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Let $K\subset L$ be number fields. A prime ideal $\mathfrak{p}$ of $\mathcal{O}_K$ is totally ramified in $\mathcal{O}_L$ (or in $L$) iff $\mathfrak{p}\mathcal{O}_L = \mathfrak{P}^n$, $n=[L:K]$.

Suppose $K=\mathbb{Q}$ and $L=\mathbb{Q}(\zeta_m)$ where $\zeta_m = e^{2\pi i/m}$. Show that $[L:K]=\varphi(m)$ using the following two facts:

  1. if $m=p^k$, then $p\mathcal{O}_L = \langle 1-\zeta_m\rangle^{\varphi(m)}$.
  2. if $\mathfrak{p}$ is totally ramified in $L$ and unramified in another extension $L'$ of $K$ , then $L\cap L'=K$.

My idea:

Firstly, prove that $\langle 1-\zeta_m\rangle$ is a prime ideal of $\mathcal{O}_L$. Then we will have a totally ramified prime, and use this to somehow show that the degree of extension for $m=p^k$ is indeed $\varphi(m)$. Then somehow use the second fact to prove for all $m$.

However, the only way I could prove that $\langle 1-\zeta_m\rangle$ is a prime ideal is by using (1) if an ideal has a prime norm, then it is prime; and (2) norm of a principal ideal is equal to the norm of its generator. But this involves computing norm of $1-\zeta_m$, which requires the knowledge of the degree of extension (number of embeddings or minimal polynomial). Hence it will be a circular argument.

I have spent a full day thinking about this problem and am still clueless. I would be grateful if someone could provide hints/outline of the solution.

Edit: I was able to use the first fact and prove for the special case that $[\mathbb{Q}(\zeta_{p^k}):\mathbb{Q}]=\varphi(p^k)$. However, now I don't know how to extend it to the general case using the second fact.

rationalbeing
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    As far as I understand, the exercise is not asking you to prove 1) and 2), it's asking you to assume 1) and 2) to be true and use them to prove the statement. – Ferra Oct 13 '19 at 10:04

1 Answers1

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If $m = p^k > 2$ and $\zeta = \exp(2\pi i/p^k)$, the only prime that ramifies in $L = \mathbb Q(\zeta)$ is $p$. This is because the discriminant of $L$ is of the form $\pm p^a$, and ramified primes are those dividing the discriminant. You may compute this discriminant using the integral basis $1, \zeta, \ldots, \zeta^{p^k-1}$ for $\mathcal O_L = \mathbb Z[\zeta]$, and the formula $$ \operatorname{disc} O_L = \prod_{i < j} (\zeta^i - \zeta^j)^2 = \pm N_{L/\mathbb Q} (\Phi_{p^k}'(\zeta) ). $$ For more details, see page 96 of Milne's notes.

Your statement 2 then implies the desired result, by using the fact that $$\mathbb Q(\zeta_m) \cap \mathbb Q(\zeta_n) = \mathbb Q(\zeta_{\gcd(m,n)}) = \mathbb Q$$ for $m$ and $n$ relatively prime, and using the multiplicativity of $\varphi$.

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