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Is the ring $\displaystyle A=\mathbb{Z} \left[ \frac{1+i \sqrt{7}}{2} \right]$ euclidean?

If $N : z \mapsto z \overline{z}$, then for all $z \in \mathbb{C}$ there exists $a \in A$ such that $N(z-a)<1$, except when $z$ has the form $\displaystyle \left(n+\frac{1}{2} \right)+ \left(m+ \frac{1}{2} \right) \frac{1+i \sqrt{7}}{2}$; in this case, you can only find a large inequality. So $A$ is "almost" euclidean, but is it actually euclidean?

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up vote 5 down vote accepted

Yes, it is Euclidean. Look, for example, at $m=n=0$. You have $$z={1\over2}+{1\over2}{1+i\sqrt7\over2}={3\over4}+{1\over4}i\sqrt7$$ Let $a=1$, so $$z-a=-{1\over4}+{1\over4}i\sqrt7$$ and $$|z-a|^2={1\over16}+{7\over16}={1\over2}\lt1$$

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