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Hello all i have a question on ramification over a cyclotomic field.

Let $p$ be any prime in $\mathbb{Z}$ and $k$ be any integer $\geq 1$. Consider the cyclotomic field $\mathbb{Q}(\zeta_{p^k})$. Is it true that $p$ totally ramifies in $\mathbb{Q}(\zeta_{p^k})$?

Recall that $p$ totally ramifies in $\mathbb{Q}(\zeta_{p^k})$ if for every prime $\wp$ of the ring of integers of $\mathbb{Q}(\zeta_{p^k})$ dividing $(p)$ we have that the inertial degree $f(\wp|p) = 1$ (since the extension is Galois, the same degree works for every prime above $p$).

My attempt: I am clueless on how to prove this. Since the ring of integers of $\mathbb{Q}(\zeta_{p^k})$ is $\mathbb{Z}[\zeta_{p^k}]$, I tried working with the prime ideal $(1-\zeta_{p^k})$ but i didn't succeed.

Thanks in advance!

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The ramification index of $(p)$ is $e=\phi (p^k)=p^{k-1}(p-1)$. The Degree Theorem then gives $$ e=[\Bbb{Q}(\zeta_{p^k}):\Bbb{Q}]=\phi(p^k)=gef, $$ where $(p)$ is the product of $g$ different prime ideals in $\Bbb{Z}[\zeta_{p^k}]$, $f=f(\wp|p) $, and $(p)=(1-\zeta_{p^k})^e$ since $p=u(1-\zeta_{p^k})^e$ with some unit $u$. Cancelling $e=gef$ by $e$ yields $gf=1$, so $g=f=1$, and we are done. Hence $(p)$ is totally ramified in $\Bbb{Q}(\zeta_{p^k})/\Bbb{Q}$.

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  • $\begingroup$ But why the ramification index is exactly $\phi(p^k)$? $\endgroup$ – El.Gon.Zalo May 1 '18 at 8:56
  • $\begingroup$ For a detailed proof, see here. $\endgroup$ – Dietrich Burde May 1 '18 at 18:05

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