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Let $p \in \mathbb{Z}$ be a prime and consider the number field $k = \mathbb{Q}[x]/(x^{(p^2 -1)/2} - p)$. We shall denote by $O_k$ the ring of integers of $k$. Let $\beta \in O_k$ be such that $\beta^{(p^2-1)/2} = p$. Then it is clear that $(\beta)$ is a prime ideal of $O_k$. Let $\widehat{O}_k$ be the completion of $O_k$ with respect to $(\beta)$.

My question is what are the roots of unity in $\widehat{O}_k$, in particular, does it have the $p+1$ -st roots of unity?

PS. Since $O_k/(\beta) = \mathbb{F}_p$ it is easy to see that $\widehat{O}_k$ has all the $p-1$ -st roots of unity using Hensel's lemma, but I could not infer anything about the $p + 1$ -st roots.

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    $\begingroup$ In a $p$-adic field, the roots of unity of order coprime to $p$ reduce isomorphically to the unit group of the residue field. $\endgroup$ – Aurel Oct 28 '19 at 9:29
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Let $K = \mathbf Q_p(\sqrt[n]{p})$, where $n \geq 1$. The polynomial $x^n-p$ is Eisenstein at $p$, so $K/\mathbf Q_p$ is totally ramified at $p$, so its residue field has size $p$, as you indicate. In a local field with residue field of characteristic $p$ and size $q$, its roots of unity with order relatively prime to $p$ are exactly the $(q-1)$-th roots of unity.

In particular, in a local field with residue field of size $p$, the roots of unity in it of order prime to $p$ are the $(p-1)$-th roots of unity.

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