How to show that $\mathbb{Z}\left[\frac{1 + \sqrt{5}}{2}\right]$ is finitely generated? We can say that  $\mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ is finitely generated if minimal polynomial of $\cfrac{1 + \sqrt{5}}{2}$ is in $\mathbb{Z}[X]$. After some calculations it can be shown that $f(X) = X^2 - X - 1 \in \mathbb{Z}[X]$ is the minimal polynomial of $\cfrac{1 + \sqrt{5}}{2}$. I think that $\mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ is generated by $\left \{1, \cfrac{1 + \sqrt{5}}{2}\ \right \}$ so that any element of it can be written in the form $a + b \cfrac{1 + \sqrt 5}{2}$ or simply $\cfrac{c}{2} + \cfrac{d \sqrt{5}}{2}$ with $c, d \in \mathbb{Z}$.  
By this claim, if I take some $f\in \mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ then I should be able to write it in the form above. Now let $f = a_n \left(\cfrac{1 + \sqrt{5}}{2} \right)^n + a_{n - 1} \left(\cfrac{1 + \sqrt{5}}{2} \right)^{n - 1} + \dots + a_1 \left(\cfrac{1 + \sqrt{5}}{2} \right) + a_0$, each $a_i \in \mathbb{Z}$. The $k^{th}$ term in the partial sum above is equal to $a_k.\cfrac{n!.5^{\frac{k}{2}}}{k!(n-k)!2^k}$ . I can not see that how these terms cancel out each other and in the end we have something like $\cfrac{c}{2}+\cfrac{d\sqrt{5}}{2}$. How do we show this?
 A: Elements of $\mathbb Z[(1 + \sqrt{5})/2]$ take the form $g((1 + \sqrt{5})/2)$ for some polynomial $g(X) \in \mathbb Z[X]$. By the division theorem, you can write $g(X) = q(X)f(X) + r(X)$ where $r = 0$ or $\deg r < \deg f = 2$ and $q, r \in \mathbb{Z}[X]$. Then
$$ g\left( \frac{1 + \sqrt{5}}{2} \right) = q\left( \frac{1 + \sqrt{5}}{2} \right)f\left( \frac{1 + \sqrt{5}}{2} \right) + r\left( \frac{1 + \sqrt{5}}{2} \right) = r\left( \frac{1 + \sqrt{5}}{2} \right). $$
Since $f((1 + \sqrt{5})/2) = 0$.
A: You might want to check that
$$\left(\frac{1+\sqrt{5}}2\right)^2=\frac{3+\sqrt5}2,$$
$$\left(\frac{1+\sqrt{5}}2\right)^3=2+\sqrt5.$$
$$\left(\frac{1+\sqrt{5}}2\right)^4=\frac{7+3\sqrt5}2$$
etc.
Indeed you can prove (by induction maybe) that
$$\left(\frac{1+\sqrt{5}}2\right)^n=\frac{a_n+b_n\sqrt5}2$$
where $a_n$ and $b_n$ are integers, moreover they are either both odd
integers or both even integers.
A: You just have to show (easy induction) that for any  $n\ge0$, there exists integers $u_n, v_n$ such that 
$$\biggl(1+\frac{\sqrt 5}2\biggr)^n=u_n+v_n\biggl(1+\frac{\sqrt 5}2\biggr).$$
The case $n=2$  simply results from the existence of the minimal polynomial.
A: You've found the minimal polynomial; this gives a very simple way to rewrite the ring:
$$ \mathbb{Z}\left[ \frac{1+\sqrt{5}}{2} \right] \cong \mathbb{Z}[X] / (X^2 - X - 1) $$
so showing the left hand side is finitely generated as an abelian group is the same thing as showing the right hand side is finitely generated as an abelian group.
For the purposes of working with elements, the congruence relation generated by the ideal $(X^2 - X - 1)$ can be summarized as
$$ X^2 \equiv X + 1 $$
which makes it easy to see that the additive group of $\mathbb{Z}[X] / (X^2 - X - 1) $ is a free abelian group with basis given by (the classes of) $1$ and $X$.
