How to find all units of $\Bbb Z[e^{2\pi i/5}]$? Let $\eta = e^{2\pi i/5}$. It was easy to show that $\{\pm 1, \pm \eta, \pm \eta^2\} \subseteq U(\Bbb Z[\eta]$. I'm struggling to show that only elements of norm $1$ are units which in my opinion would then prove equality. 
 A: What about $\pm\eta^3$ and $\pm\eta^4$? But there are still more units, e.g. we have
$$(1+\eta)(1+\eta^2+\eta^4) = 1+\eta + \eta^2 + \eta^3 + \eta^4 + \eta^5 = 1$$
(using $1+\eta + \eta^2 + \eta^3 + \eta^4 = 0$).
If you know about Dirichlet's unit theorem, then this already tells you that there are in fact infinitely many units. $\mathbb Z[\eta]$ is the ring of integers of the number field $\mathbb Q(\eta)$ which has $r = 0$ real and $s = 2$ pairs of complex embeddings into $\mathbb C$, namely by sending $\eta$ to one of $e^{\pm 2\pi i/5}, e^{\pm 2\cdot 2\pi i/5}$. Dirichlet's unit theorem now tells you that the group of units of $\mathbb Z[\eta]$ is the direct product of the roots of unity (i.e. $\{\pm 1, \pm \eta, \pm \eta^2, \pm \eta^3, \pm \eta^4\}$) and a free abelian group of rank $r+s-1 = 1$.
A: There are definitely other units.
$\eta + \eta^4 = 2\cos \frac{2\pi}{5} = \frac{\sqrt 5 - 1}{2}$. So your ring contains $\mathbb Z[\sqrt 5]$. 
But $\mathbb Z[\sqrt 5]$ contains units like $\pm 2\pm \sqrt 5$.
