# Is it correct ? $p\in\mathbb{Z}$ tamely ramified in $K\implies[K:\mathbb{Q}]$ is coprime to $p$

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 $p$.

• 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? – Lubin Oct 21 '16 at 3:42

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