I have yet to see any problem, however complicated, which, when looked at in the right way did not become still more complicated, Paul Anderson
Theorem. Let f be an irreducible polynomial of degree 5 over a subfield F of the complex numbers (F ⊆ ℂ), whose Galois group is either the alternating group A5 or the symmetric group S5, then f is not solvable. Therefore, there are quintics that are not solvable.
Proposition. Complex conjugate root theorem. Let f(z) = anxn + an-1xn-1 + ··· + a1z + a0 be a polynomial in one variable with real coefficients, and α = a + bi be a root of f with a and b real numbers (α ∈ ℂ), then its complex conjugate $\bar α = a -bi$ is also a root of f.
Proof.
Let α ∈ ℂ be a root of f ⇒ f(α) = 0.
Proposition. The symmetric group Sn can be generated by a n-cycle and a transposition. In other words, if a subgroup G of Sn, G ≤ Sn, contains a 2-cycle and a n-cycle ⇒ G = Sn.
Proof.
We can assume, without losing any generality, σ = (12) ∈ G, (12···n) = τ ∈ G
τ(12)τ-1 = (12···n)(12)(n···21) = (23). Similarly, τ(23)τ-1 = (34), ··· τ(n-2,n-1)τ-1 = (n-1, n), so that (i, i+1) ∈ ⟨σ, τ⟩ = ⟨(12⟩, (12···n)⟩ for all 1 ≤ i ≤ n-1.
Next, (23)(12)(23)-1 = (23)(12)(32) = (13), (34)(13)(34)-1 = (14),···, (n-1,n)(1,n-1)(n-1,n)-1 = (1, n), so that (1, i) ∈ ⟨σ, τ⟩ = ⟨(12⟩, (12···n)⟩ for all 1 ≤ i ≤ n.
∀i, j, 1 ≤ i ≤ j ≤ n, then (i, j) = (1, i)(1, j)(1, i)-1 = (1, i)(1, j)(i, 1) ∈ [i → 1 → j, j → 1 → i] ∈ ⟨σ, τ⟩ = ⟨(12⟩, (12···n)⟩ ⇒ ⟨σ, τ⟩ contains all transpositions ⇒ [∀σ ∈Sn, σ is the product of disjoint cycles] ⟨σ, τ⟩ = Sn
Proposition. Let f ∈ ℚ[x] be an irreducible polynomial of prime degree p with exactly one pair of non-real complex conjugate roots. Then, f is not solvable by radicals.
If p = 5, this proposition gives us a non solvable quintic.
Proof. Let α1, α2, ···, αp ∈ K be the roots of f in a splitting field K of f over ℚ[x]. Let α1, α2 ∉ ℝ, Let α3, α4, ···, αp ∈ ℝ. Let G = Gal(K/ℚ) = Gal(f).
Claim. (12) ∈ G. Consider the map σ: K → K given by the complex conjugation, σ(a +ib) = a -ib, σ is an automorphism of K, σ = (12) ∈ Gal(K/ℚ) = G, σ(α1) = α2, σ(α2) = α1, σ(αi) = αi ∀i = 3,..., p, because we know [Complex conjugate root theorem] that $α_2 = \bar α_1$ and F(α3, ···, αp) ≠ K, that transposition must be in G. In other words, the complex conjugation, which swaps the complex roots and fixes the real roots, is an automorphism in G.
σ = (12) ∈ Sp, and therefore G contains a 2-cycle.
p | [K : ℚ] = |G| ⇒ [By Cauchy’s Theorem. Let G be a finite group and p be a prime. If p | |G|, then G has an element of order p] G contains an element of order p ⇒ [p prime, see next paragraph] G contains a p-cycle
Let σ be an element of order p ⇒ σ is the product of disjoint cycles, say σ1, σ2, ···, σn, order(σ) = lcm(ord(σ1), ord(σ2), ···, ord(σn)) ⇒ ord(σi) = 1 or p, therefore the cycle decomposition will consist of p-cycles and 1-cycles, and at least there will be one p-cycle.
Another way of looking at this fact is that G operates transitively on the roots, so it contains an element of order p, a p-cycle.
Therefore, seeing G as a subgroup of Sp, G contains a 2-cycle and a p-cycle ⇒ [G ≤ Sn contains a 2-cycle and a n-cycle ⇒ G = Sn] G ≋ Sp, so f is not solvable by radicals.∎
Let’s see some examples of irreducible polynomials of degree 5 in ℚ[x] with exactly two complex conjugate roots.
Besides, f’(x) = 20x4 -20x = 20x(x3 -1), there are two critical values, namely 0 and 1. f’’(x) = 80x3 -20 ⇒ There is a local maximum at (0, 5) and a local minimum at (1, -1), so taking all these facts into consideration, we can see that there are three real roots of f, in other words, f crosses the horizontal line in these intervals: (-∞, 0), (0, 1), (1, +∞). Of course, it means that there are two complex conjugate roots ⇒ [p prime, p ≥ 5. f ∈ ℚ[x] is an irreducible polynomial of degree p such that f has exactly 2 non-real roots. Then, Gal(f) ≋ Sp and f is not solvable.] Gal(f) = S5 and f is not solvable (by radicals).
The problem about g is that g is not irreducible, but f(x) = g(x) +2 = x5 -4x +2 is irreducible by Eisenstein [p=2, 2/-4, 2/2, but 2 ɫ an = 1, 22 ɫ a0=2] and yet they both share the same graph, just shifted vertically over the y-axis 2, so f has 3 real roots, and 2 non-real roots ⇒ [p prime, p ≥ 5. f ∈ ℚ[x] is an irreducible polynomial of degree p such that f has exactly 2 non-real roots. Then, Gal(f) ≋ Sp and f is not solvable.] Gal(f) = S5 and f is not solvable (by radicals).
Proposition. General quintic polynomials are not solvable by radicals.
Proof. As we have shown the polynomial x5 -4x +2 over ℚ is not solvable by radicals, so we know that there is at least one quintic polynomial not solvable by radicals∎
More generally, ∀n ≥ 5, let f(x) = xn-5(x5 -16x +2), the splitting field of f(x) is the splitting field of x5 -16x +2 which is S5, so f is not solvable, and Gal(f) = S5.
f has exactly p-2 real roots. The reason is the same as before considering (x2 +4)(x -2)(x -4)··· (x -2(p-2)). It has p-2 real roots, namely, 2, 4, ··· 2(p-2), and 2 imaginary roots $\sqrt{-4}, -\sqrt{-4}$. Therefore, f is irreducible, has p-2 real roots and 2 imaginary roots ⇒ Gal(f) ≈ Sp and f is not solvable by radicals. We have constructed a Galois extension K/ℚ such that Gal(f) = Gal(K/ℚ) ≋ Sp for any prime p ≥ 5.