$\mathbb Z[\xi_{2n}]$ is a pid

Question

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.

Answer 1

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.