# 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?

## 1 Answer

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 $\mathbb{Z}_p[\zeta_n]$.

Here are two methods of proof:

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.

• Thanks. I hadn't been aware of that fact about integral closures of directed unions. It's quite a useful fact. Perhaps it deserves a place in your commutative algebra notes! – John M Sep 4 '11 at 23:38
• @John: okay, glad to help. I think though that this fact about integral closures is pretty clear if you think about it for a little while. (Right?) – Pete L. Clark Sep 5 '11 at 0:49