# What is the group $\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^k$ for $k > 2$ isomorphic to?

One of the important pieces for constructing the Hilbert Symbol is the fact that $$\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2 = \begin{cases} \mathbb{Z}/2 & p=\infty \\ (\mathbb{Z}/2)^2 & p \text{ odd} \\ (\mathbb{Z}/2)^3 & p \text{ even} \end{cases}$$ Does there exist a nice decomposition of the groups $\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^k$ for $k > 2$? Certainly the case $p=\infty$ is trivial since $$(\mathbb{R}^*)^k = \begin{cases} \mathbb{R}_{>0} & k \text{ even} \\ \mathbb{R}^* & k \text{ odd} \end{cases}$$ but what about the other cases? Serre uses the Legendre symbol throughout his analysis for the case $p=2$, so is there a proof using higher degree symbols?

• Look at the direct product decomposition for $\Bbb{Q}_p^*$ $$\Bbb{Q}_p^* = (p)\times \mu_{p-1}\times (1+p\Bbb{Z}_p)$$ – sharding4 Jun 14 '17 at 1:20

Your question is related to local class-field theory as follows: for a local $p$-adic field $K$, i.e. a finite extension of $\mathbf Q_p$, for any $n \ge 1$, CFT asserts that $K^*/(K^*)^{n}$ is canonically isomorphic to the Galois group over $K$ of the maximal abelian extension of $K$ with exponent dividing $n$, so that an explicit description of $K^*/(K^*)^{n}$ is actually an "explicit reciprocity law" for such abelian extensions.
The decomposition for $\mathbf Q^*_p$ invoked by @sharding4 can be generalized as $K^* = (\pi) \times W\times U_1$, where $\pi$ is an uniformizer of $K$, $W$ is the subgroup of roots of unity of $K^*$ of order prime to $p$ (if $k$ is the residue field, $W$ lifts $k^*$ by Hensel's lemma) and $U_1 = 1+(\pi)$ is the group of "principal units". So your problem for $K^*/(K^*)^{n}$ reduces to the determination of $U_1/(U_1)^n$. This can be done in 3 steps, introducing the higher principal units $U_r = 1+ (\pi)^r$ for $r \ge 1$: (1) the multiplicative group $U_r/U_{r+1}$ is isomorphic to the additive group $k$ ; (2) raising to a power $m$ prime to $p$ induces an automorphism of $U_r$; (3) $U_r^{p} \subset U_{r+1}$, and raising to the power $p$ induces an isomorphism $U_r \cong U_{r+e}$ when $r > e/(p-1)$, where $e$ is the ramification degree of $K$ over $\mathbf Q_p$. See Serre's "Local Fields", chap.IV, §6 and chap.XIV, §4.
Obviously, the problem in the "tame case", i.e. when $n$ is prime to $p$ , is completely solved (e.g. for $\mathbf Q^*_p/(\mathbf Q^*_p)^2$ when $p$ is odd) by steps (1) and (2). The "wild case" is harder because step (3) is not precise enough. Various cases can actually occur according to $K$ and $v_p (n)$ (see e.g. $\mathbf Q^*_2/(\mathbf Q^*_2)^2$), but one can show, using the "Herbrand quotient", a general result giving the order of $K^*/(K^*)^{n}$ (op. cit.,chap.XIV, §4, ex.3).