I have the following exercise in Miles Reid's Undergraduate Commutative Algebra:
Exercise 0.21: Consider the ring $B′=\Bbb Z[\tau]$, where $\tau^2=\tau+1$. Show that an element $a+b\tau$ is a unit of $B′$ if and only if $N(a+b\tau)=a^2+ab−b^2=\pm 1$. Prove that the multiplicative group of units of $B′$ is generated by $\pm 1,\tau$.
I want to know if my prove is correct, and if there is a faster way to prove the result. I think the first part of the exercise is correct, but I'm not sure about the second part.
Part I
An element $a+b\tau\in B'$ is a unit $\iff N(a+b\tau)=\pm 1$
Proof:
We have $N(a+b\tau)=a^2+ab-b^2$.
$(\Rightarrow)$ Let $u\in B'$ be a unit, then $1=N(1)=N(u)N(u^{-1})$, so $N(u)\in\Bbb Z$ is a unit, that is $N(u)=\pm 1$. Now let $u=a+b\tau$, and notice that $N(a+b\tau)=a^2+ab-b^2$.
$(\Leftarrow)$ Suppose $a^2+ab-b^2=\pm 1$, we can make the factorization: $$a^2+ab-b^2=(a+b\tau)(a+b(1-\tau))=\pm 1$$ hence $a+b\tau$ is a unit.
Part II
The multiplicative group of units of $B'$ is generated by $\pm 1,\tau$
Consider the equation $$a^2+ab-b^2=\pm 1\tag{1}$$
There are four possibilities for the parities of $a,b$ shown in the picture below.
It is enough to prove the result for the first quadrant, where both $a,b>0$. To see this, first define $S=\{x+y\tau\in B'\mid x^2+xy-y^2=\pm 1\}$, then define the counterclockwise rotation $$r:a+b\tau\mapsto -b+a\tau$$ We see that we have $r^4=\text{id}_S$, which means $\text{id}_S,r,r^2,r^3$ are bijections $S\rightarrow S$.
Suppose we start with a solution $s_1+s_2\tau\in S$, then apply $r$ so many times that $r^i (s_1+s_2\tau)$ lies in the first quadrant. If we can show that if $r^i(s_1+s_2\tau)=\pm\tau^n$ then $s_1+s_2\tau=\pm\tau^m$ the result will follow. For this result we need to know:
- We can generalize the Fibonacci numbers with $F_{-n}=(-1)^{n+1}F_n$.
- $\tau^n=F_{n-1}+F_n\tau$
A rotation of a power of $\tau$, is still a power of $\tau$, that is $$r(\tau^n)=(-1)^n\tau^{1-n}$$
Proof: We have that $$\begin{align}r(\tau^n)&=r(F_{n-1}+F_n\tau)\\&=-F_n+F_{n-1}\tau\\&=(-1)^n((-1)^{-n+1}F_n+(-1)^nF_{n-1}\tau)\\&=(-1)^n(F_{-n}+F_{1-n}\tau)\\&=(-1)^n\tau^{1-n}\end{align}$$
Quadrant I
The case follows from the following results:
We have $a\leq b$ except for finitely many cases.
Proof: Suppose that $b<a$, then using eq. $(1)$ we get $$b^2+b^2-b^2<a^2+ab-b^2=\pm 1$$ therefore $b^2<1$ which gives $b=0$. Thus, the exceptions are $$(a,b)=(-1,0),(1,0)$$ which corresponds to the units $\pm 1\in B'$.
We have $b-a\leq a$ except for finitely many cases.
Proof: Suppose that $2a<b$, and then consider $-4(a^2+ab-b^2)=\pm 4$. We get $$-b^2-2b^2+4b^2<-4a^2-4ab+4b^2=\pm 4$$ therefore $b^2<4$, which gives $b=0,\pm 1$. Thus the exceptions are $$(a,b)=(-1,-1),(0,-1), (1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)$$ which corresponds to the units $-\tau^2,-\tau,-\tau^{-1},-1,1, \tau^{-1}, \tau, \tau^2$.
This shows that starting with a solution $(a,b)$ to eq. $(1)$, with $a,b>0$ we can make a smaller solution $(b-a,a)$ in the sense that $b-a\leq a$ and $a\leq b$. This mapping $$\tau^{-1}:(a,b)\mapsto (b-a,a)$$ is the same as computing $(a+b\tau)\tau^{-1}$. At some point either $a=0$ or $b=0$, which gives us a unit one of the units $\pm 1, \tau$. If $a=b$ at some point we get the units $\pm\tau^2$. Therefore $a+b\tau=\pm \tau^n$ for some integer $n$.