Finding the Units in the Ring $\mathbb{Z}[t][\sqrt{t^{2}-1}]$ 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.
 A: For any non-negative integer $n$ the element $\pm (t \pm \sqrt{t^2 - 1})^n$ has norm $1$, and there are no elements $a(t) + b(t) \sqrt{t^2 - 1}$ of norm $-1$ because $a(1)^2 - b(1)^2 \sqrt{1^2 - 1} \ge 0$.  
Suppose $a(t) + b(t) \sqrt{t^2 - 1}$ is a unit which is not one of the above units such that $\deg b$ is minimal.  Then it has norm $1$, so $a(t)^2 - b(t)^2 (t^2 - 1) = 1$.  It is not hard to see that the units $\pm 1$ are the only units for which $b = 0$, so WLOG $b$ is nonzero.  This implies that $\deg a = d+1, \deg b = d$ for some non-negative integer $d$, and moreover the leading terms of $a$ and $b$ must agree up to sign.  If the leading terms agree, then
$$( a(t) + b(t) \sqrt{t^2 - 1})(t - \sqrt{t^2 - 1}) = (ta(t) - b(t) (t^2 - 1)) + (tb(t) - a(t)) \sqrt{t^2 - 1}$$
is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, which contradicts the assumption of minimality.  Similarly, if the leading terms of $a$ and $b$ are opposite in sign, then
$$( a(t) + b(t) \sqrt{t^2 - 1})(t + \sqrt{t^2 - 1}) = (ta(t) + b(t) (t^2 - 1)) + (tb(t) + a(t)) \sqrt{t^2 - 1}$$
is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, again contradicting the assumption of minimality.  So no such units exist.
A: PROOF $\rm\;$ It's simply a typical unit group descent. Let $\rm\: U \:$ be the group generated by the units $\rm\: (\pm t,\: 1)\:$.
 If unit $\rm\; (a,\;b) \not\in U \;$ then $\rm\;  (a,\;b)(\pm t,\; 1) = (\cdots,\;a\pm bt) \;$ is a "smaller" unit $\rm\:\not\in U\;$  since one of $\rm\: a\pm bt \;$ has smaller degree than $\rm\: b\:,\:$ as is easily verified using $\rm\; (a-bt)(a+bt) = 1-b^2 \:. \:\;$ QED
A: 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 $n \in \mathbb{Z}$.
A: Here is another argument:  
Write $x = t + \sqrt{t^2 - 1}$ and $y = t - \sqrt{t^2 - 1}$.    Then $x y = 1$, i.e. $y = x^{-1}$,
and $t = (x + y)/2$ , $\sqrt{t^2 - 1} = (x - y)/2$.
Thus we see that
$\mathbb Z[t][\sqrt{t^2 - 1}] \subset \mathbb Z[1/2][x,x^{-1}].$
Now it is easy to check that if $A$ is an integral domain, then the only units
in $A[x,x^{-1}]$ are of the form $a x^n,$ where $a \in A$ is a unit and $n$ is an integer.
Since the units in $\mathbb Z[1/2]$ are precisely the elements $\pm 2^m$ (for some $m$)
we see that the units in
$\mathbb Z[1/2][x,x^{-1}]$ are of the form $\pm 2^m x^n = \pm 2^m (t + \sqrt{t^2 - 1})^n.$
It is easy to check that if such an element and its inverse both actually lie in 
$\mathbb Z[t][\sqrt{t^2 - 1}],$ then necessarily $m = 0$ (e.g. for norm reasons,
or just looking explicitly at their denominators), and so we get the answer that
the units are precisely the elements of the form $\pm (t + \sqrt{t^2 -1 })^n$.
(Here $n$ runs over $\mathbb Z$; this is the same set as $\pm (t \pm \sqrt{t^2 - 1})^n$,
where $n$ is now non-negative.)
