Euclidean norm $\mathbb{Z}[d]$

Let $d\in\mathbb{C}\backslash\mathbb{Z}$ and $d^2\in\mathbb{Z}$. Consider the function: $$\nu: \mathbb{Z}[d] \rightarrow \mathbb{Z}, \space \nu(n+md)=n^2-m^2d^2$$

Prove that for $\alpha$ and $\beta$ from $\mathbb{Z}[d]$

1. $\nu(\alpha\beta)=\nu(\alpha)\nu(\beta)$.
2. $\alpha\in\mathbb{Z}[d]^* \iff \nu(\alpha)\in\{-1,1\}$.

My idea:

Let $\alpha=a_1+a_2d, \space \beta=b_1+b_2d$ and $\alpha^*=a_1-a_2d, \space \beta^*=b_1-b_2d$.

So $\nu(\alpha)=\alpha\alpha^*$

Can I do now: $\nu(\alpha\beta)=\alpha\alpha^*\beta\beta^*=\alpha\beta\alpha^*\beta^*=\nu(\alpha)\nu(\beta)$?

How I shall start the second part? And can I use this exercise in proving that $\mathbb{Z}[\sqrt2]$ is an euclidean ring?

• The first part has been answered here. Jan 13, 2015 at 21:57

Suppose $\alpha \in \mathbb{Z}[d]^*$, then there exists $\beta \in \mathbb{Z}[d]^*$ such that $\alpha\beta=1.$ Thus $\nu(\alpha\beta)=\nu(1)=1.$ By the multiplicative property we get $\nu(\alpha\beta)=\nu(\alpha)\nu(\beta)$. But since $\nu(\alpha) \in \mathbb{Z}$ (same for $\nu(\beta)$) therefore $\nu(\alpha)=\pm 1$.
The first part is proved as you suggested, and for the second part one direction is as follows. If $a$ is a unit then there is a $b$ with $ab=1$, hence $1=v(a)v(a)^{-1}=v(a)v(b)$ in $\mathbb{Z}$, which implies $v(a)=\pm 1$, since the norm takes integer values. You can prove that $\mathbb{Z}[\sqrt{2}]$ is norm-Euclidean using the norm, but is has been done already at MSE several times, e.g., see here and here. See also here.