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We know that for any $n$ relatively prime to $p$, $\mathbb{Q}_p(\zeta_n)$ is an unramified extension over $\mathbb{Q}$. On the other hand, we know that finite unramified extensions of a field complete with respect to a discrete valuation corresponds to finite extensions of its residue field. The residue field of $\mathbb{Q}_p$ is $\mathbb{F}_p$, and we know the finite extensions of $\mathbb{F}_p$ are $\mathbb{F}_{p^r} = \mathbb{F}_p(\zeta_{p^r-1})$, so the finite unramified extensions of $\mathbb{Q}_p$ should be $\mathbb{Q}_p(\zeta_{p^r-1})$.

Is it true that for any $n$ such that $(n,p) = 1$, there is an $r$ such that $\mathbb{Q}_p(\zeta_n) = \mathbb{Q}_p(\zeta_{p^r-1})$?

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Yes, it’s true. Let the residue field of the unramified extension be $\Bbb F_{p^r}=\Bbb F_p(\zeta_{p^r-1})$. The primitive $(p^r-1)$-th root of unity in the residue field can be lifted to characteristic zero — there are any number of ways of seeing this, like Hensel, or the neat trick of lifting $\zeta_{p^r-1}\in\Bbb F_{p^r}$ to any $z$ in your unramified field, and then repeatedly taking $p^r$-th powers. The sequence you get is $p$-adically convergent.

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  • $\begingroup$ But then is it not true that $\mathbb{Q}_p(\zeta_{p^r-1})$ are all the unramified extensions of $\mathbb{Q}_p$? $\endgroup$ – Aaron Johnson Feb 5 '18 at 5:44
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    $\begingroup$ All the finite unramified extensions, yes. There’s one of each degree. $\endgroup$ – Lubin Feb 5 '18 at 5:46
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    $\begingroup$ But if it's not true that for any $(n,p) = 1$, $\mathbb{Q}_p(\zeta_n) = \mathbb{Q}_p(\zeta_{p^r-1})$ for some $r$, it seems like this gives a contradiction that all finite unramified extensions are of the form $\mathbb{Q}_p(\zeta_{p^r-1})$? $\endgroup$ – Aaron Johnson Feb 5 '18 at 6:19
  • $\begingroup$ @AaronJohnson, you may be forgetting that the $n$th cyclotomic polynomial may easily fail to be irreducible over $\mathbb Q_p$, so that $\mathbb Q(\zeta_n)$ becomes ambiguous. The question is then about the degree(s) of the irreducible factors. $\endgroup$ – paul garrett Feb 5 '18 at 19:10
  • $\begingroup$ @paulgarrett: good point! Thanks! $\endgroup$ – Aaron Johnson Feb 6 '18 at 0:23
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The proof you give is correct. Let me try to exhibit why we should expect such a thing.

The field of complex numbers $\mathbb C$ is the unique finite extension of $\mathbb R$ and that $$ \mathbb R(i) = \mathbb R(\zeta_4) = \mathbb C. $$

Roots of unity are not real except for $\pm 1$. Hence $$ \mathbb R(\zeta_n) = \mathbb R(\zeta_4) =\mathbb C \qquad \text{if $n>2$} $$ and $$ \mathbb R(\zeta_n) = \mathbb R \qquad \text{if $n\leq2$} $$ since there is no more room between $\mathbb R$ and its algebraic closure.

Similar things happen in $\mathbb Q_p$, take $p=5$. By Hensel's lemma roots of unity are not in $\mathbb Q_5$ except for $\pm 1, \pm i$. In particular $$ \mathbb Q_5(\zeta_n) = \mathbb Q_5 \quad \Longleftrightarrow\quad n\mid 4. $$ Now since $\bar{\mathbb Q}_5/\mathbb Q_5$ is an infinite extension there is more room for intermediate extensions but still diferent roots of unity might define the same extension. For instance $\mathbb Q_5(\zeta_{5^2- 1})$ contains $\mathbb Q_5(\zeta_8)$ and $\mathbb Q_5(\zeta_3)$. You can check that $$ \mathbb Q_5(\zeta_{5^2- 1})=\mathbb Q_5(\zeta_8)=\mathbb Q_5(\zeta_3). $$

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  • $\begingroup$ Good example. I was thinking of using just that, but I wouldn’t have stated it as succinctly as you have here. $\endgroup$ – Lubin Feb 5 '18 at 22:56
  • $\begingroup$ I really appreciate your comment. $\endgroup$ – eduard Feb 6 '18 at 0:07

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