Lacking some steps in a proof regarding set of units of $\mathbb{Z}[\sqrt{3}].$

Question

I am trying to prove that:

$\mathbb{Z}[\sqrt{3}]^* = \{ x \in \mathbb{Z}[\sqrt{3}]: |N(x)| = 1 \}.$

Where $N$ is the norm function defined as follows:

Define the norm function $N:\mathbb{Q}[\sqrt{3}] \rightarrow \mathbb{Q}$ by $N(a_{1} + a_{2}\sqrt{3}) = \operatorname{det} \phi (a_{1} + a_{2}\sqrt{3}) = a_{1}^2 - 3 a_{2}^2.$

My trial is below:

Let $x \in \mathbb{Z}[\sqrt{3}]^*,$ then $x = a + b\sqrt{3}$ and there exists $y = c + d\sqrt{3} \in \mathbb{Z}[\sqrt{3}]^*$ such that $$xy = (a + b\sqrt{3})(c + d\sqrt{3})= (ac + 3bd) + \sqrt{3}(bc + ad) = 1,$$ Hence, $ac + 3bd = 1$ and $bc + ad = 0$

But then I do not know how to complete, could anyone help me please in continuing the proof?

Answer 1

Assuming that $\mathbb{Z}[\sqrt{3}]^*$ is the group of units of the ring $\mathbb{Z}[\sqrt{3}]$, here is a roadmap:

• Let $U=\{ x \in \mathbb{Z}[\sqrt{3}]: |N(x)| = 1 \}$

• Prove that $\mathbb{Z}[\sqrt{3}]^* \subseteq U$ using that $N(xy)=N(x)N(y)$

• Prove that $U \subseteq \mathbb{Z}[\sqrt{3}]^*$ using that $N(x)=x\bar x$