4
$\begingroup$

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?

$\endgroup$
8
$\begingroup$

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 $$|z-a|^2={1\over16}+{7\over16}={1\over2}\lt1$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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