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

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.

• In a $p$-adic field, the roots of unity of order coprime to $p$ reduce isomorphically to the unit group of the residue field. – Aurel Oct 28 '19 at 9:29

## 1 Answer

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.