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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 $p$.

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  • $\begingroup$ If you want help understanding the situation, I think you’ll have to give more context than you have so far. What, specifically, are $K$ and $L$ here? $\endgroup$ – Lubin Oct 21 '16 at 3:42
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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 $.

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  • $\begingroup$ Thanks a lot. But can you explain the logic behind the claim that |Gal$(K|\mathbb{Q})$| is coprime to $p$ ? $\endgroup$ – learning_math Oct 16 '16 at 11:52
  • $\begingroup$ The order of the Galois group in this example is $12$, not coprime to $p=3$. $\endgroup$ – Lubin Oct 21 '16 at 3:35

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