Prove that $\mathbb{Z}[\omega]$ is an integral domain 
Prove that $\mathbb{Z}[\omega]=\{ a+b \omega | a,b \in \mathbb{Z} \}$ is an integral domain, where $\omega^3=1$.

My solution:
Let $(a+b\omega)(c+d\omega)= (ac-bd)+(ad+bc-bd)\omega=0$
$\Rightarrow ac=bd \; \& \; ad+bc=bd. $
If $a\neq  0$, then put $c= bd/a$ to solve the equation and get
\begin{equation}
\begin{split}
&\quad d(a^2+b^2-ab)=0 \\
 &\quad \Rightarrow d= 0 \; \text{or} \; a^2+b^2=ab.
\end{split}
\end{equation}
The first case $d=0$ solves the problem, but I am not able to figure out, how to reject case 2, when $a^2+b^2=ab$.

Since $ab = a^2 + b^2 \geq a^2 > 0$, we must have $b\neq 0$ and $a$ and $b$ must have the same sign. This gives
$$
a^2 + b^2 = ab\\
\frac{a}{b} + \frac{b}{a} = 1
$$
$a/b$ is positive, and therefore has a (possibly irrational) square root. We get
$$
0\leq\left(\sqrt\frac ab - \sqrt\frac ba\right)^2 = \frac{a}{b} +  \frac{b}{a}-2  = 1-2
$$
which is a contradiciton
