## Tuesday, July 19, 2005

### Eisenstein Integers

Eistenstein integers are quadratic integers of the form Z[(-1 + √-3)/2]. They are named in honor of Ferdinand Eisenstein. For those, who need a review of quadratic integers, start here.

In today's blog, I will go over some basic properties of Eisenstein Integers. Later on, I will show how they can be used to prove that Fermat's Last Theorem has no solutions for n=3.

The details of today's blog are based on an English translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

For purposes of the proof, Dorrie proposed the idea of G-numbers. There are numbers that are based on two values: J,O where

J = (1 + i√3)/2
O = (1 - i√3)/2.

NOTE: In case this is not clear to anyone, i√3 = √-3.

Lemma 1: Every integer g is also a G-number.

g = gJ + gO = (g + gi√3)/2 + (g - gi√3)/2 = 2g/2 = g.

QED

Lemma 2: The following properties are true:

(a) J + O = 1

From the previous lemma.

(b) JO = 1

JO = (1 - [(-1)(3)])/4 = 4/4=1

(c) J2 + O = 0

J2 = (1 + i√3)2/4 = (1 + 2i√3 + (-1)(3)) /4 = (-2 + 2i√3)/4 =
(-1 + i√3) /2

(d) O2 + J = 0

O2 = (1 - i√3)2/4 = (1 - 2i√3 + (-1)(3))/4 = (-2 - 2i√3)/4 =
(-1 - i√3) /2

(e) J3 = -1

J2 = -O [From (c)]
So, J3 = -JO = -1 [From (b)]

(f) O3 = -1

O2 = -J [From (d)]
So, O3 = -OJ = -1 [From (b)]

QED

Lemma 3: The norm of a G-number aJ + bO is a2 + b2 - ab

(1) aJ + bO = (a + ai√3)/2 + (b - bi√3)/2 = (a + b)/2 + [(a - b)i√3]/2

(2) Norm(aJ + bO) = [(a + b)/2 + [(a - b)i√3]/2][(a + b)/2 - [(a - b)i√3]/2] =
(a + b)2/4 - (a - b)2(-1)(3)/4 =
[a2 + 2ab + b2]/4 - (-3)[a2 - 2ab + b2]/4 =
(a2 + 3a2 + b2 + 3b2 + 2ab - 6ab)/4
= a2 + b2 - ab.

QED

Corollary 3.1: The norm is always greater or equal to 0.

In other words, we can prove that a2 + b2 ≥ ab.

Case I: a=b

a2 + b2 = 2a2 which is ≥ ab = a2.

Case II: a is greater than b

In this case: a2 is greater than ab.

Case III: a is less than b

Same argument using b instead of a

QED

Lemma 4: The following numbers are units: J,-J, O, -O, 1,-1

(a) The norm of each of these values is 1. [See here for the explanation of why the norm of a unit is one]

J = 1(J) + 0(O) = 1 + 0 + 0 = 1
-J = (-1)J + 0(O) = 1 + 0 + 0 = 1
O = 0J + 1(O) = 0 + 1 + 0 = 1
-O = 0J + (-1)O = 0 + 1 + 0 = 1
1 = (1)J + (1)O = 1 + 1 - 1 = 1
-1 = (-1)J + (-1)O = 1 + 1 -1 = 1

(b) There are no other units

Norm(Z[(-1 + i√3)/2]) =[(a + bi√3)/2][(a - bi√3)/2] = (a2 - b2(-3) )/4= ±1

This means that: (since the norm in this case must be positive)

a2 + 3b2 = 4.

We know that absolute(b) has to be less than 2. Otherwise, 3b2 is greater than 4.

This means that we only need to consider b=-1,0,1.

If b = 0, a = ±2.
If b = 1 or b = -1 a = ±1

So, the only units are:

a=2,b=0: (2+0i√3)/2 = 1
a=-2,b=0 (-2+0i√3)/2=-1
a=1,b=1 (1 + i√3)/2 = J
a=-1,b=1 (-1 + i√3)/2 = -O
a=1, b=-1 (1 - i√3)/2 = O
a=-1, b=-1 (-1 - i√3)/2 = -J

QED

Lemma 5: J-O is a prime.

(1) Norm(J-O) = (1)2 + (-1)2 - (1)(-1) = 3.
(2) Only primes have norms that are rational integer primes. [The argument is the same as the one for Gaussian Integers, see here]

QED

Lemma 6: If aJ+bO is divisible by J-O, then a + b is divisible by 3.

(1) Since aJ + bO is divisible by J-O, there exists a value mJ + nO such that:

aJ + bO = (mJ + nO)(J - O) = mJ2 - mJO + nJO - nO2

(2) From the properties of G-Numbers (see above), we know that:

J2 = -O
O2 = -J
OJ = 1
-m = -mJ -mO
n = nJ + nO

(3) Applying these properties to (1), gives us:

aJ + bO = m(-O) -mJ - mO +nJ + nO -n(-J) = (2n-m)J + (n - 2m)O

(4) So that:
a = 2n-m
b = n-2m

(5) a + b = 2n -m + n - 2m = 3(n - m)

QED

Lemma 7: If 3 divides a + b, then J - O divides aJ + bO.

(1) So there exists g such that a + b = 3g
(2) We also support that there exists m,n such that n - m = g and 2n -m = a.
(3) This gives us that b = 3g -a = 3n - 3m - 2n + m = n - 2m.
(4) aJ + bO = (2n - m)J + (n - 2m)O
(5) Applying the reverse logic from above, we get:
(2n - m)J + (n - 2m)O = m(-O) -mJ - mO +nJ + nO -n(-J) =
=
mJ2 - mJO + nJO - nO2 = (mJ + nO)(J - O)

QED

Lemma 8: If J - O doesn't divide aJ + bO, then (aJ + bO)3 ≡ ±1 (mod 9)

(1) Since J - O doesn't divide aJ + bO, we know that 3 doesn't divide a + b since:

(a) Assume that J-O doesn't divide aJ + bO
(b) Assume that 3 divides a + b
(c) Then by the Lemma above J-O divides aJ + bO
(d) But this is not true by the assumption in (a) so we have a contradiction and we reject the assumption in (b) and conclude that 3 doesn't divide a+b.

(2) This means that we have the following possible cases:

(a) a = 3g, b = 3k ±1
(b) a = 3g ±1, b = 3k
(c) a = 3g ±1, b = 3k ±1

(3) Let λ = gJ + kO. Then in each of the three cases we have:

(a) (3g)J + (3k ±1)O = 3λ ±O
(b) (3g ±1)J + (3k)O = 3λ ±J
(c) (3g ±1)J + (3k ±1)O = 3λ ±1

(4) We can rewrite this as 3λ + ε where ε = ±O, ±J, or ±1.

(5) We can cube this result to get:
(3λ + ε)3 = (9λ2 + 6λε + ε2)(3λ + ε) =
=27λ3 + 18λ2ε + 3λε2 + 9λ2ε + 6λε2 + ε3 =
= 9 (3λ3 + 3λ2ε + λε2) + ε3

(6) Additionally, we note that ε3 = ±1 since:

(a) O3 = -1
(b) (-O)3 = 1
(c) J3 = -1
(d) (-J)3 = 1
(e) 13 = 1
(f) (-1)3 = -1

(7) Hence, we have shown that:
(aJ + bO)3 ≡ ±1 (mod 9).

QED

Lemma 9: J - O divides 3.

(1) 3 = 3J + 3O
(2) Since 3 + 3 is divisible by 3, then 3J + 3O is also divisible by J - O. [See above]

QED

Corollary 9.1: (J-O)3 divides 9.

(1) (J - O)3 = (J2 - 2JO + O2)(J - O) =
J3 - 2J2O + JO2 - J2O +2JO2 - O3 =
-1 -3J2O + 3JO2 + 1 =
3(JO2 - J2O)

(2) Since O2 = -J and J2 = -O, we get:
3(O2 - J2) = 3(O - J)(O + J) = 3(-1)(1)(J - O)

QED

Lemma 10: (J - O)2 = -3

(1) (J - O)2 = J2 - 2JO + O2

(2) Since OJ = 1, -2JO = -2.

(3) Since O2 = -J and J2 = -O, we get J2 + O2 = -O + -J = -1(O + J) = -1.

QED

Scouse Rob said...

I don't understand the reasoning in the following Lemma:

Lemma: If J - O doesn't divide aJ + bO, then aJ + bO ≡ ±1 (mod 9)

Do you mean (aJ + bO)^3 ≡ ±1 (mod 9)?

In step (5) you show that:
(aJ + bO)^3 ≡ ε^3 (mod 9)

In step (6) you show that ε^3 = ±1.

I don't understand step (7) unless it should be (aJ + bO)^3 instead of (aJ + bO).

Thanks

Rob

Larry Freeman said...

Hi Rob,

You are correct in your reasoning. I've updated the proof.

Cheers,

-Larry