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 ⊆ ··· ⊆ F_r such that F_i/F_i-1 is a simple radical ∀i and α ∈ F_r where F_i = F_i-1(α_i) with α_i^n_i ∈F_i-1 or F_i= F_i-1($\sqrt[n_i]{a_i}$) and $\sqrt[n_i]{a_i}$ denotes some root of the polynomial x^n_i -a_i.
2. There exists a chain or tower of fields: F = L₀ ⊆ L₁ ⊆ ··· ⊆ L_n such that α ∈ L_n and each L_i/L_i-1 is Abelian, i.e., Galois and the Galois group is Abelian.
3. There exists a chain or tower of fields: F = K₀ ⊆ K₁ ⊆ ··· ⊆ K_m such that α ∈ K_m and each K_i/K_i-1 is cyclic, i.e., Galois and the Galois group is cyclic.
4. There exists a chain or tower of fields: F = M₀ ⊆ M₁ ⊆ ··· ⊆ M_s such that α ∈ M_s and each M_i/M_i-1 is cyclic of primer order, i.e., Galois, the Galois group is cyclic, and the order of each partial extension [M_i : M_i-1] is prime.

Proof.

(1 ⇒ 2) α is solvable over F, F = F₀ ⊆ F₁ ⊆ ··· ⊆ F_r, α ∈ F_r, F_i/F_i-1 is a simple radical, let’s suppose F_i = F_i-1(α_i) with α_i^n_i ∈ F_i-1.

Let us consider the corresponding ξ_n1, ξ_n2, ···, ξ_nr primitive n₁-th, n₂-th, ···, n_r-th roots of unity (they all live in ℂ).

Consider now the following chain of subfields (we adjoin to F the n_i-th roots of unity): F_o = F ⊆ F_o(ξ_n1) ⊆ F_o(ξ_n1, ξ_n2) ⊆ ··· ⊆ F_o(ξ_n1, ξ_n2, ···, ξ_nr) ⊆ F₁(ξ_n1, ξ_n2, ···, ξ_nr) ⊆ ··· ⊆ F_r(ξ_n1, ξ_n2, ···, ξ_nr), α ∈ F_r

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

• F_o = F ⊆ F_o(ξ_n1) 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, F_o(ξ_n1) ⊆ F_o(ξ_n1, ξ_n2) is Abelian, and so on, and so forth (F_o(ξ_n1, ξ_n2, ···, ξ_ni)/F_o(ξ_n1, ξ_n2, ···, ξ_ni-1) is Abelian), up to F_o(ξ_n1, ξ_n2, ···, ξ_nr).
• What about F_o(ξ_n1, ξ_n2, ···, ξ_nr) ⊆ F₁(ξ_n1, ξ_n2, ···, ξ_nr)?

F₁ = F₀(α₁) where α₁^n₁ ∈ F₀⊆ F_o(ξ_n1, ξ_n2, ···, ξ_nr). F₁(ξ_n1, ξ_n2, ···, ξ_nr) = F₀(ξ_n1, ξ_n2, ···, ξ_nr)(α₁) where α₁^n₁ ∈ F₀⊆ F_o(ξ_n1, ξ_n2, ···, ξ_nr) so the extension F₁(ξ_n1, ξ_n2, ···, ξ_nr) is a splitting field of x^n₁ -α₁^n₁ over F₀(ξ_n1, ξ_n2, ···, ξ_nr) and F₀(ξ_n1, ξ_n2, ···, ξ_nr) contains ξ_n1 ⇒ [F field containing a primitive nth root of unity, K splitting field of xⁿ - a over F. Then, K/F is a cyclic extension] F₁(ξ_n1, ξ_n2, ···, ξ_nr)/F₀(ξ_n1, ξ_n2, ···, ξ_nr) is cyclic ⇒ [All cyclic groups are Abelian, but an Abelian group is not necessarily cyclic] Abelian.

F₁(ξ_n1, ξ_n2, ···, ξ_nr) is a splitting field of x^n₁ -α₁^n₁ over F₀(ξ_n1, ξ_n2, ···, ξ_nr) because once that α₁ and ξ_n1 are in F₁(ξ_n1, ξ_n2, ···, ξ_nr), all the other roots of x^n₁ -α₁^n₁ are in F₁(ξ_n1, ξ_n2, ···, ξ_nr), too.

• Using the same argument, F₁(ξ_n1, ξ_n2, ···, ξ_nr) ⊆ F₂(ξ_n1, ξ_n2, ···, ξ_nr) is Abelian, too, and so on up to F_r-1(ξ_n1, ξ_n2, ···, ξ_nr) ⊆ F_r(ξ_n1, ξ_n2, ···, ξ_nr)∎

2 ⇒ 3) Objective: There exists a chain or tower of fields F = F₀ ⊆ F₁ ⊆ ··· ⊆ F_r, such that F_i/F_i-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 K^H/F. K/K^H is cyclic because K/K^H is Galois (if G is Galois, K/K^H 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 K^H/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) [K^H : F] < [K : F] (H is a proper subgroup of G), then we can induct on [K^H : F] and have a tower of cyclic extensions from F to K^H, K/K^H is also cyclic, so we are done.

F = F₀ ⊆ F₁ ⊆ ··· F_r-1 = K^H ⊆ F_r = K, such that F_i/F_i-1 is cyclic.

3 ⇒ 1) By assumption, we have a sequence, tower, or chain F = K₀ ⊆ K₁ ⊆ ··· ⊆ K_m such that α ∈ K_m and each K_i/K_i-1 is cyclic. Let n_i = [K_i : K_i-1], and let ξ_n1, ξ_n2, ···, ξ_nm ∈ ℂ be the corresponding primitive n₁-th, n₂-th, ···, n_m-th roots of unity.

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

K₀ ⊆ K₀(ξ_n1) ⊆ K₀(ξ_n1, ξ_n2) ⊆ ··· ⊆ K₀(ξ_n1, ξ_n2,···, ξ_nm) ⊆ K₁(ξ_n1, ξ_n2,···, ξ_nm) ⊆ ··· ⊆ K_m(ξ_n1, ξ_n2,···, ξ_nm)

• K₀ ⊆ K₀(ξ_n1) is a simple radical extension because it is a cyclotomic extension.
• K₀(ξ_n1) ⊆ K₀(ξ_n1, ξ_n2), it is a simple radical extension because we are adding ξ_n2, that is, a primitive nth root of unity. The same argument goes up to K₀(ξ_n1, ξ_n2,···, ξ_nm).
• K₀(ξ_n1, ξ_n2,···, ξ_nm) ⊆ K₁(ξ_n1, ξ_n2,···, ξ_nm).

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 ξ_n1. Any of these extensions, their orders, say n’₁ divides n₁ (by the same proposition -see Recall-), e.g., [K₁(ξ_n1): K₀(ξ_n1)] divides [K₁ : K₀] = n₁, [K₂ : K₁] | [K₁ : K₀] and therefore [K₂ : K₁] | n₁ and so on, and so forth, [K₁(ξ_n1, ξ_n2,···, ξ_nm): K₀(ξ_n1, ξ_n2,···, ξ_nm)] = n₁’ divides n₁.
3. [K₁(ξ_n1, ξ_n2,···, ξ_nm): K₀(ξ_n1, ξ_n2,···, ξ_nm)] = n’₁ where n’₁|n₁, ξ_n1^n₁/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] K₁(ξ_n1, ξ_n2,···, ξ_nm) is the splitting field of an irreducible polynomial x^n’₁ -a’ over K₀(ξ_n1, ξ_n2,···, ξ_nm). Futhermore, K₁(ξ_n1, ξ_n2,···, ξ_nm) = K₀(ξ_n1, ξ_n2,···, ξ_nm)($\sqrt[n’_1]{a’}$) and therefore K₁(ξ_n1, ξ_n2,···, ξ_nm)/K₀(ξ_n1, ξ_n2,···, ξ_nm) is a simple radical extension.
5. The same argument goes up to K_m(ξ_n1, ξ_n2,···, ξ_nm), all the extensions in the tower are radical extensions.
6. α ∈ K_m ⊆ K_m(ξ_n1, ξ_n2,···, ξ_nm).

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 S_n.

• 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 S_n] 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 ⊆ A_n. 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 a_i -3 and 3, 0 ≤ i < n, p does not divide a_n = 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

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.