Integral Basis of $\mathbb{Q}(\sqrt{i})$ I understand the Integral basis of  $\mathbb{Q}(\sqrt{i})$  is:  $\{1,i,\sqrt{i},i\sqrt{i}\}$ .
To find out this basis, have i each time to compute the discriminant? 
Any reference book or notes where explains it in a simple way?
Thanks
 A: $x^4+1\bmod p$ is coprime with its derivative thus separable whenever $p\ne 2$.
Thus $\Bbb{Z}[\sqrt{i}]=\Bbb{Z}[x^4]/(x^4+1)$ is ramified only at $2$. Thus every prime ideal above $p\ne 2$ is invertible. 
Above $2$ there is only one prime ideal which is $(2,\sqrt{i}-1)$ and $(2,\sqrt{i}-1)^4=(2)$.
Thus every non-zero prime ideal is invertible, it is a Dedekind domain, whence it has to be $O_{\Bbb{Q}(\sqrt{i})}$.
A: There are several papers on the "Computation of an Integral Basis of Quartic Number Fields". In particular, one by El Fadil for all quartic number field $\Bbb Q(\alpha)$, where the minimal polynomial of $\alpha$ is of the form
$$
X^4+aX+b\in \Bbb Z[X].
$$
Clearly $a=0$ and $b=1$ for $\alpha=\sqrt{i}$. The paper first computes $p$-integral bases and at the last page is is said, that one can recover a triangular integral basis from different triangular $p$-integral
basis for all p as follows - see Proposition $2.2$.
Another reference is Integral Bases for Quartic Fields with Quadratic Subfields, which also gives the result, i.e., $\mathcal{O}_{\Bbb Q(\sqrt{i})}=\Bbb Z[\sqrt{i}]$.
