{2}-, (0, 0, 1) -x_{3}-, (1, 1, 0) -x_{1}+x_{2}-, (1, 0, 1) -x_{1}+x_{3}-, (0, 1, 1) -x_{2}+x_{3}-, (1, 1, 1) -x_{1}+x_{2}+x_{3}-}

Let’s compute them: {0, x_{1}, x_{2}, x_{1}+ x_{2}}, {0, x_{1}, x_{3}, x_{1}+ x_{3}}, {0, x_{2}, x_{3}, x_{2}+ x_{3}}, {0, x_{1}, x_{2} + x_{3}, x_{1}+ x_{2} + x_{3}}, {0, x_{2}, x_{1} + x_{3}, x_{1}+ x_{2} + x_{3}}, {0, x_{3}, x_{1} + x_{2}, x_{1}+ x_{2} + x_{3}}, {0, x_{1}+x_{2}, x_{2} + x_{3}, x_{1} + x_{3}}, there are seven subgroups and intermediate fields, say L_{1}, ···, L_{7}

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

To name a few, L_{1} = K^{{0, x1}}, L_{2} = K^{{0, x2}}, L_{3} = K^{{0, x3}}, L_{4} = 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 L_{j}s, e..g, L’_{1} is contained in L_{1}, L_{2}, and L_{4} or {0, x_{1}}, {0, x_{2}}, {0, x_{1}+x_{2}} ≤ {0, x_{1}, x_{2}, x_{1}+x_{2}}

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) = ⟨σ_{2}, σ_{3}, σ_{5}⟩ with σ_{k} 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) = ⟨σ_{2}, σ_{3}, σ_{5}⟩ ≋ 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 (x^{2} -2)(x^{2} -3)(x^{2} -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 x^{2}-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

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

- $ℚ(\sqrt{2},\sqrt{3})$. This is not possible because $[ℚ(\sqrt{2},\sqrt{3}):ℚ]=4≠2=[(ℚ\sqrt{5}):ℚ]$
- ℚ ⊥ 2 = $[(ℚ\sqrt{5}):ℚ]$
- $ℚ(\sqrt{2})$ 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(id) = $\sqrt{2}→\sqrt{2}, \sqrt{3}→\sqrt{3}, \sqrt{5}→\sqrt{5}$
- σ
_{1}= $\sqrt{2}→-\sqrt{2}, \sqrt{3}→\sqrt{3}, \sqrt{5}→\sqrt{5}$ - σ
_{2}= $\sqrt{2}→\sqrt{2}, \sqrt{3}→-\sqrt{3}, \sqrt{5}→\sqrt{5}$ - σ
_{3}= $\sqrt{2}→\sqrt{2}, \sqrt{3}→\sqrt{3}, \sqrt{5}→-\sqrt{5}$ - σ
_{4}= σ_{1}σ_{2}= $\sqrt{2}→-\sqrt{2}, \sqrt{3}→-\sqrt{3}, \sqrt{5}→\sqrt{5}$ - σ
_{5}= σ_{1}σ_{3}= $\sqrt{2}→-\sqrt{2}, \sqrt{3}→\sqrt{3}, \sqrt{5}→-\sqrt{5}$ - σ
_{6}= σ_{2}σ_{3}= $\sqrt{2}→\sqrt{2}, \sqrt{3}→-\sqrt{3}, \sqrt{5}→-\sqrt{5}$ - σ
_{7}= σ_{1}σ_{2}σ_{3}= $\sqrt{2}→-\sqrt{2}, \sqrt{3}→-\sqrt{3}, \sqrt{5}→-\sqrt{5}$

(σ_{i})^{2} = 1 ∀i, (σ_{i}σ_{j}) = 1, (σ_{1}σ_{2}σ_{3})^{2} = 1.

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

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

Let’s prove it: [K : K^{{1, σ1}}] = |{1, σ_{1}}| = 2. We know that $\sqrt{3},\sqrt{5}$ ∈ K^{{1, σ1}} because σ_{1} 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:

- L
_{2}= K^{{1, σ2}}= $\mathbb{Q}(\sqrt{2},\sqrt{5})$ - L
_{3}= K^{{1, σ3}}= $\mathbb{Q}(\sqrt{3},\sqrt{5})$ - L
_{4}= K^{{1, σ1σ2}}= [$σ_1σ_2(\sqrt{2}\sqrt{3})=(-\sqrt{2})(-\sqrt{3})=(\sqrt{2}\sqrt{3})$, so σ_{1}σ_{2}fixes $\sqrt{5}$, and $\sqrt{2}\sqrt{3}=\sqrt{6}$] $\mathbb{Q}(\sqrt{5},\sqrt{6})$ - L
_{5}= K^{{1, σ1σ3}}= [$σ_1σ_2(\sqrt{2}\sqrt{5})=(-\sqrt{2})(-\sqrt{5})=(\sqrt{2}\sqrt{5})$, so σ_{1}σ_{2}fixes $\sqrt{3}$, and $\sqrt{2}\sqrt{5}=\sqrt{10}$] $\mathbb{Q}(\sqrt{3},\sqrt{10})$ - L
_{6}= K^{{1, σ2σ3}}= $\mathbb{Q}(\sqrt{2},\sqrt{15})$ - L
_{7}= K^{{1, σ1σ2σ3}}= $\mathbb{Q}(\sqrt{6},\sqrt{15}) = \mathbb{Q}(\sqrt{6},\sqrt{10}) = \mathbb{Q}(\sqrt{10},\sqrt{15})$L

_{7}fixes all products of σ_{i}, but $\sqrt{6}\sqrt{15}=\sqrt{90}=3\sqrt{10}, \sqrt{10}$ is already L_{7}. 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 x^{2} +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, σ_{1}, σ_{2}, σ_{1}σ_{2}} = $\mathbb{Q}(\sqrt{5})$
Justification:

Obviously, $\sqrt{5}∈\mathbb{Q}(\sqrt{5})$ because 1, σ_{1}, σ_{2} and σ_{1}σ_{2} fix it.

Gal(K/L’_{1}) = |L’_{1}| = |{1, σ1, σ2, σ1σ2}| = 4 ⇒ [K : ℚ] = 8 = [K : K^{L'1}] · [K^{L'1} : ℚ] = 4 · [K^{L'1} : ℚ] ⇒ [K^{L'1} : ℚ] = 2 ⇒ [K^{L'1} : $ℚ(\sqrt{5})$] = 1 ⇒ K^{L'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 ∃σ_{i}∈ H for some i, $σ_i(\sqrt{2}+\sqrt{3}+\sqrt{5}) = \sqrt{2}+\sqrt{3}+\sqrt{5}$ ⊥

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.

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