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:
(αi - α-i) = (α - α-1)(αi-1 - αi-3 + ... ± α1-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 = 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(α-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*(Xqλ) =
= aiajXr*(Xqλ - 1) + aiajXr
(e) We can now define a function Qi,j(X) = aiajXr*(Xqλ - 1)/(Xλ-1)
(f) We can see that using Qi,j(X), each term has the following value:
aiajX(λi-i+j) = aiajXr*(Xqλ - 1) + aiajXr = Qi,j(X)(Xλ-1) + aiajXr
(g) if q is greater than 1, we have:
Qi,j(X) = aiajXr*(Xqλ - 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
(c) Since E(α) = e/e, we have:
e/e = -α2s
NOTE: -α2s is a unit since e is a unit and e is a unit. [See Lemma 5, here]
This implies that:
eα-s = -eαs.
NOTE: By Lemma 1 above, α-s and αs are units since they divide -α2s which is a unit.
Further, eα-s, -eαs 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 + ... + 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) + b2(α2 - α-2) + ... + bλ-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) = 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
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