## Monday, August 21, 2006

### Kummer's Proof for Regular Primes: αr

In today's blog, I continue the proof of Fermat's Last Theorem for regular primes. In today's blog, I go over a lemma which is used in the proof for Case I. For context and definitions, please start here at the beginning of this proof.

The details of today's content is taken from Harold M. Edwards Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Lemma 1: a unit is only divisible by a unit

Proof:

(1) Assume that a unit g(α) is divisible by a nonunit h(α).

(2) Ng(α) = 1

(3) Nh(α) ≠ 1 (see Definition 1, here for definition of unit) so Nh(α) ≥ 2.

(4) But then Nh(α) ≥ 2 must divide 1 (see Lemma 6, here) which is impossible.

(5) So we reject our assumption.

QED

Lemma 2: α - α-1 is divisible by α - 1.

Proof:

(1) α - α-1 = α - αλ-1

(2) (α - αλ-1) = (α - 1)(αλ-2 - αλ-3 + ... ± α)

QED

Lemma 3: αi - α-i is divisible by α - α-1

Proof:

i - α-i) = (α - α-1)(αi-1 - αi-3 + ... ± α1-i)

QED

Examples:

2 - α-2) = (α - α-1)(α + α-1)

3 - α-3) = (α - α-1)(α2 - 1 + α-2)

4 - α-4) = (α - α-1)(α3 - α + α-1 + α-3)

Lemma 4: if e is a unit then e/e = αr for some r.

Proof:

(1) Let E(α) = e/e.

(2) E(α) = a0 + a1α + ... + aλ-1αλ-1 (See Lemma 1, here for details)

(3) Then, E(X) = a0 + a1X + ... + aλ-1Xλ-1

(4) Using the Division Algorithm for Polynomials to divide E(Xλ-1)E(X) by (Xλ-1 - 1), we know that there exists Q(X),R(X) such that:

E(Xλ-1)E(X) = Q(X)(Xλ-1 - 1) + R(X)

where R(X) is a polynomial of degree less than λ.

(5) So, we know that there exists Ai such that:

R(X) = A0 + A1X + ... + Aλ-1Xλ-1

(6) If we set X=1, then the equation in step #4 gives us:

E(1λ-1)E(1) = Q(1)*(1λ-1-1) + R(1) =

= E(1)*E(1) = (a0 + a1 + ... + aλ-1)2 =

= Q(1)*0 + R(1) = A0 + A1 + ... + Aλ-1.

(7) If we set X=α, then the equation in step #4 gives us:

E(α-1)*E(α) = [e(α-1)/e-1)][e(α)/e(α)] = [e/e][e/e] = 1 =

= Q(α)*(αλ-1-1) + R(α) = Q(α)*(1-1) + R(α) = R(α) = A0 + A1α + ... + Aλ-1αλ-1

(8) This gives us that:

(A0 - 1) + A1α + ... + Aλ-1αλ-1 = 0.

(9) Since αλ-1 + αλ-2 + ... + α + 1 = 0 [See Lemma 2, here], we can conclude that:

A0 - 1 = A1 = A2 = ... = Aλ-1.

(10) Let k = A0-1 = A1 = ... = Aλ-1

(11) Using the equation from step #6, we have:

(a0 + a1 + ... + aλ-1)2 = A0 + A1 + ... + Aλ-1 = λ*k + 1.

(12) If we subtract k from each Ai we get:

A0 + A1 + ... + Aλ-1 = 1

Under this situation, we get:

(a0 + a1 + ... + aλ-1)2 = 1

which means that

a0 + a1 + ... + aλ-1= ± 1

Further we have the following:

k=0
A0=1
A1 = A2 = ... = Aλ-1 = 0.

(13) We can now show that A0 = a02 + a12 + ... + aλ-12 since:

(a) Each term of E(Xλ-1)E(X) is of the form (aiXλi-i)(ajXj) = aiajX(λi-i+j)

(b) Using the Division Algorithm for Integers, we know that there exists q,r such that:

λi - i + j = qλ + r where r is less than λ.

(c) Since λ divides λ*i, we know that r ≡ j - i (mod λ).

(d) We now have:

aiajX(λi-i+j) = aiajXqλ+r =

= aiajXr*(X) =

= aiajXr*(X - 1) + aiajXr

(e) We can now define a function Qi,j(X) = aiajXr*(X - 1)/(Xλ-1)

(f) We can see that using Qi,j(X), each term has the following value:

aiajX(λi-i+j) = aiajXr*(X - 1) + aiajXr = Qi,j(X)(Xλ-1) + aiajXr

(g) if q is greater than 1, we have:

Qi,j(X) = aiajXr*(X - 1)/(Xλ-1) = aiajXr*(Xqλ-λ + ... + Xλ + 1)

(h) if q = 1, then Qi,j(X) = aiajXr.

(i) if q = 0, then Qi,j(X) = 0.

(j) Since E(Xλ-1)E(X) = Q(X)(Xλ-1 - 1) + R(X) is the sum of all of these terms we can conclude that:

R(X) = ∑ (r=0,λ-1) [∑ (j-i ≡ r) aiaj]Xr.

(k) For A0 where r=0, we have the following:

A0 = (∑(j-i ≡ 0) aiaj)X0 = a0*a0 + a1*a1 + ... + aλ-1*aλ-1 = a02 + a12 + ... + aλ-12.

(14) So A0 = 1 implies that one ai = ±1 and the rest = 0.

(15) Since E(α) = a0 + a1α + ... + aλ-1αλ-1, step #14 implies that:

E(α) = 0 + 0 + (±1)αr + 0 + 0 + 0 +...

So that:

E(α) = ± αr for some r.

(16) Finally, we can show that E(α) = αr

(a) Assume that E(α) = -αr

(b) Since λ is odd, either r is even or r+λ is even (since odd + odd = even) so there exists an s such that E(α) = -α2s

NOTE: Since α = 1, -αr = -αr*1 = -*αrλ = -αλ+r

(c) Since E(α) = e/e, we have:

e/e = -α2s

NOTE:2s is a unit since e is a unit and e is a unit. [See Lemma 5, here]

This implies that:

-s = -eαs.

NOTE: By Lemma 1 above, α-s and αs are units since they divide 2s which is a unit.

Further, -s, -eαs are units because the product of units is a unit (see Lemma 3, here)

(d) Since -s is a cyclotomic integer, we have:

-s = b0 + b1α + ... + bλ-1 [See Lemma 1, here for details]

(e) Let f(α) = b0 + b1α + ... + bλ-1αλ-1

(f) Then we can see that f(α) = eα-s and f(α-1) = es.

where f(α-1) = b0 + b1α-1 + ... + bλ-1α

(g) So that, f(α) = -f(α-1) [from step #16c]

(h) 2*f(α) = f(α) + -f(α-1) = (b0 - b0) + b1(α - α-1) + b22 - α-2) + ... + bλ-1-1 + α)

(i) Now it is clear that (α - α-1) divides 2*f(α) [See Lemma 3 above]

(j) We know further (α - 1) divides 2*f(α) since α - 1 divides (α - α-1) [See Lemma 2 above]

(k) But (α - 1) does not divide 2 since Norm(α-1) = λ and λ doesn't divide Norm(2) = 2λ-1 (see Lemma 6, here for details)

(l) But if (α - 1) divides f(α) then we have a contradiction by Lemma 1 above since a unit is only divisible by a unit and Norm(α - 1) = λ is not a unit (see Definition 1, here).

(m) Therefore, we reject our assumption at step #16a.

QED