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For $\alpha = (1+ \sqrt{-3})/2 \in \mathbb{C}$ and $R = \{ x +y\alpha \, | \, x,y \in \mathbb{Z} \}$ how would you prove that $R$ is an Euclidean domain?

I started with letting $v(x) = |z|^2$ but then I'm not really sure where to go from there...

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Here for a geometric proof; it was posted more recently, but I can't find it right now. – Arturo Magidin Mar 4 '12 at 22:20
up vote 0 down vote accepted

In the meantime - you've defined $v(z)=|z|^2$. So $v$ is multiplicative ($v(z)v(w)=v(zw) \; \forall z,w$). Now, given $z,w \in R$, we're looking for $q,r$ such that $z=qw+r$ and $v(r) < v(w)$.

In other words, we're looking for $q \in R$ such that $v(z-qw)<v(w)$. This is the same as saying that $|z-qw|<|w|$, which means that $|\frac{z}{w}-q|<1$ (note that we've moved out of $R$ and into $\mathbb{C}$).

Now choose $y \in \mathbb{Z}$ such that $|y-\frac{2}{\sqrt{3}}Im(\frac{z}{w})| \leq \frac{1}{2}$ and $x \in \mathbb{Z}$ such that $|x+\frac{y}{2}-Re(\frac{z}{w})| \leq \frac{1}{2}$. Then, setting $q=x+y\alpha$, we have that $|\frac{z}{w}-q|<1$ as required.

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