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Let $\mathbb{Z}_{(p)} \subseteq \mathbb{Q}$ denote the ring of rational p-adic integers (p prime), and let $\zeta \in \mathbb{C}$ be a root of unity of order $p \cdot m$, where $m$ is not divisible by $p$. Then apparently, the ring $\mathbb{Z}_{(p)}[\zeta]$ is a principal ideal domain.

In this paper, the above statement is used implicitly, as if it was true for trivial reasons.

Does somebody know the reason why this is true? Does the statement become wrong if $\zeta$ is an arbitrary root of unity? Thank you in advance!

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  • $\begingroup$ I think the case $m=1$ would go approximately as follows. Let $I\subset R=\Bbb{Z}_{(p)}[\zeta]$ be an arbitrary proper ideal. Then $\tilde{I}=I\cap\Bbb{Z}[\zeta]$ is a product of prime ideals of $\Bbb{Z}[\zeta]$. Let $\mathfrak{q}$ be one of those prime ideals, lying above the rational prime $q$. If $p\neq q$, then there are elements of $\mathfrak{q}$ with norms coprime to $p$, i.e. units of $R$, so in $R$ the ideal $\mathfrak{q}$ becomes trivial. Therefore only the case $q=p$ is interesting. But, in $\Bbb{Z}[\zeta]$ we have the factorization of ideal $(p)=(1-\zeta)^{p-1}$. $\endgroup$ – Jyrki Lahtonen Jul 14 '18 at 19:31
  • $\begingroup$ (cont'd) So it looks like all the ideals of $R$ should be powers of $(1-\zeta)$. Several unclear points. And an argument for $m>1$ is missing. I suspect we may be able to $\Bbb{Q}_p(\zeta)$, but I don't have the time to think about this (and its anything but clear to me how we should proceed). $\endgroup$ – Jyrki Lahtonen Jul 14 '18 at 19:33
  • $\begingroup$ @JyrkiLahtonen Thanks a lot for sharing your thoughts! Do you think there is any chance that $\mathbb{Z}_{(p)}[\zeta]$ is simply the intersection of $\mathbb{Q}(\zeta)$ and of the valuation ring of $\mathbb{Q}_p(\zeta)$ (and thus is a discrete valuation ring)? $\endgroup$ – Dune Jul 14 '18 at 19:59
  • $\begingroup$ Probably not. The $m$th cyclotomic polynomial $\Phi_m(x)$ factors non-trivially modulo $p$ and hence also over $\Bbb{Q}_p$ by Hensel. So there should be several ideals "above" the rational prime $p$. Sounds more like the "non-principality" disappears when rational primes other than $p$ become units. I don't see a path though. $\endgroup$ – Jyrki Lahtonen Jul 14 '18 at 20:06
  • $\begingroup$ It is useful to provide context for the question. Maybe your comment about a paper which uses this would be better visible as a part of the question. (And it would also contribute to adding the context.) $\endgroup$ – Martin Sleziak Jul 15 '18 at 10:18
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The useful fact here is that a Dedekind domain with only finitely many primes is a PID. See, for instance, this question. The proof is so simple that I’m abashed not to have seen it.

Anyhow, your ring has only the ideals above $(p)$.

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