Exercise 1.35

Proof. If $\alpha =a+\mathit{b\omega }\in \mathbb{Z}[\omega ]$, using $1+\omega +{\omega }^2=0$ and $\bar{\omega }={\omega }^2$, we obtain

Consequently, $\lambda$ is a multiplicative function.

$\text{∙}$ If $\lambda (\alpha )=1$, then $\alpha \bar{\alpha }=1$, where $\bar{\alpha }=a+b{\omega }^2=(a-b)-\mathit{b\omega }\in \mathbb{Z}[\omega ]$, so $\alpha$ is an unit.

$\text{∙}$ Conversely, if $\alpha$ is an unit, there exists $\beta \in \mathbb{Z}[\omega ]$ such that $\mathit{\alpha \beta }=1$, then $\lambda (\alpha )\lambda (\beta )=1$, where $\lambda (\alpha ),\lambda (\beta )$ are positive integers, so $\lambda (\alpha )=1$.

If $b=0$, then $a=\pm 1$, $\alpha =1$ or $\alpha =-1$

If $b=1$, then ${(2a-1)}^2=1,2a-1=\pm 1$ : $a=0$ or $a=1$, $\alpha =\omega$ or $\alpha =1+\omega =-{\omega }^2$.

If $b=-1$, then ${(2a+1)}^2=1,2a+1=\pm 1$ : $a=0$ or $a=-1$, $\alpha =-\omega$ or $\alpha =-1-\omega ={\omega }^2$.

So

The set of units of $\mathbb{Z}[\omega ]$ is the group of the roots of $x^6-1$. □