Subring of complex numbers field Let $R = \left\{ a + b \frac{1+ i \sqrt{3} }{2}: a,b \in \mathbb{Z} \right\}$ .


*

*Show that $R$ is subring of complex numbers field.

*Designate all invertible elements of ring $R$ .


In first I must show that for any $z_1,z_2 \in R$ : $z_1-z_2 \in R$, $z_1 \cdot z_2 \in R$ ?
I have no idea for 2. I will grateful for yours help.
 A: Hint for 2): The usual way to do this is using the field norm, which is defined by
$$N:R\to R,\quad \alpha\mapsto \alpha\overline{\alpha}$$ 
where $\overline{\alpha}$ is the usual complex conjugate of $\alpha$ (since $R\subset\mathbb C$).


*

*Check, that $N$ is multiplicative, that is $N(\alpha\beta)=N(\alpha)N(\beta)$ for all $\alpha,\beta\in R$

*Simplify $\alpha\overline{\alpha}$ in terms of $a,b$ where $\alpha=a+b\frac{1+\sqrt-3}{2}$.

*Deduce from 2), that $N(\alpha)\in\mathbb N_0$ for all $\alpha\in R$.

*Use 1) and 3) to prove, that  $\alpha\in R$ is invertible, iff $N(\alpha)=1$.

*Use 2) and 5) to find all invertible elements.


Result for 2):

 $N\left(a+b\frac{1+\sqrt-3}{2}\right)=a^2+ab+b^2$

A: Let

$$\frac{1+ i \sqrt{3} }{2}=-\omega \implies \omega^2=-\omega $$

where $\omega$ is the root of the unity $x^3-1=0$. Now, let $z_1 = (a+b\omega),\,z_2=(c+d\omega)\in R$, then we have 

$$ (a+b\omega)(c+d\omega)=ad\omega+bc\omega-bd\omega+ac= ac+\left( ad+bc-bd \right) \omega=g+h\omega \in R, $$

where $ g=ac, h= ad+bc-bd \in \mathbb{Z}$.
