Let $\varepsilon = - \frac{1}{2}+ i \frac{\sqrt{3}}{2}, \varepsilon^3=1$. Now consider the set $\mathbb{Z}[\varepsilon] = \left\{ a + b \varepsilon \mid (a, b) \in \mathbb{Z}^2 \right\} \subseteq \mathbb{C}$. We take for granted that this set is in fact a commutative ring (and an integral domain). Find the elements of the group $(U(\mathbb{Z}[\varepsilon]), \cdot)$, where $U(\mathbb{Z}[\varepsilon]) = \left\{ x \in \mathbb{Z}[\varepsilon] \mid \exists y \in \mathbb{Z}[\varepsilon]\colon xy=1 \right\}$. Prove that $\mathbb{Z}[\varepsilon]$ is a Euclidean domain and find all $p \in \mathbb{Z}[\varepsilon]$ such that the equation $x^2+px+1=0$ has at least one solution.
My approach:
Assume that the polynomial $x^2+px+1=0$ has a root, let's call it $a$. By Bezout's Theorem $x^2+px+1=(x-a)(x-b)$, for some $b \in \mathbb{Z}[\varepsilon]$. So $x^2+px+q=x^2-(a+b)x+ab$. So $a$ and $b$ must be invertible and $p$ can take the values of all the possible sums of invertible elements.
However, I cannot find the set $U(\mathbb{Z}[\varepsilon])$. I have tried using the same approach as with the gaussian integers, to consider the norm $N\colon \mathbb{Z}[\varepsilon] \to \mathbb{N}, \, N(a+b\varepsilon)=a^2+b^2$. And if $a+b\varepsilon$ divides $c+d\varepsilon$ then $N(a+b\varepsilon)$ divides $N(c+d\varepsilon)$, although I don't think $N(xy)=N(x)N(y)$. My intuition says that $U(\mathbb{Z}[\varepsilon])= \left\{ 1, -1, \varepsilon, -\varepsilon, \varepsilon^2, -\varepsilon^2 \right\}$, but I do not know how to formalize this. Any help would be gladly appreciated.
