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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?

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    $\begingroup$ You are asking when the ring of integers of $\mathbf Q(\sqrt[n]{d})$ is $\mathbf Z[\sqrt[n]{d}]$. Presumably you intend for $x^n - d$ to be irreducible (before you take $d$ prime for simplicity). See the theorems in kconrad.math.uconn.edu/blurbs/gradnumthy/integersradical.pdf. You might also be interested in arxiv.org/abs/1909.07184, although it is about a setting over a base ring larger than $\mathbf Z$. $\endgroup$
    – KConrad
    Commented Jul 22, 2020 at 2:46

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