# Integral closure of p-adic integers in maximal unramified extension

Let $\mathbb Q_p$ be the field of p-adic numbers, and let $\mathbb Q_p^{\text{unr}}$ be maximal unramified extension in some algebraic closure of $\mathbb Q_p$. My understanding is that $\mathbb Q_p^{\text{unr}}$ has a fairly explicit description: $$\mathbb Q_p^{\text{unr}} = \mathbb Q_p \left(\bigcup_{(n,p)=1} \mu_n \right)$$ where $\mu_n$ is a primitive $n$th root of unity, i.e. we adjoin all $n$th roots of unity with $n$ relatively prime to $p$.

My question is: Does the integral closure of $\mathbb Z_p$ in $\mathbb Q_p^{\text{unr}}$ have a similarly explicit description? For example, does it equal: $$\mathbb Z_p \left[\bigcup_{(n,p)=1} \mu_n \right]$$ perhaps?

Yes, this is true. Since the integral closure of a directed union is the union of the integral closures, it suffices to establish this at every finite level: that is, for $$n$$ prime to $$p$$, the ring of integers in $$\mathbb{Q}_p(\zeta_n)$$ is $$\mathbb{Z}_p[\zeta_n]$$.
First Proof (Local): This follows from the structure theory of unramified extensions of local fields. For instance, you can apply Proposition 4 of these notes on local fields to $$\overline{f}$$, the minimal polynomial over $$\mathbb{F}_p$$ of a primitive $$n$$th root of unity.
Second Proof (Global): Show that the discriminant of the order $$\mathcal{O} = \mathbb{Z}[\zeta_n]$$ -- or, in plainer terms, of $$(1,\zeta_n,\ldots,\zeta_n^{\varphi(n)-1})$$ -- is prime to $$p$$. Therefore the localized order $$\mathcal{O} \otimes \mathbb{Z}_p$$ is maximal.