## An Intuitive Representation of the Eisenstein Integers

${\mathbb Z} [ \omega ]$

where ω is a primitive cube root of 1 given by

$\omega = \dfrac{-1 + {\mathrm {i}} \sqrt{3}}{2} = \sqrt[3]{1}$

are often represented in the form

$\{ a + b \cdot \omega \;|\; a, b \in {\mathbb Z}\}$
(noting that $\omega^2 = - (1 + \omega)$)

and perhaps abbreviated by pair notation such as

$\{ [ a ; b ]^{*} \;|\; a, b \in {\mathbb Z}\}$

(where the asterisk is to distinguish this notation from that introduced below)

so that

$[ a ; b ]^{*} + [ c ; d ]^{*} = [a+b \;;\; c+d]^{*}$

$[ a ; b ]^{*} \times [ c ; d ]^{*} = [(a c - b d) \;;\; (b c + a d - b d) ]^{*}$

Here is a more intuitive representation that is simpler to manipulate and reason about.

Let σ, τ be primitive sixth roots of 1 given by

$\sigma = \dfrac{1 + {\mathrm i} \sqrt{3}}{2} = \sqrt[6]{1}$

$\tau = \dfrac{1 - {\mathrm i} \sqrt{3}}{2}$

and use the representation

$\{ a \cdot \sigma + b \cdot \tau \;|\; a, b \in {\mathbb Z}\}$

abbreviated by the pair notation

$\{ [ a ; b ] \;|\; a, b \in {\mathbb Z}\}$

i.e.

$[ a ; b ] = a \cdot \sigma + b \cdot \tau$

It so happens that:

$\tau = \sigma^{-1} = \dfrac{1}{\sigma} = \sigma^{5}$

$\sigma^2 = - \tau = \omega$

$\tau^2 = - \sigma$

$\sigma + \tau = 1 = \sigma \cdot \tau$

$\sigma - \tau = {\mathrm i} \sqrt{3}$

So:

$n = [ n ; n ] \in {\mathbb {Z}}$ is pure real

${\mathrm {i}} \sqrt{3} \, n = [ n ; -n ] \in {\mathrm {i}} {\mathbb {Z}}$ is pure imaginary

$[a;b] + [c;d] = [a+b \;;\; c+d]$
— as before

$[a;b] \times [c;d] = [ (ad + bc - bd) \;;\; (ad + bc - ac) ]$
— which is more symmetrical

Some special cases that result are:

$[n;n] \times [m;m] = [nm;nm]$

$[k;k] \times [n;m] = [kn;km]$

$[n;0] \times [m;0] = [0;-nm]$

$[0;n] \times [0;m] = [-nm;0]$

$[n;0] \times [0;m] = [nm;nm]$

And we have:

$[a;b] = a \dfrac{1 + {\mathrm {i}} \sqrt{3}}{2} + b \dfrac{1 - {\mathrm {i}} \sqrt{3}}{2} = \dfrac{a+b}{2} + \dfrac{a-b}{2} {\mathrm {i}} \sqrt{3}$

With results:

$|[a;b]| = \sqrt{a^2 + b^2 - ab}$

$\arg([a;b]) = {\mathrm{atan}}_{(2)} \left( \dfrac{a+b}{2} \;,\; \dfrac{a-b}{2} \sqrt{3} \right) \approx {\mathrm{atan}} \left( \dfrac{a-b}{a+b} \sqrt{3} \right)$