# Insolvability of quintics II.

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.

1. If α is wholly real (↭ b = 0) ⇒ α = $\bar α$, and therefore $\bar α$ is a root of f, too.
2. Let α ∈ ℂ not be wholly real (↭ b ≠ 0), we have that $\bar 0 = 0$ ⇒ $\overline {f(α)} = f(α)$ ⇒ [Conjugate of Polynomial is Polynomial of Conjugate, Pr∞fWiki. Let f(z) = anxn + an-1xn-1 + ··· + a1z + a0 be a polynomial in one variable with real coefficients, and α ∈ ℂ be a complex number, then $\overline {f(α)}=f(\bar α)$] $0 = f(α) = \overline {f(α)} = f(\bar α) ⇒ f(\bar α) = 0$∎

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.

• f(x) = 4x5 -10x2 + 5 is irreducible by Eisenstein’s criteria (p = 5). Futhermore, we can see that $\lim_{x \to \infin}f(x)=\infin, \lim_{x \to -\infin}f(x)=-\infin$, so by the Intermediate Value Theorem has at least one real root.

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).

• g = x5 -4x = x(x4 -4) = x(x2-2)(x2+2), g has exactly 3 real roots, namely $0, \sqrt{2}, -\sqrt{2}$ and 2 non-real roots, $\sqrt{-2}, -\sqrt{-2}∉\mathbb{R}$.

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

• f = x5 -16x +2, f is irreducible by Eisenstein (p=2 | -16, 2 | 2, but 2 ɫ an = 1 and 22 ɫ a0 = 2) = x(x4 -16) + 2 = x(x2-4)(x2+4) + 2 = x(x2-4)(x2+4) + 2. We can see clearly that g(x) = x(x2-4)(x2+4) has exactly three real roots, namely 0, 2, and -2, and 2 non-real roots. Using the same argument as before, we have just added a small constant, namely 2, without changing the number of real roots ⇒ [This is seen by just looking at the graph of the polynomial] f(x) has exactly three real roots and two non-real roots ⇒ Gal(f) = S5 and f is 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.

• Let p ≥ 5 be a prime number, f = (x2 +4)(x -2)(x -4)··· (x -2(p-2)) +2 ∈ ℚ[x], deg(f) = p, f is irreducible because f = xp + 2(ap-1xp-1 + ··· + a1x) -4·2·4···2(p-2)+2, so by Eisenstein p=2 divides all intermediate coefficients, 2 ɫ an = 1, and 22=4 ɫ -4·2·4···2(p-2)+2

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.

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