Ruffini's Proof: Field Extensions

In today's blog, I will present Paolo Ruffini's proof that if n ≥ 5 and F is a splitting field, then F/k is not a radical tower. This constitutes step two of the Abel-Ruffini proof using field extensions. For a review of splitting fields and field extensions, start here.

Below is Pierre Lauren Wantzel's version of Ruffini's proof. Niels Abel independently presented his own version of this version which I covered previously.

Today's content is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Lemma 1:

Let u,a be functions of n parameters in a field F such that u^p = a for some prime p.

Let n ≥ 5

Let σ be the following permutation:

x₁ → x₂ → x₃ → x₁ and x_i → x_i for i greater than 3.

so that:

σf(x₁, x₂, x₃, ..., x_n) = f(x₂, x₃, x₁, ..., x_n)

Let τ be the following permutation:

x₃ → x₄ → x₅ → x₃ and x_i → x_i for i=1,2 and i greater than 5

so that:

τf(x₁, x₂, x₃, x₄, x₅, ..., x_n) = f(x₁, x₂, x₄, x₅, x₃, ..., x_n)

Then:

If a is invariant under the permutations σ and τ, then so is u.

Proof:

(1) From the given, we have u^p = a where u,a are functions of n parameters.

(2) Applying the permutation σ to both sides gives us:

σ(u)^p = σ(a)

(3) Since a is invariant with regard to σ, we have:

σ(u)^p = σ(a)= a = u^p

(4) Dividing both sides by u^p gives us:

[σ(u)/u]^p = 1

(5) Taking the p-th root of each side and multiplying by u gives us:

σ(u) = ζ_σu

where ζ_σ is some p-th root of unity.

(6) Applying σ to both sides gives us:

σ²(u) = ζ_σ²u

(7) Applying σ again we get:

σ³(u) = ζ_σ³u

(8) Now, we also know from its definition that σ³(u) = u since:

σ³f(x₁, x₂, x₃, x₄, ..., x_n) = f(x₁,x₂, x₃, x₄, ..., x_n)

(9) So, that ζ_σ³ = 1

(10) We can make the same line of argument with τ to show that:

τ(u) = ζ_τu

where ζ_τ is a pth root of unity and

ζ_τ³ = 1

(11) Putting these two results together gives us:

σ*τ(u) = σ(τ(u)) = ζ_σζ_τu

σ²*τ = σ²(τ(u)) = ζ_σ²ζ_τu

(12) Now, representing each of these as permutations we get:

σ*τ:

x₁ → x₂ → x₃ → x₄ → x₅ → x₁ and x_i → x_i for i greater than 5.

σ²*τ:

x₁ → x₃ → x₄ → x₅ → x₂ → x₁ and x_i → x_i for i greater than 5

(13) Using the above permutation maps, we get that:

(σ*τ)⁵(u) = u

(σ²τ)⁵(u) = u

(14) Using step #5 and step #10 above, we can use step #13 to conclude that:

(ζ_σζ_τ)⁵ = (ζ_σ²ζ_τ)⁵ = 1

(15) Now, we also note that:

ζ_σ = ζ_σ⁶*ζ_σ^-5 = ( ζ_σ⁶*ζ_σ^-5)*(ζ_τ⁵*ζ_τ^-5)*(ζ_σ⁵*ζ_σ^-5) = ζ_σ⁶*(ζ_τ⁵*ζ_σ⁵)*(ζ_σ^-5*ζ_σ^-5*ζ_τ^-5) =

= ζ_σ⁶*(ζ_σζ_τ)⁵(ζ_σ²ζ_τ)^-5

(16) But then using step #14 with step #15, we have:

ζ_σ = ζ_σ⁶*(ζ_σζ_τ)⁵[(ζ_σ²ζ_τ)⁵]^-1 = ζ_σ⁶*(1)*(1)^-1 = ζ_σ⁶

(17) Using step #9 above, we then have:

ζ_σ =(ζ_σ³)² = (1)² = 1

(18) Now, since (ζ_σζ_τ)⁵ =1, we have:

(1*ζ_τ)⁵ =1

so that:

ζ_τ⁵ = 1

(19) We can now show that ζ_τ = 1 since:
ζ_τ = ζ_τ⁶*ζ_τ^-5 = (ζ_τ³)²*(ζ_τ⁵)^-1

Taking ζ_τ³ = 1 (step #10) and ζ_τ⁵ = 1 (step #18), we get:

ζ_τ = (ζ_τ³)²*(ζ_τ⁵)^-1 = (1)²*(1)^-1 = 1

QED


Theorem 2: Ruffini's Theorem

Let P(X) = (X - x₁)*...*(X - x_n) = Xⁿ - s₁X^n-1 + ... + (-1)ⁿs_n = 0

where s_i are coefficients in field k.

Let K be a field such that K = k(s₁, ..., s_n)

If n ≥ 5 and F is a splitting field for P(X), then F/K is not a radical tower.

Proof:

(1) Since F is a splitting field for P(X), x₁ ∈ F.

(2) Since K is defined around the coefficients of P, it is clear that all elements of K are invariant under the permutations of σ and τ. [See Lemma 2, here]

(3) Assume F/K is a tower of radicals such that:

K = F₀ ⊂ F₁ ⊂ ... ⊂ F_m-1 ⊂ F_m = F

such that for each 0 ≤ i ≤ m:



where p_i is a prime and α_i ∈ F₀ = K

(4) Then, by induction, all elements F_i are also invariant against the permutations σ and τ since:

(a) We know that all elements of F₀ = K are invariant under σ and τ (step #2 above) so we can assume that this is true up to some i where i ≥ 0.

(b) Let a = α_i, u = a^(1/p) so that:

F_i+1 = F_i(u)

and

a ∈ F_i

(c) But since a ∈ F_i, it follows that a is invariant under σ and τ.

(d) Using Lemma 1 above, we note that this implies that u is also invariant under σ and τ.

(e) But then all elements of F_i(u) = F_i+1 are also invariant under σ and τ.

(5) But now we have a contradiction since x₁ ∈ F = F_m is not invariant under σ [since σ(x₁) = x₂]

(6) So, we reject our assumption in step #3.

QED

References

• Michael I. Rosen, "Niels Hendrik Abel and Equations of the Fifth Degree", The American Mathematical Monthly, Vol. 102, No. 6 (Jun. - Jul., 1995), pp 495-595.
• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001
• Jorg Bewersdorff, Galois Theory for Beginners, American Mathematical Society, 2006.