    # Characterizations of solvability. Cubic polynomials.

Various: 1. This is proof by repeated assertion. 2. Let’s just get right to it and I’ll give you the proof… God willing. 3. The question you should always ask is, ‘Who cares?’ 4. “These matrices wear their eigenvalues on their sleeves ‘cause they love you, Anonymous.

Theorem. Let F be a subfield of ℂ, F ⊆ ℂ, and let α ∈ ℂ be algebraic over F. Then, the following statement are all equivalent:

1. α can be expressed by radicals, solved for in terms of radicals, or simply solvable over F if α can be obtained by a chain, succession, or tower of simple radical extensions, i.e., F = F0 ⊆ F1 ⊆ ··· ⊆ Fr such that Fi/Fi-1 is a simple radical ∀i and α ∈ Fr where Fi = Fi-1i) with αini ∈Fi-1 or Fi= Fi-1($\sqrt[n_i]{a_i}$) and $\sqrt[n_i]{a_i}$ denotes some root of the polynomial xni -ai.
2. There exists a chain or tower of fields: F = L0 ⊆ L1 ⊆ ··· ⊆ Ln such that α ∈ Ln and each Li/Li-1 is Abelian, i.e., Galois and the Galois group is Abelian.
3. There exists a chain or tower of fields: F = K0 ⊆ K1 ⊆ ··· ⊆ Km such that α ∈ Km and each Ki/Ki-1 is cyclic, i.e., Galois and the Galois group is cyclic.
4. There exists a chain or tower of fields: F = M0 ⊆ M1 ⊆ ··· ⊆ Ms such that α ∈ Ms and each Mi/Mi-1 is cyclic of primer order, i.e., Galois, the Galois group is cyclic, and the order of each partial extension [Mi : Mi-1] is prime. Proof.

(1 ⇒ 2) α is solvable over F, F = F0 ⊆ F1 ⊆ ··· ⊆ Fr, α ∈ Fr, Fi/Fi-1 is a simple radical, let’s suppose Fi = Fi-1i) with αini ∈ Fi-1.

Let us consider the corresponding ξn1, ξn2, ···, ξnr primitive n1-th, n2-th, ···, nr-th roots of unity (they all live in ℂ).

Consider now the following chain of subfields (we adjoin to F the ni-th roots of unity): Fo = F ⊆ Fon1) ⊆ Fon1, ξn2) ⊆ ··· ⊆ Fon1, ξn2, ···, ξnr) ⊆ F1n1, ξn2, ···, ξnr) ⊆ ··· ⊆ Frn1, ξn2, ···, ξnr), α ∈ Fr

This is a chain or tower where each extension is Abelian:

• Fo = F ⊆ Fon1) is a cyclotomic extension ⇒ Cyclotomic extensions Let K/F be the splitting field of xn-1 over F. Then, there is an injective group homomorphism between the Galois group of the cyclotomic extension and the multiplicative group (ℤ/nℤ)*, i.e., Φ: Gal(K/F) → (ℤ/nℤ)*. In particular, Cyclotomic extensions are always Abelian.
• By the same argument, Fon1) ⊆ Fon1, ξn2) is Abelian, and so on, and so forth (Fon1, ξn2, ···, ξni)/Fon1, ξn2, ···, ξni-1) is Abelian), up to Fon1, ξn2, ···, ξnr).
• What about Fon1, ξn2, ···, ξnr) ⊆ F1n1, ξn2, ···, ξnr)?

F1 = F01) where α1n1 ∈ F0⊆ Fon1, ξn2, ···, ξnr). F1n1, ξn2, ···, ξnr) = F0n1, ξn2, ···, ξnr)(α1) where α1n1 ∈ F0⊆ Fon1, ξn2, ···, ξnr) so the extension F1n1, ξn2, ···, ξnr) is a splitting field of xn11n1 over F0n1, ξn2, ···, ξnr) and F0n1, ξn2, ···, ξnr) contains ξn1 ⇒ [F field containing a primitive nth root of unity, K splitting field of xn - a over F. Then, K/F is a cyclic extension] F1n1, ξn2, ···, ξnr)/F0n1, ξn2, ···, ξnr) is cyclic ⇒ [All cyclic groups are Abelian, but an Abelian group is not necessarily cyclic] Abelian.

F1n1, ξn2, ···, ξnr) is a splitting field of xn11n1 over F0n1, ξn2, ···, ξnr) because once that α1 and ξn1 are in F1n1, ξn2, ···, ξnr), all the other roots of xn11n1 are in F1n1, ξn2, ···, ξnr), too.

• Using the same argument, F1n1, ξn2, ···, ξnr) ⊆ F2n1, ξn2, ···, ξnr) is Abelian, too, and so on up to Fr-1n1, ξn2, ···, ξnr) ⊆ Frn1, ξn2, ···, ξnr)∎

2 ⇒ 3) Objective: There exists a chain or tower of fields F = F0 ⊆ F1 ⊆ ··· ⊆ Fr, such that Fi/Fi-1 is cyclic

Let G = Gal(K/F). By assumption, G is Abelian. If G is cyclic, then we are done.

Suppose G is Abelian, but not cyclic and let H be a proper cyclic subgroup of G. Consider,

Let’s apply induction to KH/F. K/KH is cyclic because K/KH is Galois (if G is Galois, K/KH is always Galois) and its Galois group (H) is cyclic.

G is Abelian ⇒ Every subgroup of an Abelian group is normal ⇒ [(1) An intermediate field L is Galois over F ↭ Gal(K/L) is a normal subgroup of G] The intermediate field KH/F is (1) Galois, (2) Abelian (its Galois group is Abelian because G is Abelian and any subgroup of an Abelian group is also Abelian), and (3) [KH : F] < [K : F] (H is a proper subgroup of G), then we can induct on [KH : F] and have a tower of cyclic extensions from F to KH, K/KH is also cyclic, so we are done.

F = F0 ⊆ F1 ⊆ ··· Fr-1 = KH ⊆ Fr = K, such that Fi/Fi-1 is cyclic.

3 ⇒ 1) By assumption, we have a sequence, tower, or chain F = K0 ⊆ K1 ⊆ ··· ⊆ Km such that α ∈ Km and each Ki/Ki-1 is cyclic. Let ni = [Ki : Ki-1], and let ξn1, ξn2, ···, ξnm ∈ ℂ be the corresponding primitive n1-th, n2-th, ···, nm-th roots of unity.

Consider the following tower of fields (we adjoin the ni-th roots of unity),

K0 ⊆ K0n1) ⊆ K0n1, ξn2) ⊆ ··· ⊆ K0n1, ξn2,···, ξnm) ⊆ K1n1, ξn2,···, ξnm) ⊆ ··· ⊆ Kmn1, ξn2,···, ξnm)

• K0 ⊆ K0n1) is a simple radical extension because it is a cyclotomic extension.
• K0n1) ⊆ K0n1, ξn2), it is a simple radical extension because we are adding ξn2, that is, a primitive nth root of unity. The same argument goes up to K0n1, ξn2,···, ξnm).
• K0n1, ξn2,···, ξnm) ⊆ K1n1, ξn2,···, ξnm).

Recall 1. 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]

Recall 2. F is a field containing a primitive nth root of unity. K/F a cyclic extension of n = [K : F]. Then, K is the splitting field of an irreducible polynomial xn-a over F.

1. All field extensions are cyclic.
2. [K1 : K0] = n1. Their base fields contain ξn1. Any of these extensions, their orders, say n’1 divides n1 (by the same proposition -see Recall-), e.g., [K1n1): K0n1)] divides [K1 : K0] = n1, [K2 : K1] | [K1 : K0] and therefore [K2 : K1] | n1 and so on, and so forth, [K1n1, ξn2,···, ξnm): K0n1, ξn2,···, ξnm)] = n1’ divides n1.
3. [K1n1, ξn2,···, ξnm): K0n1, ξn2,···, ξnm)] = n’1 where n’1|n1, ξn1n1/n’1 is a primitive n’1-th root of unity that is contained in the base field. This is trivial, $ξ_{n_1}$ is a n1 primitive root of unity, $ξ_{n_1}^{n_1}=1$, $(ξ_{n_1}^{\frac{n_1}{n_{1’}}})^{n_1’}=ξ_{n_1}^{\frac{n_1}{n_{1’}}·n_1’}=ξ_{n_1}^{n_1}=1$
4. Therefore, [Recall 2] K1n1, ξn2,···, ξnm) is the splitting field of an irreducible polynomial xn’1 -a’ over K0n1, ξn2,···, ξnm). Futhermore, K1n1, ξn2,···, ξnm) = K0n1, ξn2,···, ξnm)($\sqrt[n’_1]{a’}$) and therefore K1n1, ξn2,···, ξnm)/K0n1, ξn2,···, ξnm) is a simple radical extension.
5. The same argument goes up to Kmn1, ξn2,···, ξnm), all the extensions in the tower are radical extensions.
6. α ∈ Km ⊆ Kmn1, ξn2,···, ξnm).

Therefore, α is solvable over F∎

At this point we step back centuries. The procedure for solving quadratic equations can be traced back to the 🏅golden🏅 age of Babylon.

Theorem. Let F be a field, F ⊆ ℂ , and let f∈ F[x] be a polynomial of degree ≤ 4. Then, f is solvable over F, i.e.,all roots of f are solvable over F. In general, this is not the case for n ≥ 5, but they do exist, e.g., xn -1 is solvable ∀n.

Proof.

Recall 1 Let F ⊆ ℂ, let f ∈ ℚ[x] be a polynomial of degree n, and let K be the splitting field for f over ℚ. Then, f is irreducible if and only if Gal(f) is a transitive subgroup of Sn.

• For the linear equation, deg(f) = 1, f = x -a, a ∈ F, the unique solution, namely -a, is a rational function of the coefficients, and therefore, f is solvable over F.

• In the case of a quadratic equation, deg(f) = 2, f = x2 +bx + c. The quadratic formula does the job, that is, the roots are $\frac{-b+\sqrt{Δ}}{2}$ and $\frac{-b-\sqrt{Δ}}{2}$ where Δ = $b^2-4c$. The number Δ is referred to as the discriminant of the equation. The roots in general belong not to ℚ, but to the extension ℚ($\sqrt{Δ}$). The extension ℚ($\sqrt{Δ}$)/ℚ is at most a two degree extension [Any degree 2 extension is radical] and ℚ($\sqrt{Δ}$) = ℚ if $\sqrt{b^2-4c}$ ∈ ℚ. The Galois group of f is S2, a cyclic group of order 2.

Recall 2: 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.

• deg(f) = 3. Let K be the splitting field of f over F(=ℚ), so K/ℚ is Galois. There are different possibilities:
1. K = F (we will use F and ℚ interchangeable). It is trivially solvable.
2. [K : F] = 2. Any degree 2 extension is radical ⇒ it is solvable.
3. [K : F] = 3, K/F is cyclic (≋ ℤ/ℤ3) ⇒ [The chain consists of only one link, F ⊆ K, K/F cyclic] K/F solvable ⇒ f is solvable over F.
4. [K : F] = 6. A degree six Galois extension K/F (char (F) ≠ 2 or 3) is solvable.
• Quartic are solvable. Let F be a field, F ⊆ ℂ, G be the Galois group of an irreducible, separable quartic polynomial f, G = Gal(K/F) = Gal(f). Then, f is solvable over F.

Cubic polynomials. Let f(x) ∈ F[x] be an irreducible, separable polynomial of degree 3 over F, and let K be the splitting field of f over F. If D is the discriminant of f, then Gal(K/F) ≋ S3 if and only if δ = $\sqrt{D}$ ∉ F (D is not a square in F), and Gal(K/F) ≋ A3 if and only if δ ∈ F.

Let’s assume f is irreducible, otherwise it is the product of a linear and a quadratic polynomials and both are solvable.

f is irreducible ⇒ [Let F ⊆ ℂ, f ∈ F[x], deg(f) = n ≥ 1, f irreducible, then G is isomorphic to a transitive subgroup of Sn] Gal(f) is a transitive subgroup of S3. There are only 2 transitive subgroups of S3, namely, S3 and A3 = {(), (123), (132)}.

Recall Let F be a field with char(F)≠2, let f(x)∈ F[x] be an irreducible, separable polynomial, and let K be the splitting field of f(x) over F (or K be a splitting field of an irreducible polynomial over ℚ). If $δ = \prod_{i < j} (α_i-α_j)$ ∈ F (D = Δ is a square in F), then G ⊆ An. Otherwise, G $\not\subset A_n$.

There are two options, considering $δ = \sqrt{D}$:

1. δ ∈ F ↭ [G ⊆ A3, and there are only two possibilities A3 and S3] G = A3, i.e., the Galois group of its splitting field over ℚ is A3 and [K : F] = 3, e.g., f = x3 -3x + 1 ∈ ℚ[x] is irreducible, Disc(f) = [-4a23a0 + a22a12 + 18a2a1a0 -4a13 -27a02] 34 is a square in ℚ (δ = 9 ∈ ℚ). Hence, [K : F] = 3 and G = A3.

x→ x-1, x3 -3x2 +3 is irreducible by Eisenstein, p = 3 divides ai -3 and 3, 0 ≤ i < n, p does not divide an = 1, p2 = 9 does not divide a0

2. δ ∉ F ↭ G [G $\not\subset A_3$] = S3 and [K : F] = 6, e.g., x3 +3x +1 ∈ ℚ[x] is irreducible, Disc(f) = [-4a23a0 + a22a12 + 18a2a1a0 -4a13 -27a02] -5·33 is not a square in ℚ (δ = $\sqrt{-135}$ ∉ ℚ). Hence, [K : F] = 6 and G = S3.

x→ x-1, f(x-1) = x3 -3x2 + 6x -3 is irreducible by Eisenstein, p = 3 divides -3, 6, and -3, but does not divide 1, and p2=9 does not divide -3.

Another example is x3 -4x + 2. It has discriminant 148 = 22·37, hence it is not a square in ℚ, so the corresponding Galois group is S3.

# Bibliography 