# Congruence properties of global units

Let $K$ be a number field, $O$ its ring of integers with global unit group $O^\times$. Further, let $M\subset O^\times$ be a subgroup of finite index.

My question is: Is there a non-zero ideal $\frak{n}$ of $O$, such that $(1+\frak{n})\cap$ $O^\times \subset M$?

By Dirichlet, $O^\times$ and hence $M$ are finitely gen. $\mathbb{Z}$-modules, even free (up to roots of unity). I feel that this should be almost enough to compare the topologies of $\widehat{O^\times}$ and $\widehat{O}^\times$, where $\widehat{}$ is profinite completion. But playing around with this hasn't amounted to anything. I feel, like I'm missing something obvious here...

P.S: Minkowski's "geometry of numbers" was also not helpful in attacking this.

Here is (roughly) the idea of the proof. It suffices to consider the case when $M = (\mathcal{O}^{\times})^m$. Suppose that $\mathfrak{n}$ is a product of distinct primes. Then there is a map: $$\mathcal{O}^{\times}/(\mathcal{O}^{\times})^m \rightarrow (\mathcal{O}/\mathfrak{n})^{\times}/(\mathcal{O}/\mathfrak{n})^{\times m} = \prod (\mathcal{O}/\mathfrak{p})^{\times}/(\mathcal{O}/\mathfrak{p})^{\times m}.$$
It's tempting to try to show this is injective, and then one is done. To do this, since the first group is finite, it would be enough to show that for each non-trivial element $\epsilon$ in the source, there is a prime (or rather infinitely many primes) $\mathfrak{p}$ so that $[\epsilon]$ is non-trivial in $(\mathcal{O}/\mathfrak{p})^{\times}/(\mathcal{O}/\mathfrak{p})^{\times m}$.
Here's a natural way to "prove" this. Let $L = K(\zeta_m, \epsilon^{1/m})$. Because $\epsilon$ is (by assumption) not an $m$th power, the extension $L/K(\zeta_m)$ "is" non-trivial. Then, by Cebotarev, there exists a prime $\mathfrak{p}$ whose Frobenius is a non-trivial element of $\mathrm{Gal}(L/K(\zeta_m))$. It's not so hard to see that this gives precisely the condition you require. The "problem" is that it's not quite correct to say that $L/K(\zeta_m)$ is non-trivial when $\epsilon$ is not an $m$th power; this is true when $m$ is odd, but there are problems at the prime $2$. This turns out to m ake the argument slightly annoying; there's a genuine issue making this strategy work because of issues related to Grunewald-Wang. So you have to fuss around a little because of this.
To give some further indication of the subtlety, if you require that the $\mathfrak{n}$ are actually of the form $\mathfrak{p}^m$ for a power of a single prime $\mathfrak{p}$, then the problem is (in general) open; this is precisely the Leopoldt conjecture for $(K,\mathfrak{p})$.