There is nothing more practical than a good theory, Leonid Brezhnev.

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)**

- $K = \mathbb{Q}(\sqrt{3},\sqrt{5})$, $σ: \sqrt{3} → -\sqrt{3}, τ: \sqrt{5} → -\sqrt{5}, μ = στ: \sqrt{3} → -\sqrt{3}, \sqrt{5} → -\sqrt{5}, \sqrt{15} → \sqrt{15}$. Gal($\mathbb{Q}(\sqrt{3},\sqrt{5})/\mathbb{Q}$) = {id, σ, τ, μ} ≋ ℤ/ℤ
_{2}xℤ/ℤ_{2}. Then, G = Gal(K/F), [K/F is Galois if F = K^{G}] Gal(K/K^{G}) = Gal($\mathbb{Q}(\sqrt{3},\sqrt{5})/ℚ$) and the extension is obviously Galois.[K: ℚ] = 4, and K as a vector space over ℚ has a basis {$1, \sqrt{3}, \sqrt{5}, \sqrt{15}$} and dim

_{ℚ}K = 4, so every element a ∈ K can be written uniquely as $a + b\sqrt{3} + c\sqrt{5} + d\sqrt{15}$. σ is defined uniquely as $a + b\sqrt{3} + c\sqrt{5} + d\sqrt{15} → a - b\sqrt{3} + c\sqrt{5} - d\sqrt{15}$, τ: $a + b\sqrt{3} + c\sqrt{5} + d\sqrt{15} → a + b\sqrt{3} - c\sqrt{5} - d\sqrt{15}$, μ: $a + b\sqrt{3} + c\sqrt{5} + d\sqrt{15} → a - b\sqrt{3} - c\sqrt{5} + d\sqrt{15}$

Why [K : ℚ] = 4? We build the tower, $\mathbb{Q} ⊆ \mathbb{Q}(\sqrt{3}) ⊆ \mathbb{Q}(\sqrt{3}, \sqrt{5})$. A) $\sqrt{3}∉\mathbb{Q}$, x^{2} -3 is irreducible in ℚ, and degree(x^{2}-3) = 2 ⇒ [$\mathbb{Q}(\sqrt{3}) : \mathbb{Q}$] = 2. B) $\sqrt{5}∉\mathbb{Q}(\sqrt{3})$, x^{2} -5 is irreducible in $\mathbb{Q}(\sqrt{3})$, and degree(x^{2}-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/ℚ is normal. Then K/ℚ is a finite normal extension if and only if it is the splitting field of some polynomial f(x) ∈ K[x]. It suffices to consider **f(x) = (x ^{2} -3)(x^{2} -5)**. Its splitting field must contain $\sqrt{3}$ and $\sqrt{5}$ and since K is the smallest field containing both of them, K must be in fact the splitting field of f(x) ⇒ K is normal over ℚ.

- A completely similar argument as a previous example shows that $K = \mathbb{Q}(\sqrt{2},\sqrt{3})$, $σ: \sqrt{2} → -\sqrt{2}, τ: \sqrt{3} → -\sqrt{3}, μ = στ: \sqrt{2} → -\sqrt{2}, \sqrt{3} → -\sqrt{3}, \sqrt{6} → \sqrt{6}$. Gal($\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$) = {id, σ, τ, μ} ≋ ℤ/ℤ
_{2}xℤ/ℤ_{2}. Then, G = Gal(K/F) = [K/F is Galois if F = K^{G}] Gal(K/K^{G}) = Gal($\mathbb{Q}(\sqrt{2},\sqrt{3})/ℚ$) and the extension is obviously Galois.

Futhermore, K has 3 subfields L such that [K : L] = 2 (⇒ [L : ℚ] = 2), namely $ℚ(\sqrt{2}),ℚ(\sqrt{3}),ℚ(\sqrt{6})$, and $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2} +\sqrt{3})$

K/ℚ is normal. Then K/ℚ is a finite normal extension if and only if it is the splitting field of some polynomial f(x) ∈ K[x]. It suffices to consider **f(x) = (x ^{2} -2)(x^{2} -3)**. Its splitting field must contain $\sqrt{2}$ and $\sqrt{3}$ and since K is the smallest field containing both of them, K must be in fact the splitting field of f(x) ⇒ K is normal over ℚ.

- Let K = $\mathbb{Q}(\sqrt[3]{2}, w)$, F = ℚ, and w is the primitive third root of unity $w = e^\frac{2πi}{3}$
Reclaim: Any degree 2 extension is normal, but it does not necessarily have to be Galois.

? = [$\mathbb{Q}(\sqrt[3]{2}, w) : \mathbb{Q}(\sqrt[3]{2})$] = 2 because (i) ? > 1 because w ∉ ℝ and $\mathbb{Q}(\sqrt[3]{2}) ⊆ \mathbb{R}$. (ii) ? ≤ 2 because w is a root of unity, x^{3} = 1 is reducible, x^{3}-1 = (x -1)(x^{2} +x +1), x^{2} +x +1 is irreducible of degree 2.

[$\mathbb{Q}(\sqrt[3]{2}, w): \mathbb{Q}$] = 2 · 3 = 6. We know K is the splitting field of x^{3} - 2 over ℚ (the roots are $\sqrt[3]{2}, \sqrt[3]{2}w, \sqrt[3]{2}w^2$, they are all in K and if you remove any of them, you don’t get K, it is generated by **all its roots**) ⇒ K/ℚ is **normal.**

The possible images of $\sqrt[3]{2}$ are $\sqrt[3]{2}, \sqrt[3]{2}w, \sqrt[3]{2}w^2$, and w can be mapped to w and w^{2}. There exists a homomorphism K → K for every choice of images $\sqrt[3]{2}$ and w, so there are 6 (3 x 2 choices) automorphism K → K, |Gal(K/ℚ)| = 6 = [K : ℚ] ⇒ K/ℚ is Galois

$\sqrt[3]{2}$ and w are linear independent, and their images are to be chosen independently (the **conjugates of a root, say α, are the images of α under the field homomorphism**) of each other because $\sqrt[3]{2}$ ∈ ℝ, and w is not real. Futhermore, there exists a homomorphism K → K for every choice of images of w and $\sqrt[3]{2}$, **it will send any root to one of its conjugates**.

Finally, what is Gal(K/ℚ)? There are only **two groups of order 6 up to isomorphism, namely the cyclic group ℤ/6ℤ (Abelian), and the symmetric group of order 3, S _{3}**.

- σ
_{0}: $\sqrt[3]{2}→\sqrt[3]{2}$, w → w. - σ
_{1}: $\sqrt[3]{2}→\sqrt[3]{2}w$, w → w. - σ
_{2}: $\sqrt[3]{2}→\sqrt[3]{2}w^2$, w → w. - σ
_{3}: $\sqrt[3]{2}→\sqrt[3]{2}$, w → w^{2}. - σ
_{4}: $\sqrt[3]{2}→\sqrt[3]{2}w$, w → w^{2}. - σ
_{5}: $\sqrt[3]{2}→\sqrt[3]{2}w^2$, w → w^{2}.

σ_{4}∘σ_{5} = σ_{2} ≠ σ_{1} = σ_{5}∘σ_{4} ⇒ Gal(K/ℚ) is not Abelian ⇒ Gal(K/ℚ) ≋ S_{3}

Let’s have a look at the subgroups. The trivial subgroups are S_{3} (whose extension is ℚ) and {σ_{0} = id} (whose extension is K). The order of the proper subgroups can only be a divisor of 6 by Lagrange’s Theorem, so we are left with 2 or 3.

It is left to the reader to check that σ_{3}^{2} = σ_{4}^{2} = σ_{5}^{2} = σ_{0}

σ_{4}^{2} (w) = σ_{4}(w^{2}) = w^{4} = w. $σ_4^{2}(\sqrt[3]{2}) = σ_4(\sqrt[3]{2}w) = \sqrt[3]{2}ww^2 = \sqrt[3]{2}w^3 = \sqrt[3]{2} ⇒ σ_4^{2} = σ_0.$

Therefore, we have the following subgroups of order 2, namely H_{3} = {σ_{0}, σ_{3}}, H_{4} = {σ_{0}, σ_{4}}, and H_{5} = {σ_{0}, σ_{5}}. None of them are normal (H◁G iff ∀g ∈ G, gNg^{-1} ⊆ N), e.g., H_{4} is not normal because σ_{5}∘σ_{4}∘σ_{5}^{-1} = [σ_{5}^{2} = σ_{0} = id] σ_{5}∘σ_{4}∘σ_{5} = σ_{1}∘σ_{5} = σ_{5} ∉ H_{4}. Therefore, the corresponding field extensions (L_{i} is the extension whose Galois group is H_{i}) are not Galois.

Notice that $σ_3(\sqrt[3]{2}) = \sqrt[3]{2} ⇒ ℚ(\sqrt[3]{2}) ⊆ L_3$

Therefore, $ℚ(\sqrt[3]{2}) ⊆ L_3$ and [L_{3} : $ℚ(\sqrt[3]{2})$] = 1 ⇒ **L _{3} = $ℚ(\sqrt[3]{2})$**. Following a similar argument, we can demonstrate that

As for subgroups of order 3 (they are isomorphic to ℤ_{3}), we just need to find non trivial elements of order 3. It is left to the reader to show that σ_{1} and σ_{2} are elements of order 3, σ_{1}∘σ_{2} = σ_{2}∘σ_{1} = σ_{0}, so we have only one subgroup of order 3, namely, L = {σ_{0}, σ_{1}, σ_{2}}. σ_{i}(w) = w, 0 ≤ i ≤ 2 ⇒ ℚ(w) ⊆ L

Therefore, L = ℚ(w). Futhermore, since H ◁ G (all subgroups of index 2 are normal) ⇒ **ℚ(w)/ℚ is a Galois extension** (the splitting field of a separable polynomial x^{2} +x +1). The following are two diagrams with the subgroups of S_{3} ≋ Gal(K/ℚ) and their corresponding fixed fields.

- 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. A far more detail explanation is given in Consequences Galois III. Exercises. ℚ(√2+√3+√5)

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 = |Gal(K/ℚ)|.

Fact: The square roots of different primes are linearly independent over the field of rationals

To show that [K : ℚ] = 8, we know that [$\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$] = 4, so by the Fundamental Theorem of Galois Theory it has either one or three intermediate fields of degree 2 over ℚ (since a group of order 4 has either a single subgroup of index 2, when the group is cyclic, or three, when it is the Klein 4-group). Since this field contains the three distinct intermediate fields $\mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{3}),~and~ \mathbb{Q}(\sqrt{6})$, it **cannot also contain the field $\mathbb{Q}(\sqrt{5})$, which is clearly distinct from those three**. Therefore, [K : ℚ] = [$\mathbb{Q}(\sqrt{2},\sqrt{3})(\sqrt{5}):\mathbb{Q}(\sqrt{2},\sqrt{3})$]·[$\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$] = 2·4 = 8.

- Let K = $\mathbb{Q}(\sqrt[4]{2})$, F = ℚ. [K : F] = 4. It is
**not normal**, f(x) = x^{4}-2 has a root but not all of them ($±\sqrt[4]{2}, ±\sqrt[4]{2}i$) since K ⊆ ℝ. Futhermore, possible images of $\sqrt[4]{2}$ in K are $\sqrt[4]{2}, -\sqrt[4]{2}$, so there are only two possibilities: identity, σ: $\sqrt[4]{2} → -\sqrt[4]{2}$ ⇒ [K : ℚ] = 4 >**|Gal(K/ℚ)| = 2 ⇒ K/ℚ is not Galois, Gal(K/ℚ)**≋ ℤ/ℤ_{2}

What is K^{Gal(K/ℚ)}. We know [K : K^{Gal(K/ℚ)}] = |Gal(K/ℚ)| = 2

Since [K : ℚ] = 4 ⇒ [K^{Gal(F/ℚ)} : ℚ] = 2.

Notice that $σ(\sqrt{2}) = σ(\sqrt[4]{2}\sqrt[4]{2})= σ(\sqrt[4]{2})^2 = (-\sqrt[4]{2})^2 = \sqrt{2} ⇒\sqrt{2}$ is fixed by 1 and σ ⇒ $\mathbb{Q}(\sqrt{2})⊆\mathbb{K}^{Gal(\mathbb{K}/\mathbb{Q})}$

$\mathbb{Q}(\sqrt{2})⊆\mathbb{K}^{Gal(\mathbb{K}/\mathbb{Q})}$ ⇒ $\mathbb{Q}(\sqrt{2})=\mathbb{K}^{Gal(\mathbb{K}/\mathbb{Q})}$

- Let p be a prime number, r a positive integer, r ≥ 1, $K=\mathbb{F}_{p^r}, F = \mathbb{F}_p,$ [K : F] = r.

F finite, p = char(F) > 0. [**Classification of finite fields.** Let F be a finite field of order q = p^{r}. Then, K is the splitting field of the polynomial x^{q}-x ∈ $\mathbb{F}_p[x]$.] **K is the splitting field of $x^{p^{r}}-x~ over~ \mathbb{F_p}$** ⇒ the extension is **normal**.

Let σ: K → K, σ(α) = α^{p}, σ is a field homomorphism (x + y)^{p} = x^{p} + y^{p} because when p is prime ⇒ p | ${p \choose k}$, injective ⇒ If a set X is finite, an injective map f: X → X is also surjective. Futhermore, σ(a) = a ∀a ∈ F_{p} ⇒ σ is an F-automorphism of K, i.e., σ ∈ Gal(K/F).

σ is injective because F field, Φ: F→ F, Ker(Φ) is an ideal of F, a field ⇒ KerΦ={e} or KerΦ = F ⇒ Φ is 1-1 because ker(Φ) is the trivial ideal, or Φ is zero (⊥ because ker(Φ) = F).

Notice σ^{2}(α) = σ(σ(α)) = σ(α^{p}) = (α^{p})^{p} = $α^{p^{2}}, σ^i(α)=α^{p^{i}}$ ∀α ∀i ≥ 1

Let’s suppose order of σ is n, σ^{n} = id, i.e, σ^{n}(α) = α ∀α ∈ K ⇒ $σ^n(α)=α^{p^{n}}=α$ ⇒ ∀α ∈ K, α is a root of f(x) = $x^{p^{n}}-x$ in K. Since deg(f(x)) = p^{n}, f(x) has at most p^{n} roots. However, every element of K is a root of f ⇒ |K| = p^{r} ≤ p^{n} ⇒ r ≤ n.

Theorem. A polynomial f of degree n over a **field** F has at most n roots in F.

Futhermore, [**Classification of finite fields.** Let F be a finite field of order q = p^{r}. Then, F is the splitting field of the polynomial x^{q}-x ∈ $\mathbb{F}_p[x]$. K is the splitting field of $x^{p^{r}}-x~ over~ \mathbb{F_p}$] $σ^r(α)=α^{p^{r}}=α$ ⇒ σ^{r} = 1 ⇒ order(σ) ≤ r ⇒ n = r ⇒ The subgroup generated by σ in $Gal(\mathbb{F_{p^r}}/\mathbb{F_p})$ = {1, σ, σ^{2},···, σ^{r-1}} has order r ⇒ $|Gal(\mathbb{F_{p^r}}/ \mathbb{F_p})|$ ≥ r.

$r = [\mathbb{F_{p^r}} : \mathbb{F_p}] ≥ [F_{p^r} : F_{p^r}^G] = |Gal(\mathbb{F_{p^r}}/ \mathbb{F_p})|≥ r ⇒ [\mathbb{F_{p^r}} : \mathbb{F_{p^r}^G}] = r~ and~ [F_{p^r}^G : \mathbb{F_p}] = 1 ⇒ \mathbb{F_{p^r}^G} = \mathbb{F_p} ⇒ \mathbb{F_{p^r}}/ \mathbb{F_p}$ is Galois.

Futhermore, $Gal(\mathbb{F}_{p^r}/ \mathbb{F}_p)≋ℤ/rℤ$ because we have already found an automorphism of order r, so it is cyclic.

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, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).