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I am trying to understand this proof of the Kronecker-Weber theorem by Franz Lemmermeyer http://arxiv.org/pdf/1108.5671.pdf.

The set-up is this: $K/\mathbb{Q}$ is a cyclic extension of prime degree $p$ where $p$ is the only ramified prime. We have $F=\mathbb{Q}(\zeta_p)$ and $L=KF$, the compositum, meaning that $L=F(\sqrt[p]{\mu})$ for $0\neq \mu \in \mathcal{O}_F$.

My question is regarding the proof of Lemma 2.2, (Let $\mathfrak{q}$ be a prime ideal in $F$ with $(\mu)=\mathfrak{q}^r\mathfrak{a}$, where $\mathfrak{q}\nmid\mathfrak{a}$. If $p\nmid r$ and $L/\mathbb{Q}$ is abelian, then $\mathfrak{q}$ splits completely in $F/\mathbb{Q}$.)

I don't understand the notation " $\mathfrak{q}^r\|\:\xi^p\mu^a$ ", and furthermore, how does this imply that $r\equiv ar \pmod{p}$?

Any help would be much appreciated.

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    $\begingroup$ I'm pretty sure that the notation $\mathfrak{q}^r\|\:\xi^p\mu^a$ means that $\mathfrak{q}^r \mid \xi^p\mu^a$ but $\mathfrak{q}^{r+1} \nmid \xi^p\mu^a$. $\endgroup$ – Greg Martin Apr 15 '15 at 21:11
  • $\begingroup$ @GregMartin Thanks for that, that may work in the context. Did you read the proof? If so would you care to shed some light as to how the implication works? $\endgroup$ – Ramified_Minds Apr 16 '15 at 10:00

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