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Let $\xi_{2n} \in \mathbb C$ a primitive $2n^{th}$ root of unity for some integer $n\ge 2 $.

Is the inclusion $\mathbb Z[\xi_{2n}] \hookrightarrow \mathbb C$ flat?

It is possible to answer this question positively by proving that $\mathbb{C}$ is a torsion-free $\mathbb Z[\xi_{2n}]$-module, but then I would need to prove that $\mathbb Z[\xi_{2n}]$ is a PID. Is it a good way to tackle this question? Thank you.

user26857
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Conjecture
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1 Answers1

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There exist values of $n$ for which this ring is not a PID. This is part of a lovely and long story in algebraic number theory -- in some sense the birth of that field was Lamé's faulty assumption that this ring is a UFD.

However, it is locally a PID, and that is (more than) enough to give you that $\mathbb{C}$ is flat. There's probably a simpler proof out there though.

hunter
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    (Simpler proof: the field of fractions is flat (https://math.stackexchange.com/questions/1259551/how-to-show-fraction-field-is-flat-without-localization), field extensions are flat, and flat over flat is flat). – hunter Dec 13 '23 at 23:27
  • Do you mean by using the composition $\mathbb Z[\xi_{2n}]\hookrightarrow\operatorname{Frac}(\mathbb Z[\xi_{2n}])=\mathbb Q[\xi_{2n}] \hookrightarrow \mathbb C$ is flat as a composition of flat maps? And for your nice original answer, you do mean that flatness is a local property and so being torsion-free over a locally PID gives locally flatness? – Conjecture Dec 14 '23 at 11:38
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    yes to both your questions – hunter Dec 15 '23 at 02:48