Exercise: Show that the smallest subdomain of complex numbers containing the element $\alpha=\frac{\sqrt{5} - 1}{2}$ is ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$.
I thought I could first show that $\mathbf{Z}[\alpha]$ is indeed an integral domain by verifying all axioms (that's easy) and then I would show that if there's a smaller domain $N$, $x \in \mathbf{Z}[\alpha]$ exists such that $x \notin N$. This $x = a + b\alpha$ for some $a,\ b \in \mathbf{Z}$ (as it is in $\mathbf{Z}[\alpha]$). But from the axioms it follows that for integral domain $N$, we have $\mathbf{Z} \in N$ and $\alpha \in N$, so $a + b\alpha \in N$, a contradiction.
It's easy to show that ${\mathbf Z}[\alpha]$ is a commutative ring, but I have trouble showing that $a,\ b \in {\mathbf Z}[\alpha],\ a,b \neq 0 \implies ab \neq 0$.
Have I chosen a sensible way or is there a better approach?
Thanks.
A subring of a field is automatically an integral domain: zero divisors in it would be zero divisors in the field.
The set $\{a+b\alpha:a,b\in\mathbb{Z}\}$ is surely contained in any subring of $\mathbb{C}$ containing $\alpha$. So you just have to prove it's a subring.
Closure under addition and opposite is clear; moreover $0=0+0\alpha$ and $1=1+0\alpha$. For closure under multiplication: $$ (a+b\alpha)(c+d\alpha)=ac+(bc+ad)\alpha+bd\alpha^2 $$ But $$ \alpha^2=\frac{5+1-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}=1-\frac{\sqrt{5}-1}{2} =1-\alpha $$ so you can conclude.