<s>[INST] 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?
 [/INST] 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 $f((1 + \sqrt{5})/2) = 0$.
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