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:

x

_{1}→ x

_{2}→ x

_{3}→ x

_{1}and x

_{i}→ x

_{i}for i greater than 3.

so that:

σf(x

_{1}, x

_{2}, x

_{3}, ..., x

_{n}) = f(x

_{2}, x

_{3}, x

_{1}, ..., x

_{n})

Let τ be the following permutation:

x

_{3}→ x

_{4}→ x

_{5}→ x

_{3}and x

_{i}→ x

_{i}for i=1,2 and i greater than 5

so that:

τf(x

_{1}, x

_{2}, x

_{3}, x

_{4}, x

_{5}, ..., x

_{n}) = f(x

_{1}, x

_{2}, x

_{4}, x

_{5}, x

_{3}, ..., 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

_{σ}u

where ζ

_{σ}is some p-th root of unity.

(6) Applying σ to both sides gives us:

σ

^{2}(u) = ζ

_{σ}

^{2}u

(7) Applying σ again we get:

σ

^{3}(u) = ζ

_{σ}

^{3}u

(8) Now, we also know from its definition that σ

^{3}(u) = u since:

σ

^{3}f(x

_{1}, x

_{2}, x

_{3}, x

_{4}, ..., x

_{n}) = f(x

_{1},x

_{2}, x

_{3}, x

_{4}, ..., x

_{n})

(9) So, that ζ

_{σ}

^{3}= 1

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

τ(u) = ζ

_{τ}u

where ζ

_{τ}is a pth root of unity and

ζ

_{τ}

^{3}= 1

(11) Putting these two results together gives us:

σ*τ(u) = σ(τ(u)) = ζ

_{σ}ζ

_{τ}u

σ

^{2}*τ = σ

^{2}(τ(u)) = ζ

_{σ}

^{2}ζ

_{τ}u

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

σ*τ:

x

_{1}→ x

_{2}→ x

_{3}→ x

_{4}→ x

_{5}→ x

_{1}and x

_{i}→ x

_{i}for i greater than 5.

σ

^{2}*τ:

x

_{1}→ x

_{3}→ x

_{4}→ x

_{5}→ x

_{2}→ x

_{1}and x

_{i}→ x

_{i}for i greater than 5

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

(σ*τ)

^{5}(u) = u

(σ

^{2}τ)

^{5}(u) = u

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

(ζ

_{σ}ζ

_{τ})

^{5}= (ζ

_{σ}

^{2}ζ

_{τ})

^{5}= 1

(15) Now, we also note that:

ζ

_{σ}= ζ

_{σ}

^{6}*ζ

_{σ}

^{-5}= ( ζ

_{σ}

^{6}*ζ

_{σ}

^{-5})*(ζ

_{τ}

^{5}*ζ

_{τ}

^{-5})*(ζ

_{σ}

^{5}*ζ

_{σ}

^{-5}) = ζ

_{σ}

^{6}*(ζ

_{τ}

^{5}*ζ

_{σ}

^{5})*(ζ

_{σ}

^{-5}*ζ

_{σ}

^{-5}*ζ

_{τ}

^{-5}) =

= ζ

_{σ}

^{6}*(ζ

_{σ}ζ

_{τ})

^{5}(ζ

_{σ}

^{2}ζ

_{τ})

^{-5}

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

ζ

_{σ}= ζ

_{σ}

^{6}*(ζ

_{σ}ζ

_{τ})

^{5}[(ζ

_{σ}

^{2}ζ

_{τ})

^{5}]

^{-1 }= ζ

_{σ}

^{6}*(1)*(1)

^{-1}= ζ

_{σ}

^{6}

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

ζ

_{σ}=(ζ

_{σ}

^{3})

^{2}= (1)

^{2}= 1

(18) Now, since (ζ

_{σ}ζ

_{τ})

^{5}=1, we have:

(1*ζ

_{τ})

^{5}=1

so that:

ζ

_{τ}

^{5}= 1

(19) We can now show that ζ

_{τ}= 1 since:

ζ

_{τ}= ζ

_{τ}

^{6}*ζ

_{τ}

^{-5}= (ζ

_{τ}

^{3})

^{2}*(ζ

_{τ}

^{5})

^{-1}

Taking ζ

_{τ}

^{3}= 1 (step #10) and ζ

_{τ}

^{5}= 1 (step #18), we get:

ζ

_{τ}= (ζ

_{τ}

^{3})

^{2}*(ζ

_{τ}

^{5})

^{-1}= (1)

^{2}*(1)

^{-1}= 1

QED

Theorem 2: Ruffini's Theorem

Let P(X) = (X - x

_{1})*...*(X - x

_{n}) = X

^{n}- s

_{1}X

^{n-1}+ ... + (-1)

^{n}s

_{n}= 0

where s

_{i}are coefficients in field k.

Let K be a field such that K = k(s

_{1}, ..., 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), x

_{1}∈ 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 = F

_{0}⊂ F

_{1}⊂ ... ⊂ F

_{m-1}⊂ F

_{m}= F

such that for each 0 ≤ i ≤ m:

where p

_{i}is a prime and α

_{i}∈ F

_{0 }= 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 F

_{0}= 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 x

_{1}∈ F = F

_{m}is not invariant under σ [since σ(x

_{1}) = x

_{2}]

(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.