Somebody who thinks logically is a nice contrast to the real world or, alternatively, common sense is the least common of the senses, it is not so common.
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.
Proof.
For the sake of contradiction, suppose f is solvable ↭ Gal(f) is solvable ⊥ S_{5} or A_{5} are not solvable.
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.
Proof. (2nd version. Based on Algebra, Second Edition by Michael Artin, Theorem 16.12.4, pg 503)
Let G = Gal(f). First, let’s suppose G = S_{5}, and K be the splitting field of f over F, so G = Gal(K/F) = Gal(f), D = Disc(f), δ = $\sqrt{D}$ ∈ K.
By assumption, G = S_{5} ⇒ G = Gal(f) ⊄ A_{5} ⇒ [Recall If δ ∈ F (Δ is a square in F), then G ⊆ An.] δ ∉ F ⇒ [F(δ) : F] = 2 ⇒ [$A_n$ is the only subgroup of $S_n$ of index 2] Gal(K/F(δ)) = A_{5}, so the Galois group of f over F(δ) is A_{5}. Therefore, if we demonstrate that if Gal(f) = A_{5}, then f is not solvable, we are done (because f is not solvable over F(δ) ⇒ f is not solvable over F)
Therefore, we may assume that G is the alternating group A_{5}, a simple group, f ∈ F[x], deg(f) = 5, Gal(f) = A_{5}.
For the sake of contradiction, let’s suppose f it is solvable over F. Let α ∈ K be a root of f, then α is solvable over F, (Recall: solvable ↭ tower of Abelian fields ↭ tower of cyclic fields ↭ tower of cyclic fields of prime order) then there exist a tower of extensions which are cyclic of prime degree.
F = F_{0} ⊆ F_{1} ⊆ F_{2} ⊆ ··· ⊆ F_{r}, α ∈ F_{r}, F_{i}/F_{i-1} is Galois, [F_{i} : F_{i-1} is prime]
Lemma. Let F’/F be a Galois extension of prime degree p. Let K and K' be the splitting field of f over F and F' respectively. Then, Gal(K'/F')≋A_{5}.
Proof.
K is the splitting field of f (irreducible) over F, α_{1}, α_{2}, α_{3}, α_{4}, α_{5} are five distinct elements, K = F(α_{1}, α_{2}, α_{3}, α_{4}, α_{5}). K’ is the splitting field of f over F'.
[F’: F] = p, Gal(F’/F) ≋ ℤ/pℤ, and it cannot have proper intermediate fields because it is a prime number.
By assumption, F’/F is Galois ⇒ F’ is the splitting field of a polynomial g ∈ F[x] ⇒ g is irreducible. For the sake of contradiction, let’s suppose that g is reducible, say g = h·h’, deg(h)>0, deg(h’)>0. However, we can assume that deg(h)>1, deg(h’)>1, and (h, h’) = 1. Otherwise, g = h(x-a), a ∈ F and the splitting field of h is the same as the splitting field of g
⊥ Therefore, g is irreducible of deg p, F’ = F(β_{1}, ···, β_{p}). K’ = F(α_{1}, α_{2}, α_{3}, α_{4}, α_{5}, β_{1}, ···, β_{p}) is the composite of F’ and K in $\bar F$, and notice that K’ is the splitting field of f over F’ and it is generated by its roots, i.e., K’ = F’(α_{1}, α_{2}, α_{3}, α_{4}, α_{5})
Each of the extension fields is a Galois extension, and the Galois groups have been labeled in the diagram, e.g., K’/F is Galois because it is the composite of K/F and F’/F, and they are both Galois extensions (K’ is the splitting field of fg over F).
Let N = Gal(K’/F), N is Galois. From Galois’ fundamental theorem we know H ◁ N and N/H ≋ G’ ≋ ℤ/pℤ (F’/F is Galois), H’ ◁ N and N/H’ ≋ G ≋ A_{5} (K’/F is Galois).
Next, the reader should notice that H ∩ H’ = {id} because σ ∈ H’ = Gal(K’/K), σ is an F-automorphism of K’ that fixes the roots α_{1}, ···, α_{5}.
σ ∈ H = Gal(K’/F’), σ is an F-automorphism of K’ that fixes the root β_{1}, ···, β_{p}. Therefore, σ ∈ H ∩ H’ ⇒ σ is an F-automorphism of K’ that fixes β_{1}, ···, β_{p}, α_{1}, ···, α_{5}, but K’ = F(α_{1}, α_{2}, α_{3}, α_{4}, α_{5}, β_{1}, ···, β_{p}) ⇒ σ = id, H ∩ H’ is the trivial group.
Consider the canonical map, Φ: N → N/H ≋ G’. Let’s restrict the canonical map to the subgroup H’ (H’◁ N), Φ|_{H’}: H’ → N/H. The Kernel of this restriction is the trivial group: Ker(Φ|_{H’}) = Ker(Φ) ∩ H’= H ∩ H’ = {id} ⇒ Φ|_{H’}: H’ → N/H ≋ G’ ≋ ℤ/pℤ, the restriction Φ|_{H’} is injective ⇒ It maps H’ isomorphically to a subgroup of G’, a cyclic group of prime order, ℤ/pℤ, so there are only two options: either H’ is the trivial group, or else H’ is cyclic of order p.
H’ is the trivial group, H’ = {id}, N → N/H’≋ G ≋ A_{5}, H’◁ N ⇒ so the map is onto (H’◁ N, then there is always a canonical surjective group homomorphism from N to the quotient group N/H’ that sends an element g∈ N to the coset determined by g), but it is also 1-1 (its kernel is H’ = [By assumption] {id}) ⇒ N ≋ A_{5} ⇒ [A_{5} is simple] N is simple.
N does have a normal subgroup, namely H ◁ N, such that N/H ≋ G’ ≋ ℤ/pℤ, N ≋ A_{5} ⇒ |N| = 60, N/H ≋ G’ ≋ ℤ/pℤ ⇒ |N/H| = p ⇒ |H| = 60/p ⇒ |H| ≠ 1 and |H| ≠ 60, so H is a proper, non-trivial normal subgroup of N, but N is supposed to be simple ⊥
H’ ≋ ℤ/pℤ, |N| = [N/H ≋ G’ ≋ ℤ/pℤ] = |G’||H| = |H|p. |N| = [N/H’≋G ≋ A5] = |G||H’| = 60p. Therefore, |H| = 60.
Consider the canonical map ψ:N → N/H’ ≋ G (it is always surjective), and restrict it to the subgroup H, ψ|_{H}.
The kernel of this restriction is the trivial group: Ker(ψ|_{H}) = Ker(ψ) ∩ H = H’ ∩ H = {id}, so ψ|_{H} is injective. Therefore, H is isomorphic to a subgroup of G. However, |H| = 60 = [G ≋ A5] |G| ⇒ Since both groups have order 60, the restriction ψ|_{H} is indeed an isomorphism ⇒ H ≋ G ≋ A5∎
[Continuing with the proof…]
Notice:
However, f is not irreducible in F_{r}[x] because it has a root, namely α ∈ F_{r} ⇒ f = (x - α)f’, f’ ∈F_{r}[x], deg(f’) = 4
Since K_{r} is the splitting field of f over F_{r} ⇒ K_{r} is the splitting field of f’ over F_{r}, deg(f’) = 4 ⇒ Gal(K_{r}/F_{r}) ≤ S_{4}, and in particular |Gal(K_{r}/F_{r}| ≤ |S_{4}| = 24, but we have already established that Gal(K_{r}/F_{r}) = A_{5}, so has order 60 ⊥ Therefore, f is not solvable over F.