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.



Roots of unity, in the completion of the ring of integers of a number field, at a prime ideal
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. 


