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We know that if $\xi$ is a primitive $n^\text{th}$-root of unity, then the ring of integers $\mathcal{O}_{\mathbb{Q}(\xi)}$ of $\mathbb{Q}(\xi)$ is $\mathbb{Z}[\xi]$.

Can we generalise this result to say much about the ring of integers $\mathcal{O}_{K(\xi)}$ of $K(\xi)$, where $K / \mathbb{Q}$ is some finite algebraic extension?

Is it the case that $\mathcal{O}_{K(\xi)} = \mathcal{O}_{K}[\xi]$?

If this is not generally true, do we have a characterisation of circumstances where this may hold?

Failing that, do we have an alternate description of $\mathcal{O}_{K(\xi)}$ in terms of $\mathcal{O}_{K}$?

I would appreciate any comments, or even just a reference for these kinds of results.

J. W. Tanner
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Adam Higgins
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    Did you also ask on https://mathoverflow.net/ ? I don't think that there is an easy answer to that question, and you might have better luck finding research specialists over there. – Dirk Oct 28 '20 at 12:11
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    I think you should look at the local question first, and these links might help: https://mathoverflow.net/questions/17289 https://math.stackexchange.com/questions/117973 https://mathoverflow.net/questions/136052 – Bart Michels Oct 28 '20 at 12:49
  • Thank you for your suggestions. It is clear to me that my question is far more involved than I thought it might be. I'll have a look through those links and consider posting this question to mathoverflow. – Adam Higgins Oct 28 '20 at 12:58

1 Answers1

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By a result in Hilbert's report, the ring of integers in a compositum KL of normal extensions is the obvious one generated by the rings of integers in $K$ and $L$ if the discriminants of $K$ and $L$ are coprime.

If the discriminants are not coprime, the result is fals in general. In particular, the ring of integers in ${\mathbb Q}(\sqrt{3},\sqrt{-1})$ is not ${\mathbb Z}[\sqrt{3}][\sqrt{-1}]$.

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    Not sure if it is the best way but coprime discriminant implies that each uniformizer of $O_{LK}$ is either in $O_L$ or $O_K$, so $O_L O_K$ contains all the uniformizers and hence is a Dedekind domain. – reuns Oct 28 '20 at 14:17
  • Thank you for this! Especially for the example. I'm going to leave the question open, at least for a little while, to encourage further discussion of the situation in which the equality may fail (i.e non-coprime case) – Adam Higgins Oct 28 '20 at 14:38