Monday, May 22, 2006

Cyclotomic Integers: x + αiy

In today's blog, I continue with a review of Harold M. Edwards' analysis of cyclotomic integers. The details for today are taken from Section 4.3 of Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

This blog continues from the previous blog that where I showed how polynomials of the form x + αiy relate to Fermat's Last Theorem.

Lemma 1:

if h(α) is a cyclotomic prime that divides x + αiy where x,y are relatively prime and αi ≠ 1,

then there exists a prime integer p divisible by h(α) such that any integer f is divisible by h(α) if and only if f is also divisible by p.


(1) Let h(α) be a cyclotomic prime that divides x + αiy

(2) Then, h(α) also divides N(x + αiy) [See here for definition of norm]

(3) Then N(x + αiy) is a rational integer (see Lemma 5 here) which is a product of primes p1*p2*pn (see Theorem 3 here).

(4) One of the prime numbers pi must be divisible by h(α) (see here for reason).

(5) We know h(α) cannot divide any number relatively prime to pi since:

(a) Let m be a number that is relatively prime to pi

(b) Then, there exists a,b such that 1 = am + bp. (see Lemma 1 here for proof)

(c) if h(α) also divided m, then h(α) would necessarily divide 1 (since it also divides p). But this is impossible since h(α) is a prime (see here for definition of prime).

(6) So, we see that the integers which are divisible by h(α) are all multiples of p.


Corollary 1.1: u ≡ v (mod h(α)) if and only if u ≡ v (mod p)


(1) Let u,v be rational integers.

(2) u ≡ v (mod p) → p divides u - v

(3) h(α) divides p so h(α) divides u - v.

(4) u ≡ v (mod h(α)) → h(α) divides u - v

(5) But by Lemma 1 above, p must also divide u-v.


Lemma 2: h(α) divides x + αiy where x,y are relatively prime implies that αj is congruent to a rational integer mod h(α)


(1) y is not congruent to 0 mod p since:

(a) Assume y ≡ 0 (mod p)

(b) Then x ≡ 0 (mod h(α)) since h(α) would divide both y and x + αiy

(c) And since x is an integer, this would imply that x ≡ 0 (mod p) [By Lemma 1 above]

(d) But then x,y are both divisible by p which is impossible since x,y are relatively prime.

(e) So, we reject our assumption in #1a.

(2) So, there exists a,b such that 1 = ay + bp (see Lemma 1 here for proof)

(3) And this means that bp = ay - 1 so that ay ≡ 1 (mod p) [See here for review of modular arithmetic]

(4) Which also means that ay ≡ 1 (mod h(α)) since h(α) would also divide ay - 1.

(5) Now since h(α) divides x + αiy, h(α) also divides a(x + αiy) = ax + ayαi

(6) Now from step #4, we know that ay ≡ 1 (mod h(α)), so we can take this fact and conclude that ax + ayαi ≡ ax + αi (mod h(α)) [See here for review of modular arithmetic]

(7) But then αi ≡ -ax (mod h(α)) where -ax is a rational integer.


Lemma 3:

if λ is a prime and αλ = 1 and j is less than α but αj ≠ 1, then there exists an integer k, i such that k ≡ (-ax)i (mod p) such that for any cyclotomic integer g(α), g(α) ≡ g(k) mod h(α)


(1) Since j is between 1 and λ - 1 and since λ is a prime, j and λ are relatively prime.

(2) Then, there exists a,b such that aj + bλ = 1. (See Lemma 1 here)

(3) So, aj ≡ 1 (mod λ) (See here for review of modular arithmetic if needed)

(4) So αaj = α

(5) From Lemma 2 above, we know that αj ≡ -ax (mod h(α))

(6) So, αji ≡ (-ax)i (mod h(α))

(7) Let k be an integer such that k ≡ (-ax)i (mod p)

(8) Since h(α) divides p, k ≡ (-ax)i (mod h(α))

(9) From step #4, step #6, and step #8, we get:

k ≡ α (mod h(α))

(10) Using Lemma here, we can conclude:

g(k) ≡ g(α) (mod h(α))



Let h(α) be a prime cyclotomic integer which divides x + αjy (where x,y are relatively prime, αj ≠ 1) and which divides p (a prime integer).

Then there is an integer k where α ≡ k (mod h(α)) such that:

f(α) ≡ g(α) mod h(α) if and only if f(k) ≡ g(k) (mod p)


(1) From Lemma 3 above, we know that there exists an integer k such that:

α ≡ k (mod h(α))

(2) From this and from Lemma 1 here, we know that:

f(α) ≡ f(k) (mod h(α))

g(α) ≡ g(k) (mod h(α))

(3) Now, if f(α) ≡ g(α) (mod h(α)), then from (#2), we have: f(k) ≡ g(k) (mod h(α))

(4) Since f(k) and g(k) are integers, using Lemma 1 above we get:

f(k) ≡ g(k) (mod p)

(5) Now if f(k) ≡ g(k) (mod p), we can also conclude that:

f(k) ≡ g(k) (mod h(α)) since h(α) divides p.

(6) So, finally, we can conclude from step #2 that:

f(α) ≡ g(α) (mod h(α))


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