Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 8 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.