GCD of two elements in $\mathbb Z \left[\frac{1 + \sqrt{-11}}{2}\right]$ I have to find $(3 + \sqrt{-11}, 2 + 4\sqrt{-11})$ in $\mathbb Z \left[\frac{1 + \sqrt{-11}}{2}\right]$.
If $\mathbb Z \left[\frac{1 + \sqrt{-11}}{2}\right]$ is an Euclidean domain, the euclidean algorithm should be acceptable for computing the GCD of any two elements of the ring.
And here is the trouble: trying to divide given elements at some step I get a remainder with norm bigger than the divider's one. I'm stuck and can't find a solution.
Now I'm not sure if i use the method correctly. Could you please explain me how to do it in the right way or suggest another method?
 A: Well, let's see: $$\frac{3 + \sqrt{-11}}{2 + 4 \sqrt{-11}} = \frac{5 - 11 \sqrt{-11}}{18}.$$ Since $$\frac{5}{18} \approx \frac{1}{2}$$ and $$\frac{-11}{18} \approx -\frac{1}{2}$$ we have $$3 + \sqrt{-11} = (2 + 4 \sqrt{-11})\left(\frac{1}{2} - \frac{\sqrt{-11}}{2}\right) - 20.$$
Yeah, that is a problem. Try $$\frac{5 - 11 \sqrt{-11}}{18} \approx \sqrt{-11}$$ instead. Then $3 + \sqrt{-11} = (2 + 4 \sqrt{-11})(\sqrt{-11}) + 47.$ Hmm, even worse.
There is one third possibility to try: $$3 + \sqrt{-11} = (2 + 4 \sqrt{-11})\left(-\frac{1}{2} - \frac{\sqrt{-11}}{2}\right) + (-18 + 2 \sqrt{-11}).$$ That's the worst one so far. Maybe this is not a Euclidean domain after all.
Wait, did we compare norms for the two GCD function arguments before doing anything else? In $\mathbb Z$ we don't need to worry about that, the algorithm winds up flipping them as needed. But maybe in this domain, $\mathcal O_{\mathbb Q(\sqrt{-11})}$, it is essential. Obviously $N(2 + 4 \sqrt{-11}) > N(3 + \sqrt{-11})$. So then we try $$\frac{2 + 4 \sqrt{-11}}{3 + \sqrt{-11}} = \frac{5}{2} + \frac{\sqrt{-11}}{2},$$ which is an integer in this domain.
That means $2 + 4 \sqrt{-11}$ is divisible by $3 + \sqrt{-11}$, surprise, huh!? Indeed, $N(2 + 4 \sqrt{-11}) = 180$ and  $N(3 + \sqrt{-11}) = 20$. So $$2 + 4 \sqrt{-11} = (3 + \sqrt{-11}) \left(\frac{5}{2} + \frac{\sqrt{-11}}{2}\right) + 0.$$ The algorithm works.
