Composite fields

Being a mathematician is a bit like being a manic depressive: you spend your life alternating between giddy elation and black despair, Steven G. Krantz.

Definition. Let L₁ and L₂ be subfields of K. The composite or composium of the fields L₁, L₂ in K, denoted by L₁L₂, is defined to be the smallest subfield of K containing both fields L₁, L₂ or the intersection of all subfields containing both subfields.

Lemma. If L is a finite extension of F, then L is algebraic and finitely generated over F.

Proposition. If L₁/F and L₂/F are both finite-degree sub-extensions of K/F, F ⊆ L₁ ⊆ K, L₁ = F(α₁, ···, α_r), F ⊆ L₂ ⊆ K, L₂ = F(β₁, ···, β_s). Then, L₁L₂ = F(α₁, ···, α_r, β₁, ···, β_s).

Proof.

F ⊆ L₁ ⊆ K, L₁ = F(α₁, ···, α_r), F ⊆ L₂ ⊆ K, L₂ = F(β₁, ···, β_s).

L₁L₂ contains F and each of α₁, ···, α_r (it contains L₁) and β₁, ···, β_s (it contains L₂), hence it contains F(α₁, ···, α_r, β₁, ···, β_s).

Conversely, F(α₁, ···, α_r, β₁, ···, β_s) contains both L₁ and L₂, hence it contains L₁L₂ (the intersection of all subfields containing both subfields, that is, the smallest subfield containing both subfields L₁ and L₂) ⇒ L₁L₂ = F(α₁, ···, α_r, β₁, ···, β_s)∎

Degrees of Composites. The degree of a composite L₁L₂ is smaller than the product of the degrees of the fields that generate the field. (i) If L₁/F and L₂/F are both finite-degree subextensions of K/F, then [L₁L₂ : F] ≤ [L₁ : F] · [L₂ : F]. Futhermore, (ii) if the degrees [L₁ : F] and [L₂ : F] are relatively prime, then equality holds.

Proof.

By the previous result, L₁L₂ = F(α₁, ···, α_r, β₁, ···, β_s) = L₁(β₁, ···, β_s) ⇒ [β₁, ···, β_s span L₁L₂/L₁] [L₁L₂ : L₁] ≤ [L₂ : F].

Futhermore, [L₁L₂ : F] = [L₁L₂ : L₁] · [L₁ : F] ≤ [L₂ : F] · [L₁ : F] ∎ (i)

Besides, [L₁L₂ : F] = [L₁L₂ : L₁] · [L₁ : F], so [L₁L₂ : F] is divisible by both [L₁ : F] and [L₂ : F], say n and m respectively. If m and n are relatively prime, then [L₁L₂ : F] is divisible by n·m, so [L₁L₂ : F] ≥ n·m ⇒ [(i) [L₁L₂ : F] ≤ n·m] [L₁L₂ : F] = n·m = [L₁ : F] · [L₂ : F] ∎ (ii)

Proposition. If L₁/F and L₂/F are both finite-degree Galois extensions of K inside a common field F, then L₁L₂/F is a Galois finite extension.

Proof.

Each L_i is the splitting field of a polynomial f_i over F. Then, L₁L₂ is the splitting field for the product of f₁ and f₂ because L₁L₂ contains L₁, so f₁(x) splits in this field. Same argument applies to f₂(x). So the polynomial f₁(x)f₂(x) will also split completely in L₁L₂. Therefore, L₁L₂ is normal.

As L₁/F is Galois, L₁ is the splitting field of a separable polynomial f(x) ∈ F[x] ⊆ L₂[x], the roots of f(x) generate L₁ over F, so they generate L₁L₂ over L₂ ⇒ L₁L₂ is the splitting field of a separable polynomial f(x) over L₂ and therefore L₁L₂/L₂ is Galois, hence separable.

L₂/F is normal ⇒ separable. Then, L₁L₂/L₂ separable, L₂/F separable ⇒ [Separable is transitive] L₁L₂/F is separable ⇒ [We have proved that it is normal, too] L₁L₂/F is Galois∎

Example: L₁ = $ℚ(\sqrt{2})$, $ℚ(\sqrt{2})/ℚ$ is Galois, L₂ = $ℚ(\sqrt{3})$, $ℚ(\sqrt{3})/ℚ$ is Galois, the composite of L₁, L₂ is L₁L₂ = $ℚ(\sqrt{2}, \sqrt{3})$ and $ℚ(\sqrt{2}, \sqrt{3})/ℚ$ is Galois.

Proposition. The composite of two radical extensions is radical. L₁/F, L₂/F radical extensions ⇒ the composite L₁L₂/F is radical, too.

Observation. We need an ambient field K containing both fields, say L₁ and L₂, in order to be able to make sense or talk about the composite L₁L₂. Typically, K = $\bar F$.

Proof.

L₁/F radical ⇒ ∃ F ⊆ [F(a₁)/F is simple radical where a₁^n₁ ∈ F] F(a₁) ⊆ [F(a₂)/F(a₁) is a simple radical where a₂^n₂ ∈ F(a₁)] F(a₁, a₂) ⊆ ··· F(a₁, a₂, ···, a_r) = L₁

L₂/F radical ⇒ F ⊆ [simple radical, b₁^m₁ ∈ F] F(b₁) ⊆ [simple radical] F(b₁, b₂) ⊆ ··· F(b₁, b₂, ···, b_s-1) ⊆ [s.r., b_s^m_s ∈ F(b₁, b₂, ···, b_s-1)] F(b₁, b₂, ···, b_s) = L₂

We know L₁L₂ = F(a₁, a₂, ···, a_r, b₁, b₂, ···, b_s)

Consider F(a₁, a₂, ···, a_r, b₁, b₂, ···, b_s-1) ⊆ F(a₁, a₂, ···, a_r, b₁, b₂, ···, b_s). This extension is generated by b_s, but we know that b_s^m_s ∈ F(b₁, b₂, ···, b_s-1) ⊆ F(a₁, a₂, ···, a_r, b₁, b₂, ···, b_s-1), so it is a simple radical extension.

So we can continue using the same argument… and we are adding/subtracting (depending on how -the direction- you see it) b₁, …, b_s to every field of the chain.

F(a₁, a₂, ···, a_r, b₁) ⊆ [s.r., b₂^m₂ ∈ F(b₁) ⊆ F(a₁, a₂, ···, a_r, b₁)] F(a₁, a₂, ···, a_r, b₁, b₂)··· ⊆ F(a₁, a₂, ···, a_r, b₁, b₂, ···, b_s-2) ⊆ F(a₁, a₂, ···, a_r, b₁, b₂, ···, b_s-1).

We claim that F(a₁, a₂, ···, a_r) ⊆ F(a₁, a₂, ···, a_r, b₁). This is also a simple radical extension because b₁^m₁ ∈ F ⊆ F(a₁, a₂, ···, a_r)

And then, we can conclude, F(a₁) ⊆ F(a₁, a₂) ⊆ ··· ⊆ F(a₁, a₂, ···, a_r) and we are adding/subtracting (depending on how you see it) a₁, …, a_r to every field of the chain.

Putting all together, F(a₁) ⊆ F(a₁, a₂) ⊆ ··· ⊆ F(a₁, a₂, ···, a_r) ⊆ F(a₁, a₂, ···, a_r, b₁) ⊆ ··· ⊆ F(a₁, a₂, ···, a_r, b₁, b₂, ···, b_s-2) ⊆ F(a₁, a₂, ···, a_r, b₁, b₂, ···, b_s-1). Therefore, L₁L₂/F is a radical extension ∎

Proposition. Let F be field, F ⊆ ℂ (characteristic(F) = 0) Let L/F be a radical extension. Then, there exist a K/L extension such that K/F is both Galois and radical.

Proof.

Let L/F be a radical extension.

F ⊆ [s.r.] F(α₁) ⊆ [s.r.] F(α₁, α₂) ⊆ [s.r.] ··· ⊆ [s.r.] F(α₁, α₂, ···, α_n) = L.

Let S be the set of all F-conjugates of α₁, α₂, ···, α_n in some extension of L, say $\bar L$

F-conjugates of α₁ are the roots of the irreducible polynomial of α₁ over F.

Let K = F(S). Of course, α₁, α₂, ···, α_n ⊆ S, F(S) ⊇ L, but we claim that K is the required extension, i.e., K/F is Galois and radical. K is said to be the Galois closure of L/F.

Let f_i be the irreducible polynomial of α_i over F. Then, K is the splitting field of f₁f₂···f_n over F and we are in characteristic zero ⇒ K/F is Galois.

Let F ⊆ F(α) be a s.r. extension, αⁿ ∈ F and let σ: F(α) → L be an F-homomorphism. Let’s apply σ to F ⊆ F(α) ⇒ [σ is an F-homomorphism] σ(F) (= F) ⊆ σ(F(α)) ( = F(σ(α)) ) ⇒ F ⊆ F(σ(α)). Futhermore, σ(α)ⁿ = σ(αⁿ) = [By assumption, α_n ∈ F, σ fixes F] αⁿ ∈ F, that is, F ⊆ F(σ(α)) is simple radical, too.

F ⊆ F(α₁) ⊆ F(α₁, α₂) ⊆ ··· ⊆ F(α₁, α₂, ···, α_n) = L (*). Consider G = [We have already demonstrated that K/F is Galois] Gal(K/F) = {σ₁ = id, σ₂, ···, σ_r}

apply σ₁ to (*): F ⊆ F(α₁) ⊆ F(α₁, α₂) ⊆ ··· ⊆ F(α₁, α₂, ···, α_n) = L

apply σ₂ to (*): F ⊆ F(σ₂α₁) ⊆ F(σ₂α₁, σ₂α₂) ⊆ ··· ⊆ F(σ₂α₁, σ₂α₂, ···, σ₂α_n) = σ₂L, and all of these are simple radicals because of the previous proposition.

[…]

apply σ_r to (*): F ⊆ F(σ_rα₁) ⊆ F(σ_rα₁, σ_rα₂) ⊆ ··· ⊆ F(σ_rα₁, σ_rα₂, ···, σ_rα_n) = σ_rL, and all of these are simple radicals.

K = F(S) is the composite of L, σ₂L, σ₃L, ···, σ_rL and they are all radical extensions ⇒ [ L₁/F, L₂/F radical extensions ⇒ the composite L₁L₂/F is radical, too.] K/F is radical ∎

Recall K = Split_F(f₁·f₂···f_m) = F(S) = F(σL, σ ∈ Gal(K/F)), S is the set of all F-conjugates of α₁, α₂, ···, α_n in some extension of L, say $\bar L$

Bibliography

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. Field and Galois Theory, by Patrick Morandi. Springer.