Integral Basis of $\mathbb{Q}(\sqrt{i})$

Question

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

Answer 1

$x^4+1\bmod p$ 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 $O_{\Bbb{Q}(\sqrt{i})}$.

Answer 2

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., $\mathcal{O}_{\Bbb Q(\sqrt{i})}=\Bbb Z[\sqrt{i}]$.