    # 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, ···, αr), F ⊆ L2 ⊆ K, L2 = F(β1, ···, βs). Then, L1L2 = F(α1, ···, αr, β1, ···, βs).

Proof.

F ⊆ L1 ⊆ K, L1 = F(α1, ···, αr), F ⊆ L2 ⊆ K, L2 = F(β1, ···, βs).

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

Conversely, F(α1, ···, αr, β1, ···, βs) 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, ···, αr, β1, ···, βs)∎

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, ···, αr, β1, ···, βs) = L11, ···, βs) ⇒ [β1, ···, βs 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 Li is the splitting field of a polynomial fi 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.

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 a1n1 ∈ F] F(a1) ⊆ [F(a2)/F(a1) is a simple radical where a2n2 ∈ F(a1)] F(a1, a2) ⊆ ··· F(a1, a2, ···, ar) = L1

L2/F radical ⇒ F ⊆ [simple radical, b1m1 ∈ F] F(b1) ⊆ [simple radical] F(b1, b2) ⊆ ··· F(b1, b2, ···, bs-1) ⊆ [s.r., bsms ∈ F(b1, b2, ···, bs-1)] F(b1, b2, ···, bs) = L2

We know L1L2 = F(a1, a2, ···, ar, b1, b2, ···, bs)

Consider F(a1, a2, ···, ar, b1, b2, ···, bs-1) ⊆ F(a1, a2, ···, ar, b1, b2, ···, bs). This extension is generated by bs, but we know that bsms ∈ F(b1, b2, ···, bs-1) ⊆ F(a1, a2, ···, ar, b1, b2, ···, bs-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, …, bs to every field of the chain.

F(a1, a2, ···, ar, b1) ⊆ [s.r., b2m2 ∈ F(b1) ⊆ F(a1, a2, ···, ar, b1)] F(a1, a2, ···, ar, b1, b2)··· ⊆ F(a1, a2, ···, ar, b1, b2, ···, bs-2) ⊆ F(a1, a2, ···, ar, b1, b2, ···, bs-1).

We claim that F(a1, a2, ···, ar) ⊆ F(a1, a2, ···, ar, b1). This is also a simple radical extension because b1m1 ∈ F ⊆ F(a1, a2, ···, ar)

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

Putting all together, F(a1) ⊆ F(a1, a2) ⊆ ··· ⊆ F(a1, a2, ···, ar) ⊆ F(a1, a2, ···, ar, b1) ⊆ ··· ⊆ F(a1, a2, ···, ar, b1, b2, ···, bs-2) ⊆ F(a1, a2, ···, ar, b1, b2, ···, bs-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, ···, αn) = L.

Let S be the set of all F-conjugates of α1, α2, ···, αn 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, ···, αn ⊆ 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 fi be the irreducible polynomial of αi over F. Then, K is the splitting field of f1f2···fn 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, αn ∈ 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, ···, αn) = L (*). Consider G = [We have already demonstrated that K/F is Galois] Gal(K/F) = {σ1 = id, σ2, ···, σr}

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

apply σ2 to (*): F ⊆ F(σ2α1) ⊆ F(σ2α1, σ2α2) ⊆ ··· ⊆ F(σ2α1, σ2α2, ···, σ2αn) = σ2L, and all of these are simple radicals because of the previous proposition.

[…]

apply σr to (*): F ⊆ F(σrα1) ⊆ F(σrα1, σrα2) ⊆ ··· ⊆ F(σrα1, σrα2, ···, σrαn) = σrL, and all of these are simple radicals.

K = F(S) is the composite of L, σ2L, σ3L, ···, σrL 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 = SplitF(f1·f2···fm) = F(S) = F(σL, σ ∈ Gal(K/F)), S is the set of all F-conjugates of α1, α2, ···, αn in some extension of L, say $\bar L$

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