# Why $\mathbb Q_p(\xi_n) = \mathbb Q_p$ implies $n|(p-1)$.

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.

• I think you mean primitive $n$-th root of unity. – user10354138 Oct 1 '18 at 15:28
• @user10354138 Thanks! – CYC Oct 1 '18 at 15:32
• 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 $n\mid (p-1)$. – user10354138 Oct 1 '18 at 15:47
• @user10354138 What if $n \geq p$? – CYC Oct 24 '18 at 8:20