Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $$|z-a|^2={1\over16}+{7\over16}={1\over2}\lt1$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.