TITLE: Finding the Units in the Ring $\mathbb{Z}[t][\sqrt{t^{2}-1}]$
QUESTION [7 upvotes]: This is problem taken from Problem 4.
I couldn't find the solution anywhere and I am curious to see a solution for this problem, as I can at least comprehend the question and it seems that the mechanism for the solution involved will be somewhat understandable.

REPLY [2 votes]: Just a "geometric translation" of Matt's "algebraic" proof:
It is clear that the ring $\mathbb{Z}[t, \sqrt{t^2 - 1}]$ is equal to $A = \mathbb{Z}[x,y] / (x^2 - y^2 + 1)$. Consider the ring $B = A \otimes_\mathbb{Z} \mathbb{C} = \mathbb{C}[x,y] / (x^2 - y^2 + 1)$.
$B$ is the ring of regular functions of the hyperbola $X \subseteq \mathbb{A}^2_\mathbb{C}$ defined by the equation $x^2 - y^2 + 1 = 0$. The projection of $X$ into one of its asymptotes gives an isomorphism
$$
(x,y) \mapsto x-y
$$
which maps $X$ onto $\mathbb{A}^1_\mathbb{C} \setminus \{ 0 \}$. Therefore $B$ is isomorphic, as a $\mathbb{C}$-algebra, to $\mathbb{C}[u,u^{-1}]$, where $u = x-y$ is trascendental over $\mathbb{C}$. So every unit in $B$ is of the form $\lambda u^n$, for some $\lambda \in \mathbb{C}^*$ and $n \in \mathbb{Z}$. Since $A$ is flat over $\mathbb{Z}$, $A \subseteq B$. Therefore every unit in $A$ is of the form $\pm  (x-y)^n$, for some $n \in \mathbb{Z}$.