    # Fundamental Theorem of Galois Theory II.

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

# Recall

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) 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(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.

Solution:

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.

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.

Proof.

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.

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