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

Suppose $K$ is a $p$-adic field (finite extension of the $p$-adics), and let $n$ be any integer (independent of what $p$ is). Define $U$ to be the set of all $x$ in $K$ such that $|x| = 1$ and such that $x = y^n$ for some $y$ in $K$. I would like to show that $U$ is an open set and that as a multiplicative group $U$ has finite index in the group of elements of $K$ of norm $1$. What's the best way of seeing why this is true (assuming it is)?

I pretty much have an idea why this holds.. in the $p$-adic case you can prove the $n$th powers are of bounded index in ${\bf Z}_{p^l}$ for each $l$ and then use an inverse limiting-type argument as $l$ goes to infinity to get this for $K = {\mathbb Q_p}$, and I think an analogous argument using powers of a uniformizer in place of powers of $p$ should work for a general $K$. But I keep thinking that this should be some well-known result or something that follows quickly from a well-known result. So I thought I'd throw this out.

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

There are lots of ways to see this. My preferred method is as follows:

What you want to show is that if $x \equiv 1 \bmod \pi^N$ for sufficiently large $N$ then $x$ is an $n$th power; here $\pi$ is uniformizer. (This shows that $U$ contains all $x$ that are $\equiv 1 \bmod \pi^N$, and hence is open, as desired.)

Well, just use the classical binomial formula: writing $x = 1 + \pi^N y,$ we have $$x^{1/n} = (1 + \pi^N y)^{1/n} = \sum_{i=0}^{\infty} \frac{1}{n}(\frac{1}{n} - 1) \cdots (\frac{1}{n} - i + 1) \frac{\pi^{N i} y^i}{i !}.$$ The denominator of the $i$th term is (bounded above by) $n^i i!$, so provided that $N$ is large enough, the ratio $\dfrac{\pi^{N i}}{n^i i!}$ will tend to zero $\pi$-adically, and thus so will our series. It's then easy to argue that this series in fact converges to an $n$th root of $x$, as required.

A good example to think about is the case $n = 2$ and $K = \mathbb Q_p$, first when $p$ is odd, and then when $p = 2$. In the former case you should find that any $x \equiv 1 \bmod p$ is a square, while in the latter case, you will find that the condition $x \equiv 1 \bmod 8$ is required.

share|cite|improve this answer
Thanks, this makes a lot of sense. – Zarrax Jan 16 '11 at 20:45

I like Matt E's answer as an elementary and appealing way to see that $U^N$ has finite index in $U$ for all $N$. (Here I am writing $U$ for the full unit group $\mathcal{O}_K^{\times}$ of $K$ and $U^N = \{x^N \ | \ x \in U\}$. I just want to remark that it is not so much harder to give a general formula for the index $[U:U^N]$ in the general case (including local fields of positive characteristic $p$, so long as $p \nmid N$).

The answer is that if $v$ is the normalized (i.e., $\mathbb{Z}$-valued) valuation on $K$ and $q$ is the cardinality of the residue field, then

$$[U:U^N] = q^{v(N)} \ \# \mu_N(K).$$

This is Theorem 12 in these notes. The treatment follows Lang's Algebraic Number Theory. (Perhaps it is worth mentioning that the argument is a bit tricky but completely elementary.)

Let $U_n = \{x \in U \ | \ x \equiv 1 \pmod{\mathfrak{p}^n} \}$, so that the $U_n$'s are a cofinal system of open subgroups of $U$. In other words, for a subgroup of $U$ to be open, it is necessary and sufficient that it contain $U_n$ for some $n$. We want to show that $U^N$ is open. But the proof of the above theorem proceeds by showing that for all sufficiently large $r$,

$$ U_{r+v(N)} = U_r^N \subset U^N,$$

so indeed $U^N$ is open. Moreover, since every subgroup $H$ of $U$ of finite index $N$ contains $U^N$, it follows that every finite index subgroup of $U$ is open.

Yet another approach is to develop the theory of the logarithm as in Exercise 5.3 of loc. cit. to give, for all sufficiently large $n$, an isomorphism of topological groups from $U_n$ to the additive group $(\mathcal{O}_K,+)$. This also implies that each $U^N$ is open and of finite index in $U$.

share|cite|improve this answer

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.