On the Norm of Eisenstein integer

Question

Let us start with the set $\mathbb{Z[w]}=\{a+bw: a,b\in\mathbb{Z}\}$ where $w=(-1+i\sqrt{3})/2$. This set lies inside the set of complex numbers. Let us define the function $$N:\mathbb{Z[w]}\to\mathbb{Z}^+\cup\{0\}$$ by $N(a+bw)=a^2+b^2-ab$. It is easy to check that for any $\alpha,\beta\in\mathbb{Z}[w]$ we have $$N(\alpha\cdot\beta)=N(\alpha)\cdot N(\beta).$$

Now, let $\alpha=a+bw$ and $\beta=c+dw$ be two elements of $\mathbb{Z}[w]$ with $\beta\neq 0$. Then $$\frac{\alpha}{\beta}=t+sw$$ for some $t,s\in \mathbb{Q}$. Let us choose two integers $p$ and $q$ such that $|t-p|\leq\frac{1}{2}$ and $|s-q|\leq\frac{1}{2}$. Write $$\phi=(t-p)+(s-q)w.$$

In here (please see Proposition 1.1), he wrote that $N(\phi)=(t-p)^2+(s-q)^2-(t-p)(s-q)$. I got confused with this because $t-p$ and $s-q$ may not be integers. Is there any reason how he come up with $N(\phi)$?

The norm in this case is just multiplying $\phi$ by its conjugate, which will give you the same formula for rational coefficients since they are invariant under conjugation. — Tob Ernack, Nov 15 '17 at 15:14
@Tob Ermack So, you mean that the function $N$ can be extended to $\mathbb{Q}[w]$ so that if $r+s\in \mathbb{Q}[w]$ then $$N(r+sw)=r^2+s^2-rs?$$ — Juniven Acapulco, Nov 15 '17 at 15:19
$p+q\omega$ is just the quotient of the Euclidean division of $\alpha$ by $\beta$ (Eisenstein's integers are a Euclidean domain for this norm). — Bernard, Nov 15 '17 at 15:49

score 3 · Accepted Answer · answered Nov 16 '17 at 19:48

Presumably Prof. Valmir Buçaj is still alive, so you could ask him. My guess is that he likes to avoid fractions as much as possible, and though it greatly obscures things for specific examples, it perhaps simplifies the process of working out the proof of the theorem.

Declare $$\alpha = a + b \omega = \frac{2a - b}{2} + \frac{b \sqrt{-3}}{2} = \frac{v}{2} + \frac{w \sqrt{-3}}{2}$$ and $$\beta = c + d \omega = \frac{2c - d}{2} + \frac{d \sqrt{-3}}{2} = \frac{x}{2} + \frac{y \sqrt{-3}}{2}.$$ I think I did that correctly... someone will loudly point it out whatever mistake I've made in there. (Also, there is a slight inefficiency in defining $b = w$ and $d = y$, but I wanted to make the point that $\omega \neq w$).

It is assumed that $N(\alpha) \geq N(\beta)$, but if not, the Euclidean algorithm ought to switch them so that they satisfy that loose inequality. And $N(\beta)$ is easily calculated thus: $$N(\beta) = \frac{x^2}{4} + \frac{3y^2}{4}.$$ It's definitely possible that neither of those summands are integers from $\mathbb Z$. But if $x, y \in \mathbb Z$, $N(\beta)$ will work out to an integer just the same. e.g., $$N\left(\frac{5}{2} + \frac{\sqrt{-3}}{2}\right) = \frac{5^2}{4} + \frac{3 \times 1^2}{4} = \frac{25}{4} + \frac{3}{4} = \frac{28}{4} = 7.$$ This is, in my opinion, easier than calculating $N(3 + \omega)$. It also makes it much easier to locate $\beta$ in the complex plane.

Now we wish to compute $$\frac{\alpha}{\beta}.$$ Suppose $$\alpha = \frac{3}{2} + \frac{5 \sqrt{-3}}{2},$$ and $\beta$ is the same as in the previous example. Then $$\frac{\alpha}{\beta} = \frac{\frac{3}{2} + \frac{5 \sqrt{-3}}{2}}{\frac{5}{2} + \frac{\sqrt{-3}}{2}} = \ldots$$ I'm sorry, I don't feel like working out that computation, I'd rather go with $$\frac{\alpha}{\beta} = \frac{4 + 5 \omega}{3 + \omega} = \ldots$$ Yeah, that looks a bit more manageable. I'll worry about locating them in the complex plane on another occasion.

On the Norm of Eisenstein integer

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