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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 $0$. enter image description here

I divide all the distances from $0$ to neighboring Eisenstein integers by $2$, then mark them.

enter image description here

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.

enter image description here

According to the book, the shaded region should be:

enter image description here

What am I doing wrong?

Edit: I now see that tiling with parallelograms works far better, and that my star had gaps.

enter image description here

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?

cansomeonehelpmeout
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    "What am I doing wrong?". You are drawing complicated pictures rather than attending to the problem? I'd look at the parallelogram with vertices $\frac12(\pm1\pm\omega)$. – Angina Seng Jul 07 '20 at 14:21
  • @AnginaSeng What gives you the impression that I'm not "attending to the problem"? For $\frac{\rho}{\beta}$ to have $\gamma$ as the closest Eisenstein integer, it has to lie inside the star. I don't see why I should look at the parallelogram instead, that's essentially what I'm asking. – cansomeonehelpmeout Jul 07 '20 at 15:10
  • You are drawing complicated pictures..... Anyway the locus of the points in $\Bbb C$ that are closer to $0$ than any other Eisenstein integer is a convex hexagon, not a star. – Angina Seng Jul 07 '20 at 16:12
  • @AnginaSeng You're right about the hexagon, thank you! Also, is the parallellogram chosen because it tiles the plane? – cansomeonehelpmeout Jul 07 '20 at 17:09
  • Would the downvoter care to explain how I should improve this post? – cansomeonehelpmeout Jul 07 '20 at 17:11
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    See https://en.wikipedia.org/wiki/Voronoi_diagram for a more general set-up. Every point in the plane has a coset representative in my parallellogram, and every point therein is at distance $<1$ from zero. – Angina Seng Jul 07 '20 at 17:19
  • @AnginaSeng Thank you! – cansomeonehelpmeout Jul 07 '20 at 17:22

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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 $|a - bx|^2 < |b|^2$.

WhatsUp
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  • Is $||$ the norm, or the Euclidian metric? – cansomeonehelpmeout Jul 08 '20 at 10:36
  • The Euclidean metric, i.e. the usual absolute value in $\Bbb C $. – WhatsUp Jul 08 '20 at 10:59
  • @cansomeonehelpmeout You do realize that the norm is the square of the absolute value in $\Bbb{C}$? Minimizing one minimizes the other. This holds in all imaginary quadratic rings. In the real quadratic rings such as $\Bbb{Z}[\sqrt2]$ the norm is not so simple to describe geometrically, because in that case the algebraic conjugates have very different absolute values. – Jyrki Lahtonen Jul 08 '20 at 13:20
  • I know that $N(a+bi)=|a+bi|^2=a^2+b^2$, yes. I was't sure whether the norm in this answer was the $a^2-ab+b^2$ or $a^2+b^2$. – cansomeonehelpmeout Jul 08 '20 at 13:37
  • @WhatsUp Thank you! With $a^2+b^2\leq \frac{1}{3}$ I get that $N(\alpha)\leq\frac{1}{3}-ab$. The point inside the equilateral triangle gives the largest distance with coordinates $\frac{1}{\sqrt{3}}(-\frac{\sqrt{3}}{2},\frac{1}{2})$ which gives $$N(\alpha)\leq\frac{1}{3}+\frac{1}{4\sqrt{3}}<1$$ Does this mean that I really don't need to consider parallelograms, rather equilateral triangles? (I'm confused why the author drew a parallellogram) – cansomeonehelpmeout Aug 09 '20 at 13:07
  • Yes, triangles are enough here. They drew a parallelogram because it represents the quotient $\Bbb C/\Bbb Z[\omega]$, but obviously the parallelogram can be cut into two triangles. – WhatsUp Aug 09 '20 at 15:00