$K_S$ modulo $K_S^n$, where $K_S$ is the group of $S$-units Let $K$ be a field containing the $n$th roots of unity, $S$ a finite set of places containing all the archimedean ones, and $K_S$ the group of $S$-units, i.e. those $x \in K^{\ast}$ which are units at all places outside $S$.  It is a consequence of the unit theorem that there exist $c_1, ... , c_{s-1} \in K_S$ such that every $x \in K_S$ can be uniquely expressed as $$\zeta c_1^{m_1} \cdots c_{s-1}^{m_{s-1}}$$ where $m_i$ are integers and $\zeta$ is a root of unity in $K$.  Serge Lang claims (Algebraic Number Theory, page 216):

$K_S$ modulo the $n$th roots of unity is a free abelian group on $s-1$ generators.

and uses this to conclude that $K_S/K_S^n$ has $n^s$ elements.  However, is his statement correct?  $K_S$ modulo the group of roots of unity in $K$ is free abelian of rank $s-1$.  There could be other roots of unity in $K$ besides $n$th roots of unity.

I don't think that D_S's argument is correct, because if $H$ is the group of units, its torsion subgroup consists of the roots of 1 in K, and its  Z-rank is equal to r + c - 1, where r is the number of real embeddings of K, and 2c the number of complex embeddings (Dirichlet's theorem). So $H/H^n$ should have $n^{r+c}$ elements (*). Actually, S here is supposed to contain the archimedean primes, so that a straightforward generalization of Dirichlet's theorem says that indeed $K_S$ modulo torsion  has Z-rank equal to s - 1.
(*) The point is that the roots of 1 in a field form a cyclic group $W$, so $W/W^n$ is cyclic (but could be trivial if the order of $W$ is coprime to n - which is not the case here).
