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Let $p$ be a prime number, $\mathbb Q_p$ be the completion of $\mathbb Q$ w.r.t $\,p$, denote $\xi_n$ primitive $n$-th root of unity in a fixed algebraic closure of $\mathbb Q_p$ with $p\nmid n$, we have:

If $\mathbb Q_p(\xi_n) = \mathbb Q_p$, then $n|(p-1)$.

Why is the above property true?

I know a property that if $L$ unramified over $\mathbb Q_p$, then $L\simeq \mathbb Q_p(\xi_{p^m-1})$ where $m= [L:\mathbb Q_p]$, using this we get $\mathbb Q_p =\mathbb Q_p(\xi_n)\simeq \mathbb Q_p (\xi_{p-1}).$ Then I'm not sure how to proceed.

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  • $\begingroup$ I think you mean primitive $n$-th root of unity. $\endgroup$ – user10354138 Oct 1 '18 at 15:28
  • $\begingroup$ @user10354138 Thanks! $\endgroup$ – CYC Oct 1 '18 at 15:32
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    $\begingroup$ Fairly quickly, we have $\zeta_n\in\mathbb{Z}_p^\times$ because it must have $v_p=0$ in $\mathbb{Q}_p$. So we get $\mathbb{F}_p(\zeta_n)=\mathbb{F}_p$ and hence $n\mid (p-1)$. $\endgroup$ – user10354138 Oct 1 '18 at 15:47
  • $\begingroup$ @user10354138 What if $n \geq p$? $\endgroup$ – CYC Oct 24 '18 at 8:20

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