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 L_{1} and L_{2} be subfields of K. The composite or composium of the fields L_{1}, L_{2} in K, denoted by L_{1}L_{2}, is defined to be the smallest subfield of K containing both fields L_{1}, L_{2} 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 L_{1}/F and L_{2}/F are both finite-degree sub-extensions of K/F, F ⊆ L_{1} ⊆ K, L_{1} = F(α_{1}, ···, α_{r}), F ⊆ L_{2} ⊆ K, L_{2} = F(β_{1}, ···, β_{s}). Then, L_{1}L_{2} = F(α_{1}, ···, α_{r}, β_{1}, ···, β_{s}).

Proof.

F ⊆ L_{1} ⊆ K, L_{1} = F(α_{1}, ···, α_{r}), F ⊆ L_{2} ⊆ K, L_{2} = F(β_{1}, ···, β_{s}).

L_{1}L_{2} contains F and each of α_{1}, ···, α_{r} (it contains L_{1}) and β_{1}, ···, β_{s} (it contains L_{2}), hence it contains F(α_{1}, ···, α_{r}, β_{1}, ···, β_{s}).

Conversely, F(α_{1}, ···, α_{r}, β_{1}, ···, β_{s}) contains both L_{1} and L_{2}, hence it contains L_{1}L_{2} (the intersection of all subfields containing both subfields, that is, the smallest subfield containing both subfields L_{1} and L_{2}) ⇒ L_{1}L_{2} = F(α_{1}, ···, α_{r}, β_{1}, ···, β_{s})∎

**Degrees of Composites.** The degree of a composite L_{1}L_{2} is smaller than the product of the degrees of the fields that generate the field. (i) If L_{1}/F and L_{2}/F are both finite-degree subextensions of K/F, then [L_{1}L_{2} : F] ≤ [L_{1} : F] · [L_{2} : F]. Futhermore, (ii) if the degrees [L_{1} : F] and [L_{2} : F] are relatively prime, then equality holds.

Proof.

By the previous result, L_{1}L_{2} = F(α_{1}, ···, α_{r}, β_{1}, ···, β_{s}) = L_{1}(β_{1}, ···, β_{s}) ⇒ [**β _{1}, ···, β_{s} span L_{1}L_{2}/L_{1}**] [L

Futhermore, [L_{1}L_{2} : F] = [L_{1}L_{2} : L_{1}] · [L_{1} : F] ≤ [L_{2} : F] · [L_{1} : F] ∎ (i)

Besides, [L_{1}L_{2} : F] = [L_{1}L_{2} : L_{1}] · [L_{1} : F], so [L_{1}L_{2} : F] is divisible by both [L_{1} : F] and [L_{2} : F], say n and m respectively. If m and n are relatively prime, then [L_{1}L_{2} : F] is divisible by n·m, so [L_{1}L_{2} : F] ≥ n·m ⇒ [(i) **[L _{1}L_{2} : F] ≤ n·m**] [L

Proposition. If L_{1}/F and L_{2}/F are both finite-degree Galois extensions of K inside a common field F, then L_{1}L_{2}/F is a Galois finite extension.

Proof.

Each L_{i} is the splitting field of a polynomial f_{i} over F. Then, L_{1}L_{2} is the splitting field for the product of f_{1} and f_{2} because L_{1}L_{2} contains L_{1}, so f_{1}(x) splits in this field. Same argument applies to f_{2}(x). So the polynomial f_{1}(x)f_{2}(x) will also split completely in L_{1}L_{2}. Therefore, L_{1}L_{2} is normal.

As L_{1}/F is Galois, L_{1} is the splitting field of a separable polynomial f(x) ∈ F[x] ⊆ L_{2}[x], the roots of f(x) generate L_{1} over F, so they generate L_{1}L_{2} over L_{2} ⇒ L_{1}L_{2} is the splitting field of a separable polynomial f(x) over L_{2} and therefore L_{1}L_{2}/L_{2} is Galois, hence separable.

L_{2}/F is normal ⇒ separable. Then, L_{1}L_{2}/L_{2} separable, L_{2}/F separable ⇒ [Separable is transitive] L_{1}L_{2}/F is separable ⇒ [We have proved that it is normal, too] L_{1}L_{2}/F is Galois∎

Example: L_{1} = $ℚ(\sqrt{2})$, $ℚ(\sqrt{2})/ℚ$ is Galois, L_{2} = $ℚ(\sqrt{3})$, $ℚ(\sqrt{3})/ℚ$ is Galois, the composite of L_{1}, L_{2} is L_{1}L_{2} = $ℚ(\sqrt{2}, \sqrt{3})$ and $ℚ(\sqrt{2}, \sqrt{3})/ℚ$ is Galois.

Proposition. **The composite of two radical extensions is radical.** L_{1}/F, L_{2}/F radical extensions ⇒ the composite L_{1}L_{2}/F is radical, too.

Observation. We need an ambient field K containing both fields, say L_{1} and L_{2}, in order to be able to make sense or talk about the composite L_{1}L_{2}. Typically, K = $\bar F$.

Proof.

L_{1}/F radical ⇒ ∃ F ⊆ [F(a_{1})/F is simple radical where a_{1}^{n1} ∈ F] F(a_{1}) ⊆ [F(a_{2})/F(a_{1}) is a simple radical where a_{2}^{n2} ∈ F(a_{1})] F(a_{1}, a_{2}) ⊆ ··· F(a_{1}, a_{2}, ···, a_{r}) = L_{1}

L_{2}/F radical ⇒ F ⊆ [simple radical, b_{1}^{m1} ∈ F] F(b_{1}) ⊆ [simple radical] F(b_{1}, b_{2}) ⊆ ··· F(b_{1}, b_{2}, ···, b_{s-1}) ⊆ [s.r., b_{s}^{ms} ∈ F(b_{1}, b_{2}, ···, b_{s-1})] F(b_{1}, b_{2}, ···, b_{s}) = L_{2}

We know L_{1}L_{2} = F(a_{1}, a_{2}, ···, a_{r}, b_{1}, b_{2}, ···, b_{s})

Consider F(a_{1}, a_{2}, ···, a_{r}, b_{1}, b_{2}, ···, b_{s-1}) ⊆ F(a_{1}, a_{2}, ···, a_{r}, b_{1}, b_{2}, ···, b_{s}). This extension is generated by b_{s}, but we know that b_{s}^{ms} ∈ F(b_{1}, b_{2}, ···, b_{s-1}) ⊆ F(a_{1}, a_{2}, ···, a_{r}, b_{1}, b_{2}, ···, 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) b_{1}, …, b_{s} to every field of the chain.

F(a_{1}, a_{2}, ···, a_{r}, b_{1}) ⊆ [s.r., b_{2}^{m2} ∈ F(b_{1}) ⊆ F(a_{1}, a_{2}, ···, a_{r}, b_{1})] F(a_{1}, a_{2}, ···, a_{r}, b_{1}, b_{2})··· ⊆ F(a_{1}, a_{2}, ···, a_{r}, b_{1}, b_{2}, ···, b_{s-2}) ⊆ F(a_{1}, a_{2}, ···, a_{r}, b_{1}, b_{2}, ···, b_{s-1}).

We claim that F(a_{1}, a_{2}, ···, a_{r}) ⊆ F(a_{1}, a_{2}, ···, a_{r}, b_{1}). This is also a simple radical extension because b_{1}^{m1} ∈ F ⊆ F(a_{1}, a_{2}, ···, a_{r})

And then, we can conclude, F(a_{1}) ⊆ F(a_{1}, a_{2}) ⊆ ··· ⊆ F(a_{1}, a_{2}, ···, a_{r}) and we are adding/subtracting (depending on how you see it) a_{1}, …, a_{r} to every field of the chain.

Putting all together, F(a_{1}) ⊆ F(a_{1}, a_{2}) ⊆ ··· ⊆ F(a_{1}, a_{2}, ···, a_{r}) ⊆ F(a_{1}, a_{2}, ···, a_{r}, b_{1}) ⊆ ··· ⊆ F(a_{1}, a_{2}, ···, a_{r}, b_{1}, b_{2}, ···, b_{s-2}) ⊆ F(a_{1}, a_{2}, ···, a_{r}, b_{1}, b_{2}, ···, b_{s-1}). Therefore, L_{1}L_{2}/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(α_{1}) ⊆ [s.r.] F(α_{1}, α_{2}) ⊆ [s.r.] ··· ⊆ [s.r.] F(α_{1}, α_{2}, ···, α_{n}) = L.

Let S be the set of all F-conjugates of α_{1}, α_{2}, ···, α_{n} in some extension of L, say $\bar L$

F-conjugates of α_{1} are the roots of the irreducible polynomial of α_{1} over F.

Let K = F(S). Of course, α_{1}, α_{2}, ···, 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 f_{1}f_{2}···f_{n} over F and we are in characteristic zero ⇒ K/F is Galois.

Let F ⊆ F(α) be a s.r. extension, α^{n} ∈ 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(α

F ⊆ F(α_{1}) ⊆ F(α_{1}, α_{2}) ⊆ ··· ⊆ F(α_{1}, α_{2}, ···, α_{n}) = L (*). Consider G = [We have already demonstrated that K/F is Galois] Gal(K/F) = {σ_{1} = id, σ_{2}, ···, σ_{r}}

apply σ_{1} to (*): F ⊆ F(α_{1}) ⊆ F(α_{1}, α_{2}) ⊆ ··· ⊆ F(α_{1}, α_{2}, ···, α_{n}) = L

apply σ_{2} to (*): F ⊆ F(σ_{2}α_{1}) ⊆ F(σ_{2}α_{1}, σ_{2}α_{2}) ⊆ ··· ⊆ F(σ_{2}α_{1}, σ_{2}α_{2}, ···, σ_{2}α_{n}) = σ_{2}L, and all of these are simple radicals because of the previous proposition.

[…]

apply σ_{r} to (*): F ⊆ F(σ_{r}α_{1}) ⊆ F(σ_{r}α_{1}, σ_{r}α_{2}) ⊆ ··· ⊆ F(σ_{r}α_{1}, σ_{r}α_{2}, ···, σ_{r}α_{n}) = σ_{r}L, and all of these are simple radicals.

**K = F(S) is the composite of L, σ _{2}L, σ_{3}L, ···, σ_{r}L** and they are all radical extensions ⇒ [ L

Recall K = Split_{F}(f_{1}·f_{2}···f_{m}) = F(S) = F(σL, σ ∈ Gal(K/F)), S is the set of all F-conjugates of α_{1}, α_{2}, ···, α_{n} in some extension of L, say $\bar L$

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