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:
Proof.
(1 ⇒ 2) α is solvable over F, F = F_{0} ⊆ F_{1} ⊆ ··· ⊆ 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}^{ni} ∈ F_{i-1}.
Let us consider the corresponding ξ_{n1}, ξ_{n2}, ···, ξ_{nr} primitive n_{1}-th, n_{2}-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}) ⊆ F_{1}(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_{1} = F_{0}(α_{1}) where α_{1}^{n1} ∈ F_{0}⊆ F_{o}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}). F_{1}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}) = F_{0}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr})(α_{1}) where α_{1}^{n1} ∈ F_{0}⊆ F_{o}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}) so the extension F_{1}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}) is a splitting field of x^{n1} -α_{1}^{n1} over F_{0}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}) and F_{0}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}) contains ξ_{n1} ⇒ [F field containing a primitive nth root of unity, K splitting field of x^{n} - a over F. Then, K/F is a cyclic extension] F_{1}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr})/F_{0}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}) is cyclic ⇒ [All cyclic groups are Abelian, but an Abelian group is not necessarily cyclic] Abelian.
F_{1}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}) is a splitting field of x^{n1} -α_{1}^{n1} over F_{0}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}) because once that α_{1} and ξ_{n1} are in F_{1}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}), all the other roots of x^{n1} -α_{1}^{n1} are in F_{1}(ξ_{n1}, ξ_{n2}, ···, ξ_{nr}), too.
2 ⇒ 3) Objective: There exists a chain or tower of fields F = F_{0} ⊆ F_{1} ⊆ ··· ⊆ 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 = F_{0} ⊆ F_{1} ⊆ ··· 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 = K_{0} ⊆ K_{1} ⊆ ··· ⊆ 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 n_{1}-th, n_{2}-th, ···, n_{m}-th roots of unity.
Consider the following tower of fields (we adjoin the n_{i}-th roots of unity),
K_{0} ⊆ K_{0}(ξ_{n1}) ⊆ K_{0}(ξ_{n1}, ξ_{n2}) ⊆ ··· ⊆ K_{0}(ξ_{n1}, ξ_{n2},···, ξ_{nm}) ⊆ K_{1}(ξ_{n1}, ξ_{n2},···, ξ_{nm}) ⊆ ··· ⊆ K_{m}(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 x^{n}-a over F.
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., x^{n} -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 = x^{2} +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 S_{2}, 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 α^{2} ∈ F. In other words, a degree 2 extension can be obtained by adding a square root of an element of 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) ≋ S_{3} if and only if δ = $\sqrt{D}$ ∉ F (D is not a square in F), and Gal(K/F) ≋ A_{3} 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 S_{3}. There are only 2 transitive subgroups of S_{3}, namely, S_{3} and A_{3} = {(), (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 $δ = \sqrt{D}$:
x→ x-1, x^{3} -3x^{2} +3 is irreducible by Eisenstein, p = 3 divides a_{i} -3 and 3, 0 ≤ i < n, p does not divide a_{n} = 1, p^{2} = 9 does not divide a_{0}
x→ x-1, f(x-1) = x^{3} -3x^{2} + 6x -3 is irreducible by Eisenstein, p = 3 divides -3, 6, and -3, but does not divide 1, and p^{2}=9 does not divide -3.
Another example is x^{3} -4x + 2. It has discriminant 148 = 2^{2}·37, hence it is not a square in ℚ, so the corresponding Galois group is S_{3}.