Free submodule of finite index over pid I am reading Neukirch's Algebraic Number Theory and I am currently stuck on the proof of Proposition 5.7 on page 140.
Neukirch introduces the $Z_p$ module $U^{(n)} = 1 + P^n$, where $P$ is the unique maximal ideal of the ring of integers $o$ of the complete local field $K$.
The $Z_p$ action on the multiplicative group $U^{(n)}$ is defined as follows.
For $1+x\in U^{(n)}, z\in Z_p$, one has $z(x) = (1+x)^z$.
Neukirch claims that the torsion subgroup of the $Z_p$-module $U^{(1)}$ is the set of roots of unity of order powers of $p$. I.e., elements $1+x\in U^{(1)} = 1 + P$, such that $(1+x)^{m} = 1$ for some $m=p^a$ for some natural number $a$. I don't understand why $m$ must be a power of $p$. Also, if we take $U^{(1)}$'s quotient with its torsion subgroup, then we ought to get a free module. Neukirch claims that this module is $U^{(n)}$, why is that?
I do agree that since $U^{(1)}$ has $U^{(n)}$ as a submodule of finite index that we may write $U^{(1)} = U^{(n)} + (U^{(1)}/U^{(n)})$, but why should $U^{(1)}/U^{(n)}$ be isomorphic to the roots of unity of order power of $p$? I have no idea, any light shed on this matter would be highly regarded.
Thanks in advance,
Cheers!
 A: As regards the first question (why must $m$ be a power of $p$), this has been asked and answered in MS/1697097. As soon as you believe that $U^{(1)}$ is a $\mathbb Z_p$-module, it is clear that every torsion element is of $p$-power order, and then you remember that roots of unity are cyclic (and a local field cannot contain infinitely many roots of unity; or, if you even believe that $U^{(1)}$ is a finitely generated $\mathbb Z_p$-module, you know the entire torsion will be killed by some $p$-power, you can conclude from there with cyclicity that it must be exactly the group of $p^a$-th roots for some natural $a$).
Then,

if we take $U^{(1)}$'s quotient with its torsion subgroup, then we ought to
  get a free module. Neukirch claims that this module is $U^{(n)}$, why is
  that?

I do not have the book at hand, but I doubt that that is what Neukirch claims, because it is wrong in general, see counterexample at the end. What is true is that for high enough $n$, $U^{(n)}$ is isomorphic to $U^{(1)}/U^{(1)}_{tors}$ as $\mathbb Z_p$-module, namely, they are both free of rank $r = [K:\mathbb Q_p]$ (resp. $[K:\mathbb F_p((t))]$ if we are in positive characteristic).
This is shown via filtrations of $U^{(1)}$ with the $U^{(n)}$, see also link above;  certainly this is also done by Neukirch, there is a good treatment in Serre's Local Fields, and freely available on the internet is this book by Fesenko and Vostokov, where it is done in sections 1.5. and 1.6 (pp.12 to 22). The upshot is that for each $n$, there are isomorphisms
$U^{(n)}/U^{(n+1)} \simeq (k = o/P, +)$ (additive group of the residue field)
and for $n > e/(p-1)$, where $e$ is the ramification index (i.e. $(P^e) = (p)$ in $o$), raising to the $p$-th power induces an isomorphism $U^{(n)} \stackrel{\simeq}\rightarrow U^{(n+e)}$.
Now if $f$ is the inertia degree, i.e. the order of the residue field $k$ is $p^f$, this shows that for $n > e/(p-1)$, $U^{(n)}$ is a free $\mathbb Z_p$-module of rank $ef =r$. Further, since $U^{(1)}/U^{(n)}$ is finite (of order $p^{fn}$), from the structure theory of finitely generated modules (and the fact that $\mathbb Z_p$ is infinite) it follows that the free part of $U^{(1)}$ must have the same rank, i.e. be isomorphic to those $U^{(n)}$. Then it follows further from that structure theory that
$U^{(1)} \simeq U^{(1)}_{tors} \times \mathbb Z_p^r$, and of course the torsion part is $\simeq \mathbb Z_p/p^a \simeq \mathbb Z/p^a$.
So abstractly, there is an isomorphism $U^{(1)} \simeq U^{(1)}_{tors} \times U^{(n)}$ for all higher $n$, but it can happen that this iso is not given by the natural inclusion $U^{(n)} \subset U^{(1)}$ (for any $n$).

Finally, to see this, i.e. it is not necessarily true that some $U^{(n)}$ itself works as a complement to  $U^{(1)}_{tors}$ in $U^{(1)}$, look at the following counterexample: Let $p = 3$, take the unique unramified quadratic extension $F \vert \mathbb Q_3$, and let $K=F(\mu_3)$, this is the field whose principal units we consider. For $K\vert \mathbb Q_3$, we have $e=f=2, r=4$. Then there is an isomorphism of $\mathbb Z_p$-modules
$U^{(1)} \simeq \mathbb Z_3/3  \oplus \mathbb Z_3 \oplus \mathbb Z_3 \oplus \mathbb Z_3 \oplus \mathbb Z_3$
which obviously identifies the $3$rd roots of unity with the first factor $\mathbb Z_3/3 \simeq \mathbb Z/3$; but the filtration looks like this:
$U^{(2)} \simeq 0 \oplus 3\mathbb Z_3 \oplus \mathbb Z_3 \oplus \mathbb Z_3 \oplus \mathbb Z_3$
$U^{(3)} \simeq 0 \oplus 3\mathbb Z_3 \oplus 3\mathbb Z_3 \oplus 3\mathbb Z_3 \oplus \mathbb Z_3$
$U^{(4)} \simeq 0 \oplus 9\mathbb Z_3 \oplus 3\mathbb Z_3 \oplus 3\mathbb Z_3 \oplus 3\mathbb Z_3$
$U^{(5)} \simeq 0 \oplus 9\mathbb Z_3 \oplus 9\mathbb Z_3 \oplus 9\mathbb Z_3 \oplus 3\mathbb Z_3$
etc. Note that all $U^{(n\ge 2)}$ are isomorphic as $\mathbb Z_p$-modules (namely, free of rank $4$), each $U^{(n)}/U^{(n+1)}$ is isomorphic to the additive group of the residue field $\mathbb F_{3^2}$, and multiplication with $p$ gives isos $U^{(n)} \simeq U^{(n+2)}$ for all $n \ge 2$ (but not for $n=1 = e/(p-1)$); but also note that $U^{(1)}/U^{(n)}$ is never isomorphic to the torsion part.
