Kummer's Proof for Regular Primes: Case II

In today's blog, I continue the proof of Fermat's Last Theorem for regular primes. Case II makes the assumption that λ divides z and that each of the factors (x + αⁱy) are divisible by α - 1 and the resulting quotients are relatively prime. 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: (1 + α) is a unit

Proof:

(1) (α² - 1) = (α - 1)*(α + 1)

(2) Since (α² - 1) is a cyclotomic integer of the form (α^j - 1), this proves that (α + 1) is a unit. [See Lemma 3, here]

QED

Lemma 2: αⁱ is a unit

Proof:

(1) α^λ = 1 is a unit. [See Definition 1, here for definition of cyclotomic unit]

(2) αⁱ divides α^λ so therefore αⁱ is a unit. [See Lemma 5, here]

QED

Theorem: Case II of Kummer's Proof for Regular Primes

If λ divides z, (α - 1) divides all (x + αⁱy), and the resulting quotients are all relatively prime, then:

if x^λ + y^λ = z^λ where x,y,z are integers then xyz=0.

Proof:

(1) From the main proof (see here), we know that:

z^λ = x^λ + y^λ = (x + y)(x + αy)(x + α²y)*...*(x + α^λ-1y).

and we assume xyz ≠ 0.

(2) By the given, we know that each x + α^jy is divisible by α - 1, so that we have:

z^λ(α-1)^-λ = each factor divided by α - 1.

(3) Since the factors that make up z^λ(α-1)^-λ are pairwise relatively prime, it follows that the divisor of each factor is a λth power. (See Lemma 1, here)

(4) Using a property of divisors (see Corollary, here), for each x + α^jy, we have:

(α-1)^-1(x + α^jy) = e_j(t_j)^λ

where e_j is a cyclotomic unit and t_j is a cyclotomic integer.

(5) Since each factor divided by (α - 1) is relatively prime to the others, we can conclude that each t_j is relatively prime to the t_i.

(6) From #4, we know that (α - 1) can only divide one t_i and in fact, it divides t₀ since:

(a) x,y are rational integers

(b) (α - 1) divides λ so α - 1 divides x+y implies that (α-1)^λ-1 divides x + y.

This is true since (x + y) is a rational integer and the only way α - 1 can divide a rational integer is if λ=(α-1)^λ-1 (see Corollary 3.2, here) divides (x + y).

(7) Since (α-1) divides t₀, there exists k,w such that:

t₀ = (α-1)^kw where k ≥ 1 and w is not divisible by (α - 1). [In fact, we know that k is at least λ-2]

(8) We now have the following equations:

(a) x + α^-1y = (α - 1)e_-1(t_-1)^λ

The equation in step #4 gives us:

(α-1)^-1(x + α^λ-1y) = e_λ-1(t_λ-1)^λ

Multiplying (α-1) to both sides gives us:

x + α^λ-1y = (α-1)e_λ-1(t_λ-1)^λ

Since α^λ-1 = α^-1, we can also represent this as :

x + α^-1y = (α-1)e_-1(t_-1)^λ

(b) x + y = (α-1)e₀(α-1)^kλw^λ

The equation in step #4 gives us:

(α-1)^-1(x + y) = e₀(t₀)^λ

Multiplying (α-1) to both sides gives us:

x + y = (α-1)e₀(t₀)^λ

Applying step #7 gives us:

x + y = (α -1)e₀(α-1)^kλw^λ

(c) x + αy = (α-1)e₁(t₁)^λ

The equation in step #4 gives us:

(α-1)^-1(x + αy) = e_λ1(t_λ1)^λ

Multiplying (α-1) to both sides gives us:

x + αy = (α-1)e₁(t₁)^λ

(9) (α - 1)y = (α - 1)[e₁(t₁)^λ - e₀(α - 1)^kλw^λ] since:

(x + αy) - (x + y) = αy - y = (α - 1)y =

= (α - 1)([e₁(t₁)^λ] - [e₀(α-1)^kλw^λ])

(10) (α - 1)y = (α - 1)[αe₀(α - 1)^kλw^λ - αe_-1(t_-1)^λ ] since:

(x + y) - (x + α^-1y) = y - α^-1y = α^-1(α - 1)y =

= (α - 1)[e₀(α - 1)^kλw^λ - (e_-1(t_-1)^λ]

Now if we multiply α to each side, we get:

(α - 1)y = (α - 1)[αe₀(α - 1)^kλw^λ - (αe_-1(t_-1)^λ]

(11) Now if we substract the result in #10 from the result in step #9 we get:

(α -1)y - (α - 1)y = 0 =

= (α - 1) [e₁(t₁)^λ - e₀(α - 1)^kλw^λ - αe₀(α - 1)^kλw^λ + αe_-1(t_-1)^λ]

Now if we divide (α - 1) from both sides we get:

0 = e₁(t₁)^λ - e₀(α - 1)^kλw^λ - αe₀(α - 1)^kλw^λ + αe_-1(t_-1)^λ

(12) Now, let's define E₀, E_-1 such that:

E₀ = [(α + 1)e₀]/e₁

E_-1 = (α*e_-1)/e₁

(13) We can see that both E₀ and E_-1 are units since:

(a) (α+1) is a unit [See Lemma 1 above]

(b) So (α + 1)*e₀ is a unit [See Lemma 3, here]

(c) So [(α + 1)e₀]/e₁ is a unit [See Lemma 5, here]

(d) (α) is a unit [See Lemma 2 above]

(e) So α*e_-1 is a unit [See Lemma 3, here]

(f) So (α*e_-1)/e₁ is a unit [See Lemma 5, here]

(14) We can use E₀, E_-1 to simplify the equation in step #11

0 = e₁(t₁)^λ - e₀(α - 1)^kλw^λ - αe₀(α - 1)^kλw^λ + αe_-1(t_-1)^λ = e₁(t₁)^λ -(e₀ + αe₀)(α - 1)^kλw^λ + αe_-1(t_-1)^λ = e₁(t₁)^λ -(e₀)(1 + α) (α - 1)^kλw^λ + αe_-1(t_-1)^λ

Now, if we add (e₀)(1 + α) (α - 1)^kλw^λ to both sides we get:

(e₀)(1 + α) (α - 1)^kλw^λ = e₁(t₁)^λ + αe_-1(t_-1)^λ

Finally, if we divide e₁ from both sides, we get:

[(e₀)(1 + α)/e₁ ] (α - 1)^kλw^λ = (t₁)^λ + (αe_-1/e₁₎(t_-1)^λ =

= E₀(α - 1)^kλw^λ = (t₁)^λ + E_-1(t_-1)^λ

(15) Since (α-1)^λ-1 = λ * unit (see Corollary 3.2, here), we know that:

0 ≡ C₁ + E_-1C_-1 (mod λ) where C₁ ≡ (t₁)^λ (mod λ) and C_-1 ≡ (t_-1)^λ (mod λ) where C₁ and C_-1 are integers [See Corollary 3.1, here]

(16) C₁, C_-1 are not zero mod λ, since they are relatively prime to t₀ which is divisible by λ. [see step #5 above]

(17) Thus, E_-1 ≡ integer (mod λ) [This follows from step#15 and step#16]

(18) Finally, we can conclude that E_-1 = e^λ for some unit e. [This follows from Condition (B) in the definition of regular primes, see here]

(19) Let a = t₁; Let b = et_-1.

(20) Now applying step #14 to step#19 gives us:

a^λ + b^λ = E₀(α - 1)^kλw^λ

where E₀ is a unit, k is a positive integer, and a,b,(α - 1),w are pairwise relatively prime cyclotomic integers.

(21) E₀(α - 1)^kλw^λ = a^λ + b^λ = (a + b)(a + αb)(a + α²b)*...*(a + α^λ-1b) [See Lemma 1, here for details on this refactoring]

(22) By this equation is it clear that at least one of the factors (a+α^jb) is divisible by α - 1 and from this, it follows that all the factors are divisible by α - 1. [See Step #7 in main theorem, here for proof]

(23) It further follows that all the quotients are relatively prime. [See Step #6 in main theorem, here for proof]

(24) The previous argument (see step #5) that x + y is divisible by (α - 1)² no longer applies since a,b are not necessarily rational integers.

(25) But it is still true that at least one of the factors (a + α^jb) is divisible by (α - 1)² since:

(a) Using Lemma 3 here, we see that:

there exists integers c₀, c₁, d₀, d₁ such that:

a ≡ c₀ + c₁(α - 1) (mod (α - 1)²)

b ≡ d₀ + d₁(α - 1) (mod (α - 1)²)

(b) a + α^jb ≡ [c₀ + c₁(α - 1)] + [1 + (α - 1)]^j[d₀ + d₁(α - 1)] ≡

≡ c₀ + d₀ +[c₁ + d₁ + jd₀](α - 1) mod (α - 1)²

(c) Since (α - 1) divides all a + α^jb by step #22, we see that #25b implies that (α -1) divides c₀ + d₀ and further since c₀, d₀ are rational integers, this implies that λ divides c₀ + d₀.

(d) This also shows that:

(α - 1)² divides a + α^jb if and only if c₁ + d₁ + jd₀ ≡ 0 (mod λ).

This is true since (c₁ + d₁ + jd₀) is a rational integer and (α - 1) only divides a rational integer if λ divides that integer.

(e) This condition holds true for one and only value of j mod λ since (α - 1)² can only divide one since after division by (α - 1) all a + α^jb are relatively prime.

(f) This condition only holds true for one and only if one if d₀ ≡ nonzero (mod λ). The reason for this is that c₀ is shared by all a and if d₀ ≡ 0, then all a + α^jb values would be divisible by (α - 1)².

(g) Additionally, d₀ ≡ nonzero (mod λ) because otherwise (α - 1) would divide b (from step #25a) contrary to step #20 where we stated that a,b,w, and α -1 are relatively prime.

(26) From step #25, we can see that if (α - 1)² divides any factor of a + α^jb, then in the equation in step #21, k must be greater than 1.

(27) From this, we can assume that there exists a positive integer K such that k = K+1.

(28) Let B = α^jb where (α - 1)² divides a + α^jb in step #25.

This means then that we have the following equation:

a^λ + b^λ = (a + B)(a + αB)(a + α²B)*...*(a + α^λ-1B)

We can do this since if B = α^jb, then αB = α^j+1b, and further α^j-1b = α^λ-1B.

(29) Now, we note that of the (α - 1)^kλ that divide a^λ + b^λ, there are (α-1)^λ-1 that divide a + αⁱb where i ≠ j and (α-1)^{1 + Kλ} that divide a + B.

(30) Using the revised equation in step #28, we know that each factor is relatively prime so that it follows that the divisor of each factor is a λth power. (See Lemma 1, here)

(31) Using a property of divisors (see Corollary, here), for each a + α^jB, we have:

(α - 1)^-1(a + αⁱB) = e_i(t_i)^λ

(32) We can now follow the same reasoning in step #8 to get:

a + α^-1B = (α - 1)e_-1(t_-1)^λ

a + B = (α - 1)e₀(α - 1)^Kλw^λ

a + αB = (α - 1)e₁(t₁)^λ

(33) We can follow the same logic in steps #9 to #20 to get:

E(α - 1)^Kλw^λ = X^λ + Y^λ where X = t₁ and Y = et_-1.

We also note that:

X, Y, w, and (α - 1) are all pairwise relatively prime, E is a unit, and K = k-1.

(34) But this means that we are exactly in the same state as step #20 so that we can repeat this same process with a new integer K' = K-1 and so on ad infinitum with each time the positive integer K' decreasing by 1.

(35) Since K' is a positive integer, eventually we get to the state where K'=1 but this is impossible by step #26 where we saw that K' must be greater than 1.

(36) Therefore, we have an impossibility and the proof is done.

QED