$\mathbb{Z}[\alpha]$ is a euclidean ring.

Question

I want to prove, that $\mathbb{Z}[\alpha]$ with $\alpha=\frac{1+\sqrt{3}i}{2}$ is an euclidean ring with $N(a+b\alpha)=|a+b\alpha|^2$.

So far I have shown that $N(xy)=N(x)N(y)$ and therefore $N(x)\leq N(xy)$ $\forall x,y\in\mathbb{Z}[\alpha]$.

But now I am stuck with the euclidean division: I want to calculate $$\frac{a+b\alpha}{c+d\alpha}$$ where $a,b,c,d\in\mathbb{Z}[\alpha]$ in order to find the $q,r\in\mathbb{Z}[\alpha]$.

Can someone help me to calculate the above fraction?

I would be grateful for any kind of help or advice!

Thank you.

try multiplying top and bottom by $c+d\bar\alpha $ (complex conjugate of $\alpha$) — J. W. Tanner, May 14 '19 at 15:44

score 2 · Accepted Answer · answered May 15 '19 at 12:38

2

Hint: $N(x+y\alpha)=(x+y\alpha)(x+y\bar\alpha)$.

answered May 15 '19 at 12:38

lisyarus

17,012

$\mathbb{Z}[\alpha]$ is a euclidean ring.

1 Answers1