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Insolvability of quintics.

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 A5 or the symmetric group S5, 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 ⊥ S5 or A5 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 A5 or the symmetric group S5, 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 = S5, 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 = S5 ⇒ G = Gal(f) ⊄ A5 ⇒ [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)) = A5, so the Galois group of f over F(δ) is A5. Therefore, if we demonstrate that if Gal(f) = A5, then f is not solvable, we are done (because f is not solvable over F(δ) ⇒ f is not solvable over F)

K | F | F ( A δ 2 5 ) , δ F , G a l o i s

Therefore, we may assume that G is the alternating group A5, a simple group, f ∈ F[x], deg(f) = 5, Gal(f) = A5.

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 = F0 ⊆ F1 ⊆ F2 ⊆ ··· ⊆ Fr, α ∈ Fr, Fi/Fi-1 is Galois, [Fi : Fi-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')≋A5.

In other words, when we consider a Galois extension of F’/F of prime degree (or a tower as we iterate through a tower of fields of prime order), we show that no progress towards solving the equation f = 0 is made when one replaces F by F’. We do this by showing that the Galois group of f over F’ is again Gal(K’/F’) ≋ A5 and f remains irreducible over F'.

Proof.

K is the splitting field of f (irreducible) over F, α1, α2, α3, α4, α5 are five distinct elements, K = F(α1, α2, α3, α4, α5). K’ is the splitting field of f over F'.

F p ' \ K F ' K = F ( α 1 , α 2 , α 3 , α 4 , α 5 )

[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

p p r i m e F | F | F ' ( F r F = ' o ( o r F t o ( o r F o t o ( f o r o t o h f s o ) t h o ) f o f g ) h F ) , , d d e e g g ( ( h h ) ' > ) 1 > 1

⊥ Therefore, g is irreducible of deg p, F’ = F(β1, ···, βp). K’ = F(α1, α2, α3, α4, α5, β1, ···, βp) 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’(α1, α2, α3, α4, α5)

G A H 5 ' K \ K F ' N = / F F H ' G ( ' α = 1 , G F · ? ( / · β p · 1 α , 5 A · , 5 · β ? · 1 , , β · p · ) · , β p )

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).

Our plan is to show that H is isomorphic to G, i.e., H is the alternating group A5.

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 ≋ A5 (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 α1, ···, α5.

σ ∈ H = Gal(K’/F’), σ is an F-automorphism of K’ that fixes the root β1, ···, βp. Therefore, σ ∈ H ∩ H’ ⇒ σ is an F-automorphism of K’ that fixes β1, ···, βp, α1, ···, α5, but K’ = F(α1, α2, α3, α4, α5, β1, ···, βp) ⇒ 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.

  1. H’ is the trivial group, H’ = {id}, N → N/H’≋ G ≋ A5, 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 ≋ A5 ⇒ [A5 is simple] N is simple.

    N does have a normal subgroup, namely H ◁ N, such that N/H ≋ G’ ≋ ℤ/pℤ, N ≋ A5 ⇒ |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 ⊥

  2. 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.

ψ H : N N H N ψ / H H ' i s 1 G - 1 o n t o

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…]

A K 5 / \ K / 1 A F K A 5 [ 2 5 \ F . \ 1 . F / K . 2 G r A ] / G a - 5 [ a l 1 \ / . G l o F . a o i K r G . l i s r A - a ] o s , 5 1 l i , \ F G o s p r a i , p r l s r i ( o , p i m α i r m e s p i e , r m i e F p m r r e i ) m e

Notice:

  1. By assumption K is the splitting field of f over F. Gal(K/F) = Gal(f) =A5.
  2. By the previous result, let K1 be the splitting field of f over F1 ⇒ Gal(K1/F1) = A5
  3. We repeat the process up to Kr, the splitting field of f over Fr, Gal(Kr/Fr) = A5

However, f is not irreducible in Fr[x] because it has a root, namely α ∈ Fr ⇒ f = (x - α)f’, f’ ∈Fr[x], deg(f’) = 4

Since Kr is the splitting field of f over Fr ⇒ Kr is the splitting field of f’ over Fr, deg(f’) = 4 ⇒ Gal(Kr/Fr) ≤ S4, and in particular |Gal(Kr/Fr| ≤ |S4| = 24, but we have already established that Gal(Kr/Fr) = A5, so has order 60 ⊥ Therefore, f is not solvable over F.

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