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.
  

Pierre Laurent Wantzel

Pierre Larent Wantzel was born on June 5, 1814 in Paris, France. His father was a professor of applied mathematics at the Ecole speciale du Commerce. From an early age, Wantzel showed tremendous ability in mathematics. Ademar Jean Claude Barre de Saint-Venant relates an anecdote that when Wantzel was 9 years old, his teacher would send for him to help with certain difficult surveying problems.

He entered the Ecole des Arts et Metiers when he was 12. After a year, Wantzel decided that the school was not academic enough and switched to the College Charlesmagne in 1828. He would later marry the daughter of his language coach at the College Charlesmagne.

At 15, Wantzel edited a famous mathbook, Antoine-Andre-Louis Reynaud's Treatise on Arithmetic and added a proof for a method of finding square roots that had previously not had a proof. Later, in 1831, he he was awarded first prize in Latin disseration. When he applied in 1832 for the prestigious Ecole Polytechnique and the Ecole Normale, he placed first in both examinations. By 1838, he had become a lecturer in mathematics at the Ecole Polytechnique and in 1841, he also became a professor of applied mathematics at Ecole de Ponts et Chaussees.

In 1837, Wantzel became the first to prove the impossibility of duplicating the cube and trisecting an angle using only ruler and compass. The great Carl Friedrich Gauss had stated that both of these methods were impossible but had never provided proof.

In 1845, Wantzel gave a revised proof of Abel's Theorem on the impossibility of solving all equations of n ≥ 5 by radicals. In this presentation, Wantzel gave a revised proof of the one done by Paolo Ruffini. In all, he wrote over 20 works on a wide range of subjects.

By May 12, 1848, Wantzel's never-ending pattern of very hard work without break and opium-use caught up with him and he died at the very early age of 33.

Saint-Venant writes:

... one could reproach him for having been too rebellious against those counselling prudence. He usually worked during the evening, not going to bed until late in the night, then reading, and got but a few hours of agitated sleep, alternatively abusing coffee and opium, taking his meals, until his marriage, at odd and irregular hours.

One might wonder how why Wantzel doesn't rank with the greatest mathematicians. Without a doubt, he showed tremendous promise as a child and with all of his hard work, one wonders why he did not reach the highest heights in mathematics. Saint-Venant writes:

...I believe that this is mostly due ot the irregular manner in which he worked, to the excessive number of occupations in which he was engaged, to the continual movement and feverishness of his thoughts, and even to the abuse of his own facilities. Wantzel improvised more than he elaborated, he probably did not give himself the leisure nor the calm necessary to linger long on the same subject.

References

• "Pierre Laurent Wantzel", MacTutor
  

Abel's Proof: Field Extensions: Step 1: Proof

The content in today's blog is taken from the essay by Michael I. Rosen entitled "Niels Hendrik Abel and Equations of the Fifth Degree."

In today's blog, I present the key lemma that I will use to establish Step 1 of the proof using field extensions. For the proof using the ideas that Niels Abel originally presented, see here.

Lemma 1:

Let:



where q_i is prime, a_i ∈ E_i

and E_i, E_i+1 are fields that include the roots of unity

L is a field such that (E_i ∩ L) ⊂ (E_i+1 ∩ L) ⊂ L and includes the roots of unity.

M_i+1 = E_i+1 ∩ L

M_i = E_i ∩ L

γ ∈ M_i+1 but not in M_i

Then:

There exist:

γ_i = c₀ + c₁ζ^i-1β + c₂ζ^2(i-1)β² + ... + c_q-1ζ^(q-1)(i-1)β^q-1

such that:

c_i ∈ E_i

β ∈ E_i+1, β^q ∈ E_i

γ_i ∈ M_i+1

ζ is a primitive qth root of unity

Proof:

(1) From our assumptions, we know that γ^q ∈ M_i

(2) Let P(x) be a polynomial such that P(γ) = 0, that is, γ is a root for P(x).

(3) We further know that (see Corollary 3.1, here) there exists an element β ∈ E_i+1 such that β^q ∈ E_i

and there exists b₀, b₂, ..., b_q-1 ∈ E_i such that:

γ = b₀ + β + b₂β² + ... + b_q-1β^q-1

(4) We can define a function Q(γ) such that Q(γ) = P(b₀ + x + b₂x² + ... + b_q-1x^q-1)

(5) From this definition, it is clear that Q(β) = P(γ) = 0 so that β is a root for Q(γ).

(6) Let a = β^q

(7) It is clear that β is the root of Y^q - a.

(8) We also know that Y^q - a is irreducible over M_i [See Lemma 1, here]

(9) But since β is a root for both Y^q - a and Q(Y), it follows that Y^q - a divides Q(Y) and every root of Y^q - a is also a root of Q(Y). [See Theorem 3, here]

(10) By the Fundamental Theorem of Algebra [See Theorem, here], we know that there are q roots for Y^q - a and they are: &beta, ζβ, ζ²β, ..., ζ^q-1β where ζ is a primitive qth root of unity [see here for review of the primitive roots of unity].

(11) Since Y^q - a divides Q(Y) and ζⁱβ is a root of Y^q - a, it follows that Q(ζⁱβ) = 0 for all integers i.

(12) So, the numbers [see step #5 above]:

γ = γ₁ = b₀ + β + b₂β² + ... + b_q-1β^q-1

γ₂ = b₀ + ζβ + b₂ζ²β² + ... + b_q-1ζ^q-1β^q-1

...

γ_q = b₀ + ζ^q-1β + b₂ζ^2(q-1)β² + ... + b_q-1ζ^(q-1)(q-1)β^q-1

are all roots of P(x). That is, P(γ_k) = 0 for k = 1, ..., q.

(13) We also know that the numbers γ₁, γ₂, ..., γ_q are all in L since:

(a) γ ∈ M_i+1 = E_i+1 ∩ L [from step #1 above]

(b) So, γ ∈ E_i+1 and γ ∈ L.

(c) Then, using Lemma 5, here, we can see that each γ_k ∈ L.

QED

Lemma 2:

Let:



where q_i is prime, a_i ∈ E_i

and E_i, E_i+1 are fields that include the roots of unity

L is a field such that (E_i ∩ L) ⊂ (E_i+1 ∩ L) ⊂ L and includes the roots of unity.

M_i+1 = E_i+1 ∩ L

M_i = E_i ∩ L

Then:



where q_i is prime, a_i ∈ M_i

Proof:

(1) Let y be an element such that y ∈ M_i+1 but y is not an element in M_i

We know that there is at least one such element since M_i ⊂ M_i+1

(2) Using Lemma 1 above, we know that there exists y₁, ..., y_q such that each y_i ∈ M_i+1 such that:

y = y₁ = b₀ + β + b₂β² + ... + b_q-1β^q-1

y₂ = b₀ + ζβ + b₂ζ²β² + ... + b_q-1ζ^q-1β^q-1

...

y_q = b₀ + ζ^q-1β + b₂ζ^2(q-1)β² + ... + b_q-1ζ^(q-1)(q-1)β^q-1

where β ∈ E_i+1 but not in E_i

β^q ∈ E_i

b₀, ..., b_q-1 ∈ E_i

(3) Now, if we multiply each equation in step #13 above by ζ^1-i, we get:

y = y₁ = b₀ + β + b₂β² + ... + b_q-1β^q-1

ζ^-1y₂ = ζ^q-1y₂ = ζ^q-1b₀ + ζ⁰β + b₂ζ¹β² + ... + b_q-1ζ^q-2β^q-1

...

ζ^1-qy_q = ζ¹ = ζb₀ + ζ⁰β + b₂ζ^2(q-1)+1β² + ... + b_q-1ζ^(q-1)(q-1)+1β^q-1

(4) If we add up each row, then we have the following values for each column [Using Lemma 4, here]:

b₀ + ζ^-1b₀ + ... + ζ^1-qb₀ = b₀(1 + ζ^1*(q-1) + ζ^2*(q-1) + ... + ζ^(q-1)*(q-1)) = b₀*0 = 0. [Since q doesn't divide q-1.]

β + ζ⁰β + ... + ζ⁰β = qβ

b₂β² + ζ¹b₂β² + ... + ζ^q-1b₂β² = b₂β²(1 + ζ^1*1 + ζ^2*1 + ... + ζ^(q-1)*1) = b₂β²*0 = 0

...

b_q-1β^q-1 + ζ^q-2b_q-1β^q-1 + ... + ζ²b₂β² = b_q-1β^q-1(1 + ζ^1*1 + ζ^2*1 + ... + ζ^(q-1)*1) = b_q-1β^q-1*0 = 0

(5) So, adding each of the columns together gives us the following equation:

β = (1/q)∑ (i=1,q) ζ^1-iy_i ∈ L

We know that β ∈ L since all ζ^1-i,q are in L (by the given) and y_i ∈ L by step #2 above.

(6) Thus β ∈ L ∩ E_i+1 = M_i+1 and β^q = b ∈ L ∩ E = M_i

(7) Let γ ∈ M_i+1 but not M_i

(8) Then there exist (see Lemma 1 above using the same value β as before):

γ_i = c₀ + c₁ζ^i-1β + c₂ζ^2(i-1)β² + ... + c_q-1ζ^(q-1)(i-1)β^q-1 with c_i∈ E_i and each γ_i ∈ M_i+1

(9) If we multiply each γ_i by ζ^k(1-i) and add up each column in the same way that we did in steps #15 - #17), we get:

c_kβ^k = ∑ (i=1,q) ζ^k(1-i)γ_i .

(10) Now since ζ^k(1-i), γ_i ∈ M_i+1, it follows that each c_kβ^k ∈ M_i+1

(11) Since β ∈ M_i+1 [See step #6 above], it follows that:

β^k ∈ M_i+1, and using step #10 above, we get c_k ∈ M_i+1

(12) But if c_k ∈ M_i+1 = E_i+1 ∩ L, it follows that c_k ∈ L.

(13) And since c_k ∈ E_i [see step #8 above] and c_k ∈ L [see step #12 above], it follows that c_k ∈ M_i = E_i ∩ L.

(14) Thus, we have show that:



where q_i is prime, a_i ∈ M_i

QED

Theorem 2:

If E/K is a radical tower and L ⊆ E, then it follows that L/K is also a radical tower.

Proof:

(1) Since E/K is a radical tower, we have (see Definition 4, here):

K = E₀ ⊂ E₁ ⊂ E₂ ⊂ ... ⊂ E_m = E

where for each E_i we have:



such that: q_i is prime, a_i ∈ E_i

(2) Since K ⊂ L, it follows that:

K = E₀ ∩ L

(3) Further, since E_i ⊂ E_i+1, it follows that:

E_i ∩ L ⊆ E_i+1 ∩ L

(4) Since L ⊆ E, it follows that L = E ∩ L

(5) Putting steps #2, #3, and #4 together gives us:

K = (E₀ ∩ L) ⊆ (E₁ ∩ L) ⊆ ... ⊆ (E_m-1 ∩ L) ⊆ L

(6) Now, if remove all cases where E_i ∩ L = E_i+1 ∩ L and renumber, then we get the following:

K = (E₀ ∩ L) ⊂ (E₁ ∩ L) ⊂ ... ⊂ E_n = L

(7) Using Lemma 2 above, we know that for each E_i, (E_i+1 ∩ L)/(E_i ∩ L) is a field extension.

(8) Thus, we have shown that L/K is tower of radicals.

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

Abel's Proof: Field Extensions: Step 1: Lemmas

The content in today's blog is taken from the essay by Michael I. Rosen entitled "Niels Hendrik Abel and Equations of the Fifth Degree."

In today's blog, I present some initial lemmas that I will use to establish Step 1 of the proof using field extensions. For the proof using the ideas that Niels Abel originally presented, see here.

Lemma 1:

Let F be a field containing a primitive n-th root of unity.

If a ∈ F is not an nth power and q is prime, then x^q - a is irreducible over F. (see Definition 4, here).

Proof

This follows from Lemma 2, here.

QED

For Lemma 2 below, I will use the notation F(u) which is defined in Definition 3, here.

Lemma 2:

Let:

α be a root of x^q - a = 0

Then:

every γ ∈ F(α) can be written in the form:

γ = a₀ + a₁α + ... + a_q-1α^q-1

where a_i are in F.

Proof:

This follows directly from the definition of F(u) [see Definition 3, here.]

QED

Lemma 3:

Let:

q be prime

and

m,k be integers such that 1 ≤ m,k ≤ q-1

Then:

there exists integers r,s such that:

rq + sk = m

and

0 ≤ s ≤ q-1

Proof:

(1) Since q is prime and k ≤ q-1, it follows that gcd(k,q)=1. [That is, they are relatively prime]

(2) It follows that 0*k, 1*k, ..., (q-1)*k forms a complete residue system modulo q. [See Lemma 3, here]

(3) This means that for any integer m, there exists an integer s such that sk ≡ m (mod q) and 0 ≤ s ≤ q-1. [See Definition 1, here for review of a complete residue system]

(4) So that m - sk ≡ 0 (mod q). [See here for review of modular arithmetic if needed]

(5) Since q divides m -sk, it follows that there exists an integer r such that rq = m - sk

(6) Then, it follows that rq + sk = m.

QED

For Corollary 3.1 below, you will need to know that F[x] refers to the set of polynomials in field F. (See here for review of polynomials and the meaning of F[x] ). I will also use the notation F(u) which is defined in Definition 3, here.

Corollary 3.1:

Assume that x^q - a ∈ F[x] is irreducible and that α is a root (see here for review of roots of a polynomial).

Let γ be a nonzero element of F(α) with γ not in F.

Then:

There is an element β ∈ F(α) such that β^q ∈ F

and there exists b₀, b₂, ..., b_q-1 ∈ F such that:

γ = b₀ + β + b₂β² + ... + b_q-1β^q-1

Proof:

(1) From Lemma 2 above, we know that there exists a₀, a₁, ..., a_q-1 where:

γ = a₀ + a₁α + ... + a_q-1α^q-1

(2) Since γ is nonzero, we know that there exists a smallest integer k where 1 ≤ k ≤ q-1 and a_k ≠ 0.

(3) Let β = a_kα^k

(4) We know that β^q ∈ F since:

(a) a_k ∈ F [from Lemma 2 above]

(b) Since α is a root, we know that α^q - a = 0 which implies that α^q = a

(c) Since a ∈ F, it follows that α^q ∈ F.

(5) Now for each k+1 ≤ m ≤ q-1, using Lemma 3 above, we know that:

there exists r,s such that:

rq + sk = m

where 0 ≤ s ≤ q-1

(6) Then there exists c_s such that:

α^m = (α^q)^r(α^k)^s = c_sβ^s

where c_s ∈ F.

(7) Thus, if we set c₀ = a₀, we have:

γ = c₀ + β + c_k+1β^k+1 + ... + c_q-1β^q-1

where each c_i ∈ F.

QED

Lemma 4:

Let q be prime and and ζ be a primitive qth root of unity.

Then:

For each integer i:

1 + ζⁱ + ζ²ⁱ + ... + ζ^(q-1)i =



Proof:

(1) Assume that q divides i.

(2) Then there exists an integer x such that qx = i.

(3) So then:

1 + ζⁱ + ζ²ⁱ + ... + ζ^(q-1)i = 1 + (ζ^q)^x + ... + (ζ^q)^x*(q-1) = 1 + 1^x + ... + 1^x*(q-1) = q

(4) Assume q does not divide i.

(5) Then gcd(i,q)=1 and 0*i, 1*i, ..., (q-1)*i form a complete residue system modulo q. [See Lemma 3, here]

(6) Now, since ζ is a primitive q-th root of unity, we know that:

1 + ζ + ζ² + ... + ζ^q-1 = 0 [See Lemma 1, here]

(7) Now, since 0*i, 1*i, ... , (q-1)*i is a complete residue system modulo q, we know that each value is uniquely congruent to a value in the complete residue system {0, 1, 2, ..., q-1}

(8) Further, we know that if i ≡ j (mod q), then ζⁱ = ζ^j since ζ is a primitive qth root of unity [See Lemma 2, here]

(9) But then we have:

1 + ζⁱ + ζ²ⁱ + ... + ζ^(q-1)i = 1 + ζ + ζ² + ... + ζ^q-1 = 0

QED

Lemma 5:

Let y be an a root of an irreducible polynomial f(x) over K.

Let L be a field extension of K such that y ∈ L

Then:

L is a splitting field for f(x)

Proof:

(1) If one root y is an element of L, then all the roots are elements of L. [See Theorem 3, here]

(2) From the Fundamental Theorem of Algebra (see Theorem, here), we know that if y₁, ..., y_n are the roots, then:

f(x) = (x - y₁)*...*(x - y_n)

(3) It therefore follows that L is a splitting field for f(x). [See Definition 3, here]

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.

Abel-Ruffini Proof: Field Extensions

When Niels Abel prepared his impossibility proof, he used the vocabulary of the mathematics of his time. As a result of his work and the work done by Evariste Galois, the very language of mathematics changed. In 1995, Michael I. Rosen wrote an article for The American Mathematical Monthly that explained the ideas of Abel's proof in terms of modern ideas.

In today's blog, I will review the reinterpretation of Abel's proof in terms of modern mathematical concepts. The content in today's blog is taken from Michael Rosen's article.

The content in today's blog, assumes that you feel comfortable with the idea of fields. For review of this concept, start here.

Definition 1: subfield

A set B is a subfield of a set A if both A,B are fields and B is a subset of A.

Definition 2: field extension

A field A is a field extension of a field B if and only if B is a subfield of A.

Definition 3: splitting field over k

A field F is a splitting field over k for a polynomial f(x) if and only if F is a field extension of k and f(x) can be split into linear factors over F.

Definition 4: a radical tower over k

E/k is said to be a radical tower over k if there is a series of intermediate fields such that:

k = E₀ ⊂ E₁ ⊂ ... ⊂ E_m-1 ⊂ E_m = E

such that for each 0 ≤ i ≤ m:



where p_i is a prime and α_i ∈ E_i

Definition 5: solvable in radicals

An equation f(x)=0 is solvable by radicals if and only if there is a radical tower E/k such that F is a splitting field over k and F ⊂ E.

We now state Abel's proof:

Theorem: Abel's Impossibility Proof

Let f(x) = xⁿ - a₁x^n-1 + ... + (-1)ⁿa_n be the general equation of degree n.

If n ≥ 5, then this equation is not solvable in radical.

Proof:

(1) An equation f(x)=0 is solvable by radicals if and only if there is a radical tower E/k such that F is a splitting field over k and F ⊂ E. [Definition 5 above]

(2) If F is contained in a radical tower over k, then F/k is itself a radical tower. [see Theorem here for proof]

(3) If n ≥ 5 and F is a splitting field, then F/k is not a radical tower. [See Theorem, here for proof]

(4) Therefore, F is not contained in a radical tower over k.

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.
• "Splitting Field", Wikipedia