Is $\mathbb Z[\frac{1+\sqrt 5i}2]$ a ring of fractions of $\mathbb Z[\sqrt 5i]$?

Question

Does there exist a multiplicative set $S\subset \mathbb Z[\sqrt 5i]$ such that $\mathbb Z[\frac{1+\sqrt 5i}2]\cong S^{-1}\mathbb Z[\sqrt 5i]$?

Since the multiplicative structure of $\mathbb Z[\sqrt 5i]$ is rather bizarre, I have few ideas how to attack it.

My first attempt: I find that it's sufficient to disprove it if one can show that $x^2+5y^2=2^n$ has no integer solution. But I cannot prove this statement either.

Answer 1

$\mathbb Z[\sqrt{-5}]$ is a Dedekind domain with torsion class group, and for these rings every overring, that is, every ring containing it and contained in its field of fractions, is a ring of fractions.

For more details I recommend you this paper, especially Theorem 2, and this topic.

Answer 2

Let $\delta=\frac{1+\sqrt{5}i}{2}$, then $\bar{\delta}=1-\delta$, and $\delta(1-\delta)=\frac{3}{2}$. Then $\delta(1-\delta)-1=\frac{1}{2}$. If $S$ is $\{2^n:n\ge 0\}$, then $S^{-1}\mathbb{Z}[\sqrt{5}i] \subset \mathbb{Z}[\delta]$, and since $\mathbb{Z}[\delta]\subset S^{-1}\mathbb{Z}[\sqrt{5}i]$ it is in fact a ring of fractions.