Abel's Impossibility Proof: Radicals of the form R(1/m)

In today's blog, I will show an important step in the famous proof by Niels Abel that there is no general solution to the quintic equation expressible in radicals.

In today's blog, I will focus on one result:

If a solution to the general quintic is expressible in nested radicals, then for all radicals of the form R^(1/m), m = 2.

Lemma 1:

Let:

R be a symmetric function on distinct m parameters y₁, ..., y_m

If R^(1/m) is also rational function on these m parameters, then:

it takes m values when the m distinct parameters are permuted.

Proof:

(1) Let R be a symmetric function on m distinct parameters such that:

R(y₁, ..., y_m) = R(y₂, ..., y₁) = ... = R(y_m, ..., y₁)

(2) Let r be a function on y₁, ...., y_m such that r(y₁, ..., y_m) = R(y₁, ..., y_m)^(1/m) such that:

[r(y₁, ..., y_m)]^m = R(y₁, ..., y_m)

(4) By the fundamental theorem of algebra, we know that there are at most m roots that satisfy r^m - R = 0. [See here for proof of the Fundamental Theorem of Algebra]

(5) Since R is the same value for all permutations (since it is symmetric), it follows that r can only take on at most m distinct values in order r^m = R to remain true.

(6) Let α be a primitive mth root of unity.

(7) Then it follows that the possible values of r are r, αr, α²r, ..., α^m-1r.

(8) This means that there are at least m distinct return values and multiplying αⁱ to r is the same as permuting its parameters.

QED

Corollary 1.1

Let R be a rational function on distinct parameters p₁, ..., p_n that takes s distinct return values when these parameters are permuted

Then:

If R^(1/m) is also a function on those n parameters, it follows, that R^(1/m) takes exactly s*m return values when the n parameters are permuted.

Proof:

(1) Let r be a function on y₁, ...., y_m such that r(y₁, ..., y_m) = R(y₁, ..., y_m)^(1/m) such that:

[r(y₁, ..., y_m)]^m = R(y₁, ..., y_m)

(2) By the fundamental theorem of algebra, we know that there are at most m roots that satisfy r^m - R = 0. [See here for proof of the Fundamental Theorem of Algebra]

(3) And, by assumption, there are s different values of R such that we have R₁, ..., R_s

(4) So, it follows that r necessarily has s*m distinct values when its n parameters are permuted since it must satisfy the following s equations which each have m distinct values:

r^m - R₁ = 0

r^m - R₂ = 0

...

r^m - R_s = 0

QED

Lemma 2: The coefficients of a polynomial are symmetric with respect to the roots of the polynomial.

Proof:

(1) From the fundamental theorem of algebra (see here), we know that any polynomial of degree n has n roots and futher that:

a₀xⁿ + ... + a_n-1x + a_n = (x - r₁)(x - r₂)*...*(x - r_n)

where r₁, ..., r_n are the n roots.

(2) From step #1, it is clear that we can view any polynomial as a function of the n roots.

(3) But in this case, it is clear that that permuting the order of the roots will not change the value of the function and hence will not change the value of the coefficients of the polynomial.

QED

Lemma 3:

Let



where R is a function of the m parameters y₁, ..., y_m

Then:

each S, S_i, V^(1/n) are also functions of the m parameters y₁, ..., y_m

Proof:

(1) Let r(x) = S + x + S₂x² + ... + S_n-1x^n-1

(2) Now, it is clear that R = r(V^(1/n))

(3) Let α be a primitive n-th root of unity (such that αⁿ = 1 but α ≠ 1)

(4) We can also define:

R₂ = r(αV^(1/n)) =

= S + αV^(1/n) + S₂α²V^(2/n) + ... + S_n-1α^n-1V^(n-1)/n

...

R_n = r(α^n-1V^(1/n)) =

= S + α^n-1V^(1/n) + S₂α^2(n-1)V^(2/n) + ... + S_n-1α^(n-1)(n-1)V^(n-1)/n

(5) Now, we can use the reasoning in Lemma 2 (see here), Lemma 3 (see here), and Lemma 4 (see here), to conclude that:

S = (1/n)(R + R₂ + ... + R_n)

V^(1/n) = (1/n)(R + α^n-1R₂ + ... + αR_n)

S_iV^(i/n) = (1/n)(R + α^n-iR₂ + ... + αⁱR_n)

(6) Since R, R₂, ..., R_n are all functions of α and y₁, ..., y_m, and since S, V^(1/n), S_i are functions of α, R, R₂, ..., R_n, it follows that S, V^(1/n), S_i are all functions of α and y₁, ..., y_m.

QED

Corollary 3.1:

If a solution to the general mth equation is expressible in the form:



where m is a prime number and R, p₂, ..., p_m-1 are all functions of this same form finitely nested and at the deepest level each p,p_i,R is a function of the coefficients of the general mth equation.

Then:

at all levels, R is a function of the m roots of the the general mth equation

Proof:

(1) At the top level, we know that this is true from Corollary 5.1 here.

(2) Assume it is true up to level i-1 where i-1 ≥ 1 (where 1 = the top level).

(3) But, then using Lemma 3 above, it is true at the level i too.

(4) Therefore, by induction, it is true at all levels of the expression.

QED

Lemma 4:

Let r be a function on m parameters where m is prime and r^m is a symmetric function (that is, it only takes on 1 return value when the m parameters are permuted)

Then:

r can take 1, 2, or m return values when the m parameters are permuted

Proof:

(1) Now, we know that there are m! ways to permute m parameters. [See Lemma 1, here for a review of factorials if needed]

(2) Using Lagrange's Theorem, we know that the number of return values that r takes when the parameters are permuted must divide m!. [See Theorem, here]

(3) From Lemma 1 above, we know that it can take at most m distinct return values.

(4) Using Cauchy's Theorem on Permutations (see Theorem 2, here), we know that r can only take 1, 2, or m different return values when the parameters are permuted.

QED

Lemma 5:

If:

a function v on (y₁, ..., y₅), is symmetric on 4 of the parameters

and the a polynomial y⁵ - ay⁴ + by³ - cy² + dy - e is a quintic equation that takes y₁, ... , y₅ as its roots

Then:

we can assume that there exists functions p,q,r,s such that:

p+y₁ = a

q + py₁ = b

r + qy₁ = c

s + ry₁ = d

and

v is a function on p,q,r,s,y₁

Proof:

(1) Let f be a function on 5 parameters that is symmetric where 4 of the parameters are permuted such that:

f(y₁, y₂, ..., y₅) = f(y₁, y₃, ...) = f(y₁, y₅, ....)

(2) Using the Fundamental Theorem of Algebra (See here for proof of the Fundamental Theorem of Algebra), we know that there exists p,q,r,s such that:

(y - y₂)*...(y - y₅) = y⁴ - py³ + qy² -ry + s.

(3) Since this is a quartic equation and quartic equations are solvable (see Theorem, here), we know that we can define y₂, y₃, y₄, y₅ as functions of p,q,r,s.

(4) Now, from the Fundamental Theorem of Algebra, we also know that there exists a,b,c,d,e such that:

(y - y₁)*...*(y - y₅) = y⁵ - ay⁴ + by³ - cy² + dy - e.

(5) So we have:

(y - y₁)(y⁴ - py³ + qy² -ry + s) = y⁵ - ay⁴ + by³ - cy² + dy - e.

(6) So we can conclude that p,q,r,s can be expressed in terms of a,b,c,d, and y₁ since:

(a) (y - y₁)(y⁴ - py³ + qy² -ry + s)=

y⁵ - (p + y₁)y⁴ + (q + py₁)y³ - (r + qy₁)y² + (s + ry₁)y - sy₁

(b) So that we have:

p+y₁ = a

q + py₁ = b

r + qy₁ = c

s + ry₁ = d

(c) So solving for each above, we get:

p = a - y₁

q = b - py₁ = b - ay₁ + y₁²

r = c - qy₁ = c - by₁ + ay₁² - y₁³

s = d - ry₁ = d - cy₁ + by₁² - ay₁³ + y₁⁴

(7) Since f is a function of y₁, ..., y₅ and y₂, ..., y₅ are functions of p,q,r,s, then it follows that f is a function of y₁, p, q, r, s

QED

Lemma 6:

Let:

R be a rational function of the coefficients of a general quintic equation and also a rational function of the 5 roots of that quintic equation

and

m be a prime number such that R^(1/5) is not a rational function.

Then:

R^(1/5) does not return 5 values when its parameters are permuted.

Proof:

(1) Assume that R^(1/5) is a function on the 5 roots of a quintic equation that returns 5 different values when its parameters are permuted.

(2) Since it returns 5 distinct values, we know that it is not symmetric on all 5 parameters.

(3) On the other hand, it must be symmetric on at least 4, otherwise, it would return more than 5 return values (it would return at least 5*4 = 20 return values)

(4) So, we can conclude that it is symmetric on four and the parameters which can label y₂, ..., y₅ and we label the other parameter y₁.

(5) From Lemma 5 above, we know that there exists p,q,r,s which are functions of the coefficients of the quintic equation such that:

R^(1/5) is a function of p,q,r,s,y₁

(6) Therefore, there exists p, p₁, ..., p_n where each is a function of the coefficients of the quintic equation such that:

R^(1/5) = p₀ + p₁y₁ + p₂ + y₁² + ... + p_ny₁ⁿ

(7) Now, we can also assume that n ≤ 4 since:

(a) Each root is a solution to the quintic equation and we can assume that:

y₁⁵ - ay₁⁴ + by₁³ - cy₁² + dy₁ - e =0

(b) Then:

y₁⁵ = ay₁⁴ - by₁³ + cy₁² - dy₁ + e

(8) So, we have:

R^(1/5) = p₀ + p₁y₁ + p₂ y₁² + ... + p₄y₁⁴

(9) Now, we know that the 5 values that R^(1/5) can take on are:

R^(1/5), αR^(1/5), ..., α⁴R^(1/5)

(10) So, let's assume that switching y₁, y₂ results in R^(1/m) taking on the value α⁴R^(1/5)

(11) Now, this gives us that:

α⁴R^(1/5) = p₀ + p₁y₂ + p₂y₂² + ... + p₄y₂⁴

(12) So, multiplying each side by α gives us:

R^(1/5) = αp₀ + αp₁y₂ + αp₂y₂² + ... + αp₄y₂⁴

(13) But then combining step #8 with step #12, we get:

p₀ + p₁y₁ + p₂ y₁² + ... + p₄y₁⁴ = αp₀ + αp₁y₂ + αp₂y₂² + ... + αp₄y₂⁴

(14) But this is impossible unless y₁ = y₂ or α = 1.

(15) But α is a primitive 5th root of unity so α ≠ 1.

(16) Further, y₁ ≠ y₂ since we assume that swapping these two parameters resulted in a change of value from R^(1/m) to α⁴R^(1/m)

(17) So, we have a contradiction and we are forced to reject our assumption in step #1.

QED

Corollary 6.1:

Let r be a function on m parameters where m is prime and r^m is a symmetric function (that is, it only takes 1 return value when the m parameters are permuted)

Then:

m = 2

Proof:

(1) By Lemma 4 above, the function r can take 1, 2, or m return values.

(2) Since by assumption, m is prime and m divides 5! (See Lagrange Theorem, here for details] it follows that m = 2 or 5.

(4) From Lemma 6 above, we know that m cannot return 5 distinct values so m returns 1 or 2 distinct values when its parameters are permuted.

(5) Since its parameters are distinct, it follows that R^(1/m) must return 2 different values when its parameters are permuted.

(6) Assume that m = 5.

(7) Then we have a contradiction from Lemma 1 since that imples that R^(1/5) would take on 5 distinct return values which is impossible.

(8) So reject our assumption in step #6 and conclude that m = 2.

QED

References

• Peter Pesic, Abel's Proof: An Essay on the Sources and Meaning of Mathematical Unsolvability, Appendix B, The MIT Press, 2004