Show the following is a subring of $\mathbb{C}$ Let $m \in \mathbb{Z}$ which is no square and $\alpha = \frac{1+\sqrt{m}}{2} \in \mathbb{C}$. For which $m$ is $\mathbb{Z}[\alpha] = \{a + b \alpha \ | \ a,b \in \mathbb{Z} \}$ a subring.
All of the following properties are obvious:
$\bullet \ 1 \in \mathbb{Z}[\alpha] \\ \bullet \ 0 \in \mathbb{Z}[\alpha] \\ \bullet \ x-y \in \mathbb{Z}[\alpha]$
But then I started with the property that $\forall x,y \in \mathbb{Z}[\alpha] \rightarrow x \cdot y \in \mathbb{Z}[\alpha]$. Which did not work for me yet.
What I tried:
say $x = (a+b \alpha), y = (c + d \alpha) \\ x \cdot y = ac + (ad + bc)\alpha + bd\alpha^2$
I have to show that $bd\alpha \in \mathbb{Z}$ or $bd\alpha^2 \in \mathbb{Z}$.
$bd\alpha = \frac{bd + bd\sqrt{m}}{2} \ \ \ bd\alpha^2 = \frac{bd+2bd\sqrt{m}+m}{4}$
How can I show that one of those two are elements in $\mathbb{Z}$?
 A: Let's start with $a = c = 0, b = d = 1$, just to make things easier. Is $\alpha^2\in \Bbb Z[\alpha]$?
We have
$$
\alpha^2 = \frac{1 + 2\sqrt m + m}{4} = \frac{1+m}4 + \frac{\sqrt m}2
$$
and we want this to be $a + b\alpha$ for some $a, b\in \Bbb Z$:
$$
a + b\alpha = a + \frac b2 + \frac{b\sqrt m}2 = \frac{1+m}4 + \frac{\sqrt m}2
$$
Clearly we must have $b = 1$. This forces
$$
a + \frac 12 = \frac{1+m}4\\
a = \frac{m-1}{4}
$$
which means that $\alpha^2\in\Bbb Z[\alpha]$ iff $m-1$ is divisible by $4$.
And if $\alpha^2\in \Bbb Z[\alpha]$, then $$(a+b\alpha)(c+d\alpha) = ac + (ad + bc)\alpha + bd\alpha^2\in \Bbb Z[\alpha]$$
So as long as $m-1$ is divisible by $4$, $\Bbb Z[\alpha]$ is closed under multiplication and thus a subring of $\Bbb C$.
Note that neither $bd\alpha\in \Bbb Z$ or $bd\alpha^2\in \Bbb Z$ is required.
A: Here $bd\alpha\in{\Bbb Z}[\alpha]$ as $b,d\in{\Bbb Z}$.
But generally $bd\alpha^2\not\in{\Bbb Z}[\alpha]$ since $\alpha^2=\frac{1}{4}(1+\sqrt m)^2 = \frac{1}{4}(1+m + 2\sqrt m)$ and $\frac{1}{4}\not\in{\Bbb Z}[\alpha]$.
