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 - α^λ-1

(2) (α - α^λ-1) = (α - 1)(α^λ-2 - α^λ-3 + ... ± α)

QED

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

Proof:

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

QED

Examples:

(α² - α^-2) = (α - α^-1)(α + α^-1)

(α³ - α^-3) = (α - α^-1)(α² - 1 + α^-2)

(α⁴ - α^-4) = (α - α^-1)(α³ - α + α^-1 + α^-3)

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

Proof:

(1) Let E(α) = e/e.

(2) E(α) = a₀ + a₁α + ... + a_λ-1α^λ-1 (See Lemma 1, here for details)

(3) Then, E(X) = a₀ + a₁X + ... + a_λ-1X^λ-1

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

E(X^λ-1)E(X) = Q(X)(X^λ-1 - 1) + R(X)

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

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

R(X) = A₀ + A₁X + ... + A_λ-1X^λ-1

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

E(1^λ-1)E(1) = Q(1)*(1^λ-1-1) + R(1) =

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

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

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

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

= Q(α)*(α^λ-1-1) + R(α) = Q(α)*(1-1) + R(α) = R(α) = A₀ + A₁α + ... + A_λ-1α^λ-1

(8) This gives us that:

(A₀ - 1) + A₁α + ... + A_λ-1α^λ-1 = 0.

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

A₀ - 1 = A₁ = A₂ = ... = A_λ-1.

(10) Let k = A₀-1 = A₁ = ... = A_λ-1

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

(a₀ + a₁ + ... + a_λ-1)² = A₀ + A₁ + ... + A_λ-1 = λ*k + 1.

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

A₀ + A₁ + ... + A_λ-1 = 1

Under this situation, we get:

(a₀ + a₁ + ... + a_λ-1)² = 1

which means that

a₀ + a₁ + ... + a_λ-1= ± 1

Further we have the following:

k=0
A₀=1
A₁ = A₂ = ... = A_λ-1 = 0.

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

(a) Each term of E(X^λ-1)E(X) is of the form (a_iX^λi-i)(a_jX^j) = a_ia_jX^(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:

a_ia_jX^(λi-i+j) = a_ia_jX^qλ+r =

= a_ia_jX^r*(X^qλ) =

= a_ia_jX^r*(X^qλ - 1) + a_ia_jX^r

(e) We can now define a function Q_i,j(X) = a_ia_jX^r*(X^qλ - 1)/(X^λ-1)

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

a_ia_jX^(λi-i+j) = a_ia_jX^r*(X^qλ - 1) + a_ia_jX^r = Q_i,j(X)(X^λ-1) + a_ia_jX^r

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

Q_i,j(X) = a_ia_jX^r*(X^qλ - 1)/(X^λ-1) = a_ia_jX^r*(X^qλ-λ + ... + X^λ + 1)

(h) if q = 1, then Q_i,j(X) = a_ia_jX^r.

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

(j) Since E(X^λ-1)E(X) = Q(X)(X^λ-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) a_ia_j]X^r.

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

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

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

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

E(α) = 0 + 0 + (±1)α^r + 0 + 0 + 0 +...

So that:

E(α) = ± α^r for some r.

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

(a) Assume that E(α) = -α^r

(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) = -α^2s

NOTE: Since α = 1, -α^r = -α^r*1 = -*α^r*α^λ = -α^λ+r

(c) Since E(α) = e/e, we have:

e/e = -α^2s

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

This implies that:

eα^-s = -eα^s.

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

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

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

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

(e) Let f(α) = b₀ + b₁α + ... + b_λ-1α^λ-1

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

where f(α^-1) = b₀ + b₁α^-1 + ... + b_λ-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) + b₂(α² - α^-2) + ... + b_λ-1(α^-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) = 2^λ-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