Division of Eisenstein Integers

Question

I'm having difficulty figuring out how to divide two Eisenstein integers so I can execute the Euclidean algorithm.

The integers are as follows: $a=-3+13\omega$ and $b=-2+10\omega$. I assume I have to multiply by the conjugate but when I try I'm only left with garbage terms like $\omega^2$.

$ \omega^2 = -1 - \omega $ – Will Jagy Apr 07 '20 at 22:11 — Will Jagy, Apr 07 '20 at 22:11
$ (c+d \omega)(c + d \omega^2) = c^2 - cd + d^2$ – Will Jagy Apr 07 '20 at 22:27 — Will Jagy, Apr 07 '20 at 22:27
There is a very nice discussion in this answer – uhoh Dec 01 '20 at 01:02 — uhoh, Dec 01 '20 at 01:02

GreginGre · Answer 1 · 2022-12-02T17:14:27.343

Let $a=a_0+a_1\omega, b=b_0+b_1\omega, b\neq 0$.

Note $\omega^2=\bar{\omega}=-1-\omega$, so that the conjugate of $b$ is $\bar{b}=b_0+b_1\omega^2=b_0-b_1(1+\omega)$ and that $\vert b\vert^2=b_0^2-b_0b_1+b_1^2$.

The recipe to do Euclidean division is as follows:

Using the relations above, write $\dfrac{a}{b}=u+b\omega,u,v\in \mathbb{Q}$.
Choose $m,n\in\mathbb{Z}$ such that $\vert u-m\vert\leq \dfrac{1}{2},\vert v-n\vert\leq \dfrac{1}{2}$ (in thecase where you have two choices for $m$ or $n$, pick any of the possible choices)
Set $q=m+n\omega$ and $r=a-qb$.

A little algebra shows that $\vert r\vert^2<\vert b\vert^2$, so $q$ is a quotient of $a$ by $b$ and $r$ is a remainder.

Where did the $1/2$ come from? – V. Elizabeth Sep 14 '23 at 23:55 — V. Elizabeth, Sep 14 '23 at 23:55

Division of Eisenstein Integers

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