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

  1. Show that $R$ is subring of complex numbers field.
  2. 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.

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For 1. you must also show that $1 \in R$. – Najib Idrissi Jul 30 '13 at 11:05
Ok, but it is clear because for $b=0$ we have that $1 \in R$. – Thomas Jul 30 '13 at 11:40
It is indeed clear, but you need to say it to prove that $R$ is actually is subring. – Najib Idrissi Jul 30 '13 at 12:06
up vote 2 down vote accepted

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 $R\subset\mathbb C$).

  1. Check, that $N$ is multiplicative, that is $N(\alpha\beta)=N(\alpha)N(\beta)$ for all $\alpha,\beta\in R$
  2. Simplify $\alpha\overline{\alpha}$ in terms of $a,b$ where $\alpha=a+b\frac{1+\sqrt-3}{2}$.
  3. Deduce from 2), that $N(\alpha)\in\mathbb N_0$ for all $\alpha\in R$.
  4. Use 1) and 3) to prove, that $\alpha\in R$ is invertible, iff $N(\alpha)=1$.
  5. Use 2) and 5) to find all invertible elements.

Result for 2):


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So the invertible elements have to satisfy the condition $a^2 +ab + b^2 = \pm 1$ ? – Thomas Jul 30 '13 at 10:44
Yes. Can you prove that, following the steps? – Tomas Jul 30 '13 at 10:46
So we have $(a+ \frac{1}{2} b)^2 + \frac{3}{4} b^2 = 1$ . So it is an ellipse. And what now? – Thomas Jul 30 '13 at 11:38
For example, this implies $\frac{3}{4}b^2\leq 1$, which gives you an upper bound on $|b|$. Remember, that $a$ and $b$ are supposed to be integers, so this leaves only little possibilities to check. – Tomas Jul 30 '13 at 11:43
So we have $a= \mp 1 , b=\pm 1$ or $a = 0 , b= \pm 1$ or $a = \pm 1, b=0$. – Thomas Jul 30 '13 at 12:03


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

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}$.

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