## Sunday, September 07, 2008

### Abel's Proof: Step Two

In today's blog, I will cover step 2 of the original proof by Niels Abel on the quintic equation. The content in today's blog is taken from Peter Pesic's excellent book Abel's Proof.

In Step 2, Abel shows that if the general quintic equation has a solution expressible in radicals, then all irrational functions in this formula are expressible as rational functions of the roots.

This step was the gap in Paolo Ruffini's proof.

Lemma 1:

The equation:

[p + R(1/m) + ... + pm-1R(m-1)/m]5 -a[p + R(1/m) + ... + pm-1R(m-1)/m]4 + ... + e = 0

can be reduced to:

0 = q + q1R(1/m) + q2(R(2/m)) + ... + qm-1R(m-1)/m

where q, q1, q2, ... are rational functions based on the quantities a,b,c,d,e,p,p2, ... and R.

Proof:

[p + R(1/m) + ... + pm-1R(m-1)/m]5 -
a[p + R(1/m) + ... + pm-1R(m-1)/m]4 + ... + e = 0

where a,b,c,d,e are rational coefficients.

(2) Since the equation does not involve any additional radicals, we can see that it can be ordered around sums of xR(u/m) where x is a rational function of p,R,a,b,c,d,e and u is an integer.

(3) If u is greater m, then there exists q,r such that u=qm + r where r ≤ m-1.

(4) xR(u/m) = xR(qm+r)/m = xRq*R(r/m)

(5) If we set x' = x*Rq, then we have:

xR(u/m)=x'R(r/m)
where r is less than m.

(6) So, if we number each of the x', then we are left with:

[p + R(1/m) + ... + pm-1R(m-1)/m]5 - a[p + R(1/m) + ... + pm-1R(m-1)/m]4 + ... + e =x'0R(0/m) + x'1R(1/m) + ... + x'm-1R(m-1)/m

where xi is a rational function of a,b,c,d,e,p,pi,R

QED

Corollary 1.1:

If R1/m is not expressible in rationals, then q, q1, q2, ... all = 0.

Proof:

(1) Let z = R(1/m)

(2) So, we have two equations:

zm - R = 0

and

q + q1z + ... + qm-1zm-1 = 0

(3) Using Abel's Lemma (see Lemma 2, here), we can conclude that zm - R=0 is not reducible in rationals.

(4) Now, since R(1/m) is a root for both equations, we can use Abel's Irreducibility Theorem (See Thereom 3 here) to conclude that q, q1, ..., qm-1 must all equal 0.

QED

Lemma 2:

If:  ... Then:

p = (1/m)(y1 + y2 + ... + ym)

Proof:

(1) y1 + y2 + ... + ym =

mp + (1 + α + α2 + ... + αm-1)R(1/m) + p2(1 + α + α2 + ... + αm-1)R(2/m) + ... + pm-1(1 + α + α2 + ... + αm-1)R(m-1)/m

(2) Since (1 + α + α2 + ... + αm-1)=0 (see Lemma 2, here), we are left with:

mp = y1 + y2 + ... + ym

(3) So that we have:

p = (1/m)(y1 + y2 + ... + ym)

QED

Lemma 3:

If:  ... Then:

R(1/m) = (1/m)(y1 + αm-1y2 + ... + αym)

Proof:

(1) y1 + αm-1y2 + αm-2y3 + ... + αym =

= (1 + α + α2 + ... + αm-1)p + mαmR(1/m) + p2αm(1 + α + α2 + ... + αm-1)R(2/m) + .... + αmpm-1(1 + α + α2 + ... + αm-1)R(m-1)/m

(2) Since αm = 1 and (1 + α + α2 + ... + αm-1)=0 (see Lemma 2, here), we are left with:

mR(1/m) = y1 + αm-1y2 + αm-2y3 + ... + αym

QED

Lemma 4:

If:  ... Then:

piR(i/m) = (1/m)(y1 + αm-iy2 + ... + αiym)

Proof:

(1) For any i, we have:

y1 + αm-iy2 + ... + αiym =

= (1 + α + α2 + ... + αm-1)p + mαm(1 + α + α2 + ... + αm-1)R(1/m) + ... + piαmR(i/m) + .... + αmpm-1(1 + α + α2 + ... + αm-1)R(m-1)/m

(2) Since αm = 1 and (1 + α + α2 + ... + αm-1)=0 (see Lemma 2, here), we are left with:

mpiR(i/m) = y1 + αm-iy2 + αm-(i+1)y3 + ... + αiym

QED

Theorem 5:

Let : be a solution to the general quintic equation:

y5 - ay4 + by3 - cy2 + dy - e =0

where p,p2,..., pm-1, R are expressible in radicals, m is a prime, and R(1/m) is irrational.

Then the m roots are:  ... Proof:

(1) We can represent the general quintic equation as follows:

y5 + ay4 + by3 + cy2 + dy - e = 0

(2) If we now insert this solution into the equation at step #1, we are left with:

[p + R(1/m) + ... + pm-1R(m-1)/m]5 -
a[p + R(1/m) + ... + pm-1R(m-1)/m]4 + ... + e = 0

(3) Using Lemma 1 above, we can reduce the above result to get:

0 = q0 + q1R(1/m) + q2R(2/m) + ... + qm-1R(m-1)/m

where q0, q1, q2, ... are rational functions based on the quantities a,b,c,d,e,p,p2, ... and R.

(4) Using Corollary 1.1 above, we know that:

q0, q1, ..., qm-1 all equal 0.

(5) Now, it is also clear that R(1/m) has m different solutions where if R(1/m) is one solution, the solutions are:

R, αR, α2R,..., αm-1R where α is a m-th root of unity.

(6) So, if we use our equation for y, we are left with m roots:  ... QED

Corollary 5.1:

Let : be a solution to the general quintic equation:

y5 - ay4 + by3 - cy2 + dy - e =0

where p,p2,..., pm-1, R are expressible in radicals, m is a prime, and R(1/m) is irrational.

Then:

p,p2, ..., pm-1, R(1/m) are rational functions of α, and the roots: y1, y2, ..., y5

Proof:

(1) From Theorem 5 above, we have the m roots as:  ... (2) Now, we complete this proof using Lemma 2, Lemma 3, and Lemma 4, since now we have:

p = (1/m)(y1 + y2 + ... + ym)

R(1/m) = (1/m)(y1 + αm-1y2 + ... + αym)

piR(i/m) = (1/m)(y1 + αm-iy2 + ... + αiym)

QED

Corollary 5.2:

Let : be a solution to the general quintic equation:

y5 - ay4 + by3 - cy2 + dy - e =0

where each R is itself expressible in the same form such as: Then:

there exists t, t1,1, ... t5,4 such that:

v(1/n) = t + t1,1y1 + ... + t1,4y14 + ... + t5,1y5 + ... + t5,4y54

where v is any nested element of the above form.

Proof:

(1) If v is at the top level, then from Corollary 5.1 above, we know that:

v(1/m) = (1/m)(y1 + αm-1y2 + ... + αym)

(2) Likewise, if v is at the first nested level with R at the top level, then: (3) Using the same logic as Corollary 5.1 above, we where treat R1 = R, R2 = αR, ..., R5 = α4R, then we have:

v(1/n) = (1/n)(R1 + αn-1R2 + ... + αRn)

(4) Then substituting the equation in step #1 above gives us:

v(1/n) = (1/n)({(1/m)[y1 + ... + αym]}5 + ... + α*α4{(1/m)[y1 + ... + αym]}5)

(5) We can keep doing this substitution as far as needed so that we can assume that any nested form of v(1/n) is a function of y1, y2, ..., y5

(6) Finally, we can assume that no power is greater than m-1 since each root is a solution to the quintic equation and we can assume that:

yi5 - ayi4 + byi3 - cyi2 + dyi - e =0

(7) And further that:

yi5 = ayi4 - byi3 + cyi2 - dyi + e

QED

References