An Intuitive Representation of the Eisenstein Integers

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)

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