Let $K$ be an abelian number field. Let $p\in\mathbb{Z}$ be a prime which is tamely ramified in $K$. Is it true that $[K:\mathbb{Q}]$ is coprime to $p$ ?
How this question came to my mind:
I am reading this proof of the Kronecker-Weber Theorem. On page 5, there is a proposition which tells about eliminating tame ramification.
As you go through the proof, you can see that the author claims that $U$ is tamely ramified over $p$; the reason being, $[K:\mathbb{Q}]$ and $[L:\mathbb{Q}]$ are coprime to $p$. I don't understand why $[K:\mathbb{Q}]$ is coprime to $p$.
No. Take $ K = \mathbf Q(\zeta_{21}) $. $ 3 $ is totally ramified in $ \mathbf Q(\zeta_3) $ and inert in $ \mathbf Q(\zeta_7) $, so it tamely ramifies as $ (3) = \mathfrak p^2 $ in $ K $. However, $ [K : \mathbf Q] = \varphi(21) = 12 $, which is not prime to $ 3 $.