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