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 A_{5} or the symmetric group S_{5}, then f is not solvable. Therefore, there are **quintics that are not solvable**.

Proposition. **Complex conjugate root theorem**. Let f(z) = a_{n}x^{n} + a_{n-1}x^{n-1} + ··· + a_{1}z + a_{0} 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.

- If α is wholly real (↭ b = 0) ⇒ α = $\bar α$, and therefore $\bar α$ is a root of f, too.
- 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) = a
_{n}x^{n}+ a_{n-1}x^{n-1}+ ··· + a_{1}z + a_{0}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 S_{n} can be generated by a n-cycle and a transposition. In other words, if a subgroup G of S_{n}, G ≤ S_{n}, contains a 2-cycle and a n-cycle ⇒ G = S_{n}.

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 ∈S_{n}, σ is the product of disjoint cycles] ⟨σ, τ⟩ = S_{n}

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 α_{1}, α_{2}, ···, α_{p} ∈ K be the roots of f in a splitting field K of f over ℚ[x]. Let α_{1}, α_{2} ∉ ℝ, Let α_{3}, α_{4}, ···, α_{p} ∈ 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(α_{1}) = α_{2}, σ(α_{2}) = α_{1}, σ(α_{i}) = α_{i} ∀i = 3,..., p, because we know [**Complex conjugate root theorem**] that $α_2 = \bar α_1$ and F(α_{3}, ···, α_{p}) ≠ 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) ∈ S_{p}, 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 σ_{1}, σ_{2}, ···, σ_{n}, order(σ) = lcm(ord(σ_{1}), ord(σ_{2}), ···, ord(σ_{n})) ⇒ ord(σ_{i}) = 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 S_{p}, G contains a 2-cycle and a p-cycle ⇒ [**G ≤ S _{n} contains a 2-cycle and a n-cycle ⇒ G = S_{n}**] G ≋ S

Let’s see some examples of irreducible polynomials of degree 5 in ℚ[x] with exactly two complex conjugate roots.

- f(x) = 4x
^{5}-10x^{2}+ 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) = 20x^{4} -20x = 20x(x^{3} -1), there are two critical values, namely 0 and 1. f’’(x) = 80x^{3} -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) ≋ S_{p} and f is not solvable.] **Gal(f) = S _{5} and f is not solvable (by radicals).**

- g = x
^{5}-4x = x(x^{4}-4) = x(x^{2}-2)(x^{2}+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 = x^{5} -4x +2 is irreducible by Eisenstein [p=2, 2/-4, 2/2, but 2 ɫ a_{n} = 1, 2^{2} ɫ a_{0}=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) ≋ S_{p} and f is not solvable.] **Gal(f) = S _{5} and f is not solvable (by radicals).**

Proposition. General quintic polynomials are not solvable by radicals.

Proof. As we have shown the polynomial x^{5} -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 = x
^{5}-16x +2, f is irreducible by Eisenstein (p=2 | -16, 2 | 2, but 2 ɫ a^{n}= 1 and 2^{2}ɫ a_{0}= 2) = x(x^{4}-16) + 2 = x(x^{2}-4)(x^{2}+4) + 2 = x(x^{2}-4)(x^{2}+4) + 2. We can see clearly that g(x) = x(x^{2}-4)(x^{2}+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) = S**_{5}and f is not solvable by radicals.

More generally, ∀n ≥ 5, let f(x) = x^{n-5}(x^{5} -16x +2), the splitting field of f(x) is the splitting field of x^{5} -16x +2 which is S_{5}, so f is not solvable, and Gal(f) = S_{5}.

- Let p ≥ 5 be a prime number, f = (x
^{2}+4)(x -2)(x -4)··· (x -2(p-2)) +2 ∈ ℚ[x], deg(f) = p, f is irreducible because f = x^{p}+ 2(a_{p-1}x^{p-1}+ ··· + a_{1}x) -4·2·4···2(p-2)+2, so by Eisenstein p=2 divides all intermediate coefficients, 2 ɫ a_{n}= 1, and 2^{2}=4 ɫ -4·2·4···2(p-2)+2

f has exactly p-2 real roots. The reason is the same as before considering (x^{2} +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) ≈ S_{p} and f is not solvable by radicals. We have constructed a Galois extension K/ℚ such that Gal(f) = Gal(K/ℚ) ≋ S_{p} for any prime p ≥ 5.