Proof check for the smallest residues with respect to the norm over $\Bbb Z[\omega]$

Question

In the book Algebraic Theory of Quadratic Numbers by Mak Trifkovic, there is an exercise on page 20 as follows:

Find the smallest constant $E\in \mathbb{R}$ for which the following version of the division algorithm holds: for any $\alpha, \beta \in \mathbb{Z}[\omega]$, $\beta \neq 0$, there exist $\gamma, \delta \in \mathbb{Z}[\omega]$ such that $\alpha = \gamma\beta+\delta$ and $N\delta \le E·N\beta$.

Note that, in this book, $\omega = \dfrac{1+\sqrt{-3}}{2}$, and the norm of $x=a+b\omega$ is defined to be $Nx = a^2+ab+b^2$.

I tried to prove this, and here is my attempt:

Since $\alpha = \gamma\beta+\delta$ then $\delta = \alpha -\gamma \beta$. The inequality $N\delta \le E·N\beta$ then can be written as $N(\alpha -\gamma \beta)\le E.N\beta$. Since $N$ is multiplicative then $N(\alpha -\gamma \beta) = N\beta\times N \left(\dfrac{\alpha}{\beta}-\gamma\right)\le E.N\beta$, which is equivalent to $N \left(\dfrac{\alpha}{\beta}-\gamma\right)\le E$.

Then we need to find the smallest $E$ such that for any $\alpha, \beta \in \mathbb{Z}[\omega]$, there exist $\gamma$ such that $N \left(\dfrac{\alpha}{\beta}-\gamma\right)\le E$.

Up to this point I want to use $\gamma$ as an anchor point in the complex plane tessellated by the equilateral triangle form by Eisenstein integers, and I think we can expect $\dfrac{\alpha}{\beta}$ lies near by $\gamma$ so we can induce a value for E (as in the case of integer, $\gamma= \lfloor\dfrac{\alpha}{\beta}\rfloor$). But I can't find anyway to express this in this case. I wonder whether $\dfrac{\alpha}{\beta}$ falls into the triangle $\gamma, \gamma + \omega, \gamma + 1$?

Sorry for my poor English, I can't find any better way to describe my problem with this exercise. Please help me to overcome this. Thanks.

Answer 1

You have the right idea.

Hint: The squared distance from $0$ to $a+b\omega$ is given by $N(a+b\omega)=a^2+ab+b^2$. Divide the plane into equilateral triangles, then a point furthest away from any Eisenstein integer is in the center of a triangle. Can you take it from here?

In the picture above (from Google), $\omega=\frac{-1+\sqrt{-3}}{2}$, so $\omega$ in your case should be $1$ unit to the right.

Answer 2

This is not an answer but a modest contribution consisting in a graphical display (see below) of the set of numbers under consideration that should lead to a certain intuition.

Let $$w:=\tfrac12(1+i \sqrt{3}).$$

The set $\frak{E}$ of Eisenstein integers :

$$\alpha=a+bw, \ \ a,b \ \text{integers}$$ is a triangular lattice represented by red points on fig. 1 below.

The set $\frak{F}$ of values $\tfrac{\alpha}{\beta}$ where $\alpha=a+bw, \ \beta=c+dw$ for a certain range of integer values $a,b,c,d$ is superimposed on this lattice under the form of a "cloud" of blue points. Of course $\frak{E} \subset \frak{F}$.

In an unexpected way, points of $\frak{F}$ are aggregated around points of $\frak{E}$ forming starlike or more or less circular patterns.

As I understand it, your objective minimizing $E$ such that

$\ \ \ \ \ \ \ \ \ \ \ \ \ \forall \alpha, \beta \in \frak{E}, $ $\exists \gamma \in \frak{E}$ $ \ s.t. \ N\left(\dfrac{\alpha}{\beta}-\gamma\right)\le E$

is equivalent to look for a certain radius $\sqrt{E}$ around each element of $\frak{E}$ where there are no elements of $\frak{F} \setminus \frak{E}$ (a disk of "security")

Fig. 1

Answer 3

The lattice $\mathbb{Z}[\omega]$ has points at the vertices of equilateral triangles of side $1$. The Voroni cells are regular hexagons of radius $\frac{1}{\sqrt{3}}$, the required number.