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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)**

Proposition. Let K/F be a finite field separable extension. Then, there are only a finitely many intermediate fields for the extension K/F.

Proof.

Because of this lemma, Every finite separable field extension K/F can be extended to a Galois extension, we can extend K/F to an extension L/F which is Galois as follows: K = F(α_{1}, α_{2}, ···, α_{n}), ∀i α_{i} is separable because the extension is separable. Let take their distinct irreducible polynomial f_{i}, and consider their product f = f_{1}···f_{n}.

Let L be the splitting field of f, a separable polynomial, and therefore L/F is Galois ⇒ [By the Fundamental Theorem of Galois Theory] Gal(L/F) is a finite group, so it has finitely many subgroups ⇒ there are only finitely many intermediate fields for the extension L/F ⇒ [**Every intermediate field of K/F, say H, L ⊆ K ⊆ H ⊆ F is also an intermediate field of L/F**] there are only finitely many intermediate fields for the extension K/F∎

**Counterexample of a finite extension of fields K/F such that there are infinitely many intermediate fields**.

Recall that a separable extension has finitely many intermediate subfields. Therefore, K/F has to be inseparable.

Solution:

F = $\mathbb{F_2}(t, u)$ quotient field of $\mathbb{F_2}[t, u]$ where t and u are independent variables. Consider f = x^{2} -t, g = x^{2} -u ∈ F[x], α and β be roots of f, g respectively in a splitting field of fg. Let K = F(α, β) ⊆ the splitting field of fg.

Claim: K/F has finitely many intermediate subfields.

- It is a finite extension. [K : F] = 4.

α ∉ F because “t” is an independent variable and it has no square root in F = $\mathbb{F_2}(t, u)$ ⇒ [F(α) : F] > 1. Besides, α satisfies x^{2} -t ∈ F[x] ⇒ [α ∉ F, [F(α) : F] > 1] [F(α) : F] = 2

β^{2} = u ∈ F ⊆ F(α), but β ∉ F(α) ⇒ [K : F(α)] = 2 ⇒ [K : F] = 4 and, in particular, it is a finite extension.

Suppose for the sake of contradiction that β ∈ F(α) ⇒ Because [F(α) : F] = 2, a basis for F(α) is 1 and α, so β ∈ F(α) ⇒ β = a + bα, a, b ∈F ⇒ β^{2} = (a + bα)^{2} ⇒ [We are working in fields of characteristic 2] β^{2} = a^{2} + b^{2}α^{2} ⇒ u = a^{2} + b^{2}t ⇒ u and t are dependent variables ⊥

For any a ∈ F, let L_{a} = F(α + aβ). Claim: L_{a} ≠ L_{b} if a ≠ b and then we are done because F has infinitely many elements (arbitrary high degrees polynomials) so there are infinitely many intermediate subfields

Claim: L_{a} ≠ L_{b} if a ≠ b.

Proof.

Suppose for the sake of contradiction a ≠ b and yet L_{a} = L_{b} ⇒ α +aβ ∈ L_{a} = L_{b} ⇒ [L_{b} = F(α + bβ)] α +aβ -(α +bβ) ∈ L_{b} ⇒ (a-b)β ∈ L_{b} [a ≠ b ⇒ a-b ≠ 0, a -b is an element of F, a-b has multiplicative inverse] β ∈ L_{b} ⇒ aβ ∈ L_{b}, too.

β ∈ L_{b} ⇒ [α +aβ ∈ L_{b}, aβ ∈ L_{b}] (α + aβ) -aβ ∈ L_{b} ⇒ α ∈ L_{b}. Therefore, L_{b} ⊇ F(α, β) = K ⇒ [L_{b} is an intermediate field between K and F] K = L_{b} ⇒ [L_{b} : F] = 4

However, L_{b} = F(α + bβ) ⇒ (α +bβ)^{2} = [We are working in characteristic 2] α^{2} +b^{2}β^{2} = t + b^{2}u ∈ F = $\mathbb{F_2}(t, u)$ ⇒ deg_{F}(α + bβ) ≤ 2 since (α + bβ) satisfies a degree 2 polynomial over F, namely x^{2} -t -b^{2}u ⊥ ([L_{b} (=F(α+bβ)):F]=4) Therefore, L_{a} ≠ L_{b} and [L_{b} : F] ≠ 4. {L_{a}| a ∈ F} is an infinite set of intermediate fields.

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).
- Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.
- Fields and Galois Theory. Morandi. P., Springer.
- Fields and Galois Theory. By Evan Dummit, 2020.