Solvability by radicals

It’s not that I’m so smart, it’s just that I stay with problems longer, Albert Einstein

Various: 1. This is not very clear, even to me. 2. Babies are almost uniformly ugly. 3. We’ll do that by the celebrated method of brute force. 4. It takes years of practice to make easy things seem difficult. I’ve had that experience/skill, Anonymous.

A complex number α ∈ ℂ is, loosely speaking, expressible by radical over a field F ⊆ ℂ if α can be written or expressed exclusively using elements of F by the standards operations (+, -, x, ÷) and taking radicals, e.g., $\sqrt{-2}, \frac{3+4\sqrt{-3}}{7},\frac{2\sqrt{8}}{23+\sqrt{-2}}$ are expressible by radical over ℚ.

Definition. Let K/F be a finite field extension and let α ∈ K.

1. A simple radical extension is a simple extension K/F generated by a single element α satisfying αⁿ = b, i.e.,K = F(α) for some positive integer n. In other words, K is obtained from F by adjoining a radical, e.g., $ℚ(\sqrt{2})/ℚ, ℚ(w)/ℚ, ℚ(\sqrt[3]{2})/ℚ$
2. K/F is a radical extension if there is a tower of field extensions: F = F_o ⊆ F₁ ⊆ F₂ ⊆ ··· ⊆ F_r = K such that each extension F_i/F_i-1 is a simple radical extension, that is, F_i=F_i-1(α_i) with α_i^n_i ∈ F_i-1 or, alternatively, α_i is the root of an irreducible polynomial in F_i-1[x] of the form x^n_i -c_i.
3. An element α ∈ K which is algebraic over F is said to be solvable by radicals or simply solvable over F if exist a radical extension L/F such that α ∈ L, F = F_o ⊆ F₁ ⊆ F₂ ⊆ ··· ⊆ F_r = L, where F_i/F_i+1 is a simply radical extension.

4. An algebraic extension K/F is solvable over F if every element of K is solvable over F.

   The elements of K can be obtained from those of F by means of rational operations together with the taking of roots, e.g. F = ℚ, $(3+\sqrt{2})^{\frac{1}{5}}$ lies in a field F₂, where F₁ = ℚ(α₀), $α_0^2=2$, F₂ = F₁(α₁), $α_1^5=3+\sqrt{2}$.

5. A polynomial f ∈F[x] is solvable over F if all its roots in a splitting field are solvable over F.

Examples:

• Examples of simple radical extensions of ℚ are ℚ($\sqrt{2}$), ℚ($\sqrt[3]{2}$), and more generally, ℚ($\sqrt[n]{2}$).
• (char F ≠ 2) Any extension K/F of degree [K : F] = 2 is a simple radical.

Theorem Suppose F is a field, char(F) ≠ 2. Let K/F be a degree 2 extension. Then, K/F is Galois and K = F(α) where α² ∈ F. In other words, a degree 2 extension can be obtained by adding a square root of an element of F.

K/F is a degree 2 extension ⇒ K ≠ F ⇒ ∃α ∈ K \ F. Then, K = F(α). Let δ be the discriminant of the irreducible polynomial, δ = $\sqrt{b^2-4c}$. K = F(δ) and δ² ∈ F.

• Any cyclotomic extension is simple radical: F ⊆ F(ξ), ξⁿ = 1 ∈ F.
• $\mathbb{Q(\sqrt{2},\sqrt{3})}/\mathbb{Q}$ is a radical extension, ℚ ⊆ [deg 2 ⇒ simple radical] $\mathbb{Q(\sqrt{2})}$ ⊆ [$\mathbb{Q(\sqrt{2},\sqrt{3})}=(\mathbb{Q(\sqrt{2})})(\sqrt{3})$, deg 2 ⇒ simple radical] $\mathbb{Q(\sqrt{2},\sqrt{3})}/\mathbb{Q}$ but $\mathbb{Q(\sqrt{2},\sqrt{3})}/\mathbb{Q}$ is not a simple radical extension

K = $\mathbb{Q(\sqrt{2},\sqrt{3})}$Suppose for the sake of contradiction K = ℚ(α) with αⁿ ∈ ℚ. K is the splitting field of (x²-2)(x²-3) over ℚ (char(ℚ) = 0) ⇒ the extension is Galois, $\sqrt{2}, \sqrt{3}$ are linearly independent (there are four choices $\sqrt{2} →±\sqrt{2}, \sqrt{3}→±\sqrt{3}$) ⇒ [K : ℚ] = 4 ⇒ αⁿ ∈ ℚ, for some n ≥ 4 because the irreducible polynomial of α over ℚ has degree four.

σ ∈ Gal(K/Q), σ ≠ id ⇒ σ(α)ⁿ = [σ is an homomorphism: K → K] σ(αⁿ) = [αⁿ ∈ ℚ, σ fixes ℚ] αⁿ ⇒ σ(α) is also an nth root of αⁿ = {ξα | ξⁿ=1} ↭ σ(α) = ξα where ξ is a nth root of unity

K/Q is Galois ⇒ K/Q is normal [Every irreducible polynomial – xⁿ-α =0 – over F having a root in K – α ∈ K is obviously a root – splits completely over K] σα ∈ K ⊥ K = $\mathbb{Q(\sqrt{2},\sqrt{3})}⊆\mathbb{R}$, σ ∈ G, σα = {α, -α, α₃, α₄} [[K: ℚ] = |Gal(K/ℚ)| = 4, since K = ℚ(α), σa defines completely σ. n ≥ 4 implies σα = ξα for some σ], but α₃ and α₄ are of the form ξα ⇒ [every element, α in particular, other than zero has a multiplicative inverse in K] ξα ∈ K, α ∈ K ⇒ ξ ∈ K ⊆ ℝ but ξ ∉ ℝ since n ≥ 4.

• Let f ∈ ℚ[x] be an irreducible cubic polynomial which has three real root (e.g, x³ -3x +1), and K be the splitting field of f over ℚ (f has three real roots, K ⊆ ℝ). Let’s show that K/ℚ is not simple radical.

For the sake of contradiction, let’s suppose K = ℚ(α) ⇒ [K : ℚ] = 3 or 6. αⁿ ∈ ℚ, n ≥ 3. Let G = Gal(K/ℚ), σ ∈ G, σα = [|G| ≥ 3] {α, -α, ξα -for ξ an nth root of unity- ⊥, …}

ξα ∈ K, α ∈ K ⇒ ξ ∈ K. ξⁿ = 1, n ≥ 3, ξ ∉ ℝ, and K ⊆ ℝ ⊥

Recall. K/F is a cyclic extension if K/F is Galois and Gal(K/F) is cyclic.

Proposition. Let K/F be a cyclic extension. Let α ∈ L where L is an extension of K. Then, K(α)/F(α) is also a cyclic extension and [K(α) : F(α)] divides [K : F]

Proof.

K/F is a cyclic extension ⇒ K/F is Galois ⇒ K is the splitting field, say K = F(α₁, α₂, ···, α_r), of a separable polynomial f ∈ F[x] where α_i ∈ K are distinct roots of f ⇒ K(α) = F(α)(α₁, α₂, ···, α_r), f ∈ F[x] ⊆ F(α)[x], so K(α) is a splitting field of a separable polynomial over F(α) ⇒ K(α)/F(α) is Galois.

Let σ ∈ Gal(K(α)/F(α)), σ(α) = α. Consider σ|_K: K → K and that is the case (In general, σ|_K: K →K(α)) because K/F is Galois ⇒ K/F is normal ⇒ An F-automorphism of K which fixes F lands on K, i.e., σ(K) ⊆ K. Therefore, we get a group homomorphism: Gal(K(α)/F(α)) → Gal(K/F), σ → σ|_K.

It is obviously a group homomorphism, and it is also one-to-one, because σ∈ Ker(Gal(K(α)/F(α))) ⇒ σ|_K = id, but we know σ(α) = α ⇒ σ = id on K(α) ⇒ Gal(K(α)/F(α)) is isomorphic to a subgroup of Gal(K/F). By assumption, 🔑Gal(K/F) is cyclic and a subgroup of a cyclic group is cyclic (⇒ Gal(K(α)/F(α)) is cyclic) and its order divides the order of the group, that is, |Gal(K(α)/F(α))| = [K(α)/F(α) is Galois] [K(α) : F(α)] divides |Gal(K/F)| = [K : F].

Theorem. A degree six Galois extension K/F (char(F) ≠ 2 or 3) is solvable.

Proof.

[K : F] = 6, G = Gal(K/F), |G| = [By assumption, K/F is Galois] [K : F] = 6 ⇒ G ≋ ℤ/6ℤ or S₃, in either case, G has a normal subgroup H ◁ G of order 3, [G : H] = 2. We could apply the fundamental theorem of Galois theory,

K/F is Galois ⇒ K/L is Galois ⇒ |Gal(K/L)| = [K : K^H] = 3 ⇒ Gal(K/L) ≋ ℤ/3ℤ

Let w be a primitive 3rd root of unity in an extension of L.

K/L is not only Galois, but cyclic (Gal(K/L) ≋ ℤ/3ℤ) ⇒ [Proposition. Let K/F be a cyclic extension. Let α ∈ L where L is an extension of K. Then, K(α)/F(α) is also a cyclic extension and [K(α) : F(α)] divides [K : F]] K(w)/L(w) is cyclic.

K(w)/L(w) is cyclic and L(w) does contain a primitive 3rd of unity by construction, namely w ⇒ [Cyclic ⇒ Kummer. Let F field containing a primitive nth root of unity, K/F is a cyclic extension of degree n ⇒ K is the splitting field of an irreducible polynomial xⁿ-a over F, a ∈ F. Futhermore, K = F(α) where a = αⁿ] In other words, K/F is a Kummer extension ⇒ ∃a ∈ L(w): xⁿ-a is irreducible, K(w) = L(w)($\sqrt[n]{a}$) ⇒ K(w)/L(w) is a simple radical extension.

Now, we only need to consider the following tower, F ⊆ [s.r. because [L:F] = 2] L ⊆ [s.r. because w³ = 1 ∈ L, Cyclotomic extension] L(w) ⊆ [s.r. by the previous argument] K(w) ⇒ K(w)/F is a radical extension (there is a tower of simple radical extensions ending in K(w)) and K ⊆ K(w) ⇒ Every element of K is solvable over F ⇒ K is solvable over F ∎

Theorem. Let F be a field of characteristic zero. Let K/F be a cyclic extension. Then, the extension K/F is solvable.

Proof.

Let [K : F] = n, let ξ be a primitive nth root of unity in an extension of K.

K/F cyclic ⇒ K(ξ)/F(ξ) cyclic ⇒ [cyclic and ξ ∈ F(ξ), by the same argument as before] K(ξ)/F(ξ) is simple radical. Therefore, F ⊆ [This is a cyclotomic extension, so it is simple radical, ξⁿ=1] F(ξ) ⊆ [Kummer ⇒ simple radical] K(ξ) ⇒ K(ξ)/F is radical ⇒ K/F is solvable∎

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).