## Wednesday, August 23, 2006

### 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) 2 - 1) = (α - 1)*(α + 1)

(2) Since2 - 1) is a cyclotomic integer of the form j - 1), this proves that (α + 1) is a unit. [See Lemma 3, here]

QED

Lemma 2: αi is a unit

Proof:

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

(2) αi divides αλ 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 xλ + yλ = zλ where x,y,z are integers then xyz=0.

Proof:

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

zλ = xλ + yλ = (x + y)(x + αy)(x + α2y)*...*(x + αλ-1y).

and we assume xyz ≠ 0.

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

zλ(α-1) = each factor divided by α - 1.

(3) Since the factors that make up zλ(α-1) 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 + αjy, we have:

(α-1)-1(x + αjy) = ej(tj)λ

where ej is a cyclotomic unit and tj is a cyclotomic integer.

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

(6) From #4, we know that (α - 1) can only divide one ti 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 + αλ-1y) = eλ-1(tλ-1)λ

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

x + αλ-1y = (α-1)eλ-1(tλ-1)λ

Since αλ-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)wλ

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)wλ

(c) x + αy = (α-1)e1(t1)λ

The equation in step #4 gives us:

(α-1)-1(x + αy) = eλ1(tλ1)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)wλ] since:

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

= (α - 1)([e1(t1)λ] - [e0(α-1)wλ])

(10) (α - 1)y = (α - 1)[αe0(α - 1)wλ - αe-1(t-1)λ ] since:

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

= (α - 1)[e0(α - 1)wλ - (e-1(t-1)λ]

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

(α - 1)y = (α - 1)[αe0(α - 1)wλ - (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)wλ - αe0(α - 1)wλ + αe-1(t-1)λ]

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

0 = e1(t1)λ - e0(α - 1)wλ - αe0(α - 1)wλ + 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)wλ - αe0(α - 1)wλ + αe-1(t-1)λ = e1(t1)λ -(e0 + αe0)(α - 1)wλ + αe-1(t-1)λ = e1(t1)λ -(e0)(1 + α) (α - 1)wλ + αe-1(t-1)λ

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

(e0)(1 + α) (α - 1)wλ = e1(t1)λ + αe-1(t-1)λ

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

[(e0)(1 + α)/e1 ] (α - 1)wλ = (t1)λ + (αe-1/e1)(t-1)λ =

= E0(α - 1)wλ = (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 = eλ 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:

aλ + bλ = E0(α - 1)wλ

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)wλ = aλ + bλ = (a + b)(a + αb)(a + α2b)*...*(a + αλ-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+αjb) 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 + αjb) 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 + αjb ≡ [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 + αjb 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 + αjb 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 + αjb 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 + αjb 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 + αjb, 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 = αjb where (α - 1)2 divides a + αjb in step #25.

This means then that we have the following equation:

aλ + bλ = (a + B)(a + αB)(a + α2B)*...*(a + αλ-1B)

We can do this since if B = αjb, then αB = αj+1b, and further αj-1b = αλ-1B.

(29) Now, we note that of the (α - 1) that divide aλ + bλ, 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 + αjB, we have:

(α - 1)-1(a + αiB) = ei(ti)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)wλ

a + αB = (α - 1)e1(t1)λ

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

E(α - 1)wλ = Xλ + Yλ 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