It’s well known that the structure of the maximal order of $mathbb{Q}(sqrt{d})$ depends on $d$ modulo $4$: (assuming $d$ is squarefree), the maximal order is $mathbb{Z}left(frac{1+sqrt{d}}{2}right)$ if $d equiv 1 mod{4}$ and $mathbb{Z}(sqrt{d})$ otherwise. The structure of the biquadratic extension $mathbb{Q}(sqrt{d},sqrt{e})$ has a similar but more complicated description.

Now let’s consider $K=mathbb{Q}(sqrt(n){d})$. For simplicity, let’s consider only $d$ prime. Based on some SAGE computation, it seems that there is a degree $n$ extension $M$ of $mathbb{Q}(mu_{n^2})$ such that $mathcal{O}_K$ contains more than just $mathbb{Z}(sqrt(n){d})$ iff $d$ splits in $M$. (Note that if $n$ is prime, then $M$ is unique.) For $n=2$, one sees that $M=mathbb{Q}(i)$.

Is this known? It doesn’t seem too hard to prove that if $d$ is split, then there are more algebraic integers in $K$. But does this appear anywhere?