Prove that $\mathbb{Z}[\omega]$ is an integral domain

Question

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 $a^2+b^2=ab$.

Answer 1

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

Answer 2

The ring $\Bbb Z[\omega]$ can be embedded into the field $\Bbb C$ of complex numbers. Every subring of a field is an integral domain.