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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.

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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.

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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 $U$.

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