I'm asked to prove that $\Bbb Z[\omega]$, where $\omega^2+\omega+1=0$, is a Euclidean domain. The norm is $N(a+b\omega)=(a+b\omega)(a+b\omega^2)$.
My strategy is to write $\alpha=\beta\gamma+\rho$, then look at $$\frac{\alpha}{\beta}=\gamma+\frac{\rho}{\beta}$$ for $\alpha,\beta,\gamma,\delta\in\Bbb Z[\omega]$, and sketch the area where $\gamma$ is the closest Eisenstein integer, in order to bound $N(\rho)$ by $N(\beta)$. Sketching this area is where my confusion starts.
I start by doing this with $0$.
I divide all the distances from $0$ to neighboring Eisenstein integers by $2$, then mark them.
These are the red dots. Then i draw lines between them. $\frac{\rho}{\beta}$ needs to stay inside the star for $\gamma$ to be the closest Eisenstein integer.
According to the book, the shaded region should be:
What am I doing wrong?
Edit: I now see that tiling with parallelograms works far better, and that my star had gaps.
Let $\frac{\rho}{\beta}=a+b\omega$. The longest straight line inside an equilateral triangle, with side length $1$, is the height of value $\frac{\sqrt 3}{2}$. Therefore $$\left|\frac{\rho}{\beta}\right|^2=a^2+b^2\leq\frac{3}{4}\\N\left(\frac{\rho}{\beta}\right)=a^2-ab+b^2\leq\frac{3}{4}-ab$$
If the blue dot hits the upper left corner it will have the coordinates $$\frac{\sqrt 3}{2}(-\cos(30),\sin(30))=\frac{\sqrt 3}{2}(-\frac{\sqrt 3}{2},\frac{1}{2})$$ This gives $$N\left(\frac{\rho}{\beta}\right)\leq\frac{3}{4}-\left(-\frac{\sqrt 3\sqrt 3}{2\cdot 2}\right)\left(\frac{\sqrt 3}{2\cdot 2}\right)= \frac{3}{4}+\frac{3\sqrt{3}}{16}>1$$ This gives a norm greater than $1$, how could this happen? Is it not a good idea to consider the parallelogram?
You only need to show that, for any complex number $z$, there exists an Eisenstein integer $x$ such that $|z - x| < 1$.
This is quite obvious from your picture: if $z$ lies in a triangle, then at least one vertex of the triangle has distance $\leq \frac 1 {\sqrt 3}$ to $z$.
Once you know that, you can perform Euclidean division as follows.
For any pair of elements $a, b\in \Bbb Z[\omega]$ with $b \neq 0$, let $z$ be the quotient $a / b$ and let $x$ be an Eisenstein integer such that $|z - x| < 1$.
We then have $|a - bx| = |b|\cdot |z - x| < |b|$, which gives $|a - bx|^2 < |b|^2$.