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Fundamental Theorem of Galois Theory II.

Math is the only place where truth and beauty mean the same thing, Danica McKellar


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]

K | K F ^ H | [ H G | : = H [ ] K = : [ K K ^ ^ H 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.


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(α1, α2, ···, αn), ∀i αi is separable because the extension is separable. Let take their distinct irreducible polynomial fi, and consider their product f = f1···fn.

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.


F = $\mathbb{F_2}(t, u)$ quotient field of $\mathbb{F_2}[t, u]$ where t and u are independent variables. Consider f = x2 -t, g = x2 -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.

  1. It is a finite extension. [K : F] = 4.
K | F | F ( = 2 α 2 ) , F ( α α , β F )

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 x2 -t ∈ F[x] ⇒ [α ∉ F, [F(α) : F] > 1] [F(α) : F] = 2

β2 = 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 ⇒ β2 = (a + bα)2 ⇒ [We are working in fields of characteristic 2] β2 = a2 + b2α2 ⇒ u = a2 + b2t ⇒ u and t are dependent variables ⊥

For any a ∈ F, let La = F(α + aβ). Claim: La ≠ Lb 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: La ≠ Lb if a ≠ b.


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

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

However, Lb = F(α + bβ) ⇒ (α +bβ)2 = [We are working in characteristic 2] α2 +b2β2 = t + b2u ∈ F = $\mathbb{F_2}(t, u)$ ⇒ degF(α + bβ) ≤ 2 since (α + bβ) satisfies a degree 2 polynomial over F, namely x2 -t -b2u ⊥ ([Lb (=F(α+bβ)):F]=4) Therefore, La ≠ Lb and [Lb : F] ≠ 4. {La| 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.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.
  5. Fields and Galois Theory. Morandi. P., Springer.
  6. Fields and Galois Theory. By Evan Dummit, 2020.
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