# pth root of unity in $p$-adic field

It is well known that $\mathbb{Q}_p(\mu_n)$ is a totally ramified extension of degree $(p-1)p^n$ if $\mu_n$ is a primitive $p^n$th root of unity.

However how true is this statement for a finite extension $K/\mathbb{Q}_p$?

I remember seeing it as an exercise (I have no idea where nor whether it is actually the correct statement) that for each $n$ there exist $p$-adic field $K_n$ which contains $\mu_{n-1}$ (but not $\mu_n$) such that $K_n(\mu_n)$ is unramified extension of $K_n$. Firstly I want to verify if it is true and if it's true, whether it is possible to classify such fields?

• That should be $(p-1)p^{n-1}$, or do I have an off-by-one error? – Hurkyl Jun 14 '15 at 13:43
• @Hurkyl You are right! I won't bother editting it as I got all I wanted though. – Jack Yoon Jun 16 '15 at 11:05

## 1 Answer

The finite extensions of $\Bbb Q_p$ are also local fields, so you can use the usual characterization of local field extensions being unramified.

Recall: a finite extension of local fields $L/K$ is unramified iff $[L:K]=[\lambda:\kappa]$ where $\lambda,\kappa$ are the residue fields for $L,K$ respectively. This comes from the fundamental relationship that

$$e(L|K)f(L|K)=[L:K]$$

and the fact that $f(L:K)=[\lambda:\kappa]$. But then since your $K$ is a finite extension of $\Bbb Q_p$ you know the residue field is a finite extension of $\Bbb F_p$, so that unramified extensions are classified by intermediate fields $\ell$ such that

$$\kappa\subseteq \ell\subseteq \overline{\kappa}=\overline{\Bbb F_p}$$

the last equality since the algebraic closure of $\kappa$, being itself a finite extension of $\Bbb F_p$ is the same as the algebraic closure of $\Bbb F_p$. This completely classifies unramified extensions of such $K$. You even have a nice description: if $\ell\cong \Bbb F_{p^n}$ then the corresponding extension of $K$ is just $K(\mu_{p^n-1})$ with $\mu_k$ as usual the set of $k^{th}$ roots of $1$.

From this we may further deduce a nice corollary that the maximal, unramified extension of $K$ is simply $K(\mathbf{\mu})$ with

$$\mathbf{\mu}=\bigcup_{(n,p)=1,\; \mu_n\not\subseteq K} \mu_n$$

This correspondence is--as usual--due to Hensel's lemma and the characterization of finite fields as collections of roots of unity (plus $0$). Naturally the $(n,p)=1$ condition comes from the fact that all $p^{n}$ roots of unity are already in $\kappa$ since $1$ is the only one such and the Frobenius is an automorphism.

From this you can also see that since there is no $p^n$ roots of $1$ un an unramified extension, and since $\mu_{mn}=\mu_m\times\mu_n$ when $(n,m)=1$, that any extension of $K$ by $p^n$ roots of $1$ must be totally ramified, unless $K$ already contains those roots of $1$. Wedge this against the fact that you can always write any extension as an unramified, followed by a totally ramified, and you have your more general approach.

• Ok this clarified a lot. I re-read the proof of $\mathbb{Q}_p(\mu_n)/\mathbb{Q}_p$ being totally ramified (in my non-standard notation) and I swear this generalised as long as $K$ did not contain $p$th root of unity. So the exercise I supposedly saw is either non-existent or wrong I guess. – Jack Yoon Jan 29 '15 at 23:23
• @JackYoon I'm glad to hear it helped! It's a really beautiful theory, one of my favorites! – Adam Hughes Jan 29 '15 at 23:50