Finiteness of the index of $k$-th powers of units in $\mathbb{Q}_p$

Question

Let $\mathbb{Q}_p$ be the completion of the rationals with respect to the $p$-adic absolute value given by $\nu_p(x) = p^{-ord_p(x)}$.

I understand how the index $[ \mathbb{Q}_p^\times : (\mathbb{Q}_p^\times )^2 ]$ is calculated using Hensel's lemma, and an explanation is found here Quadratic extensions of p-adic rationals

What I don't understand is how one determines the index of the arbitrary $k$-th powers of the units in $\mathbb{Q}_p$, i.e.

1. How do we know (if it is true) that $[\mathbb{Q}_p^\times : (\mathbb{Q}_p^\times )^k]$ is finite, and specifically what is the index when $p\not\mid k$? The case when $k=p$ is treated here $ \Bbb Q_p^×/(\Bbb Q_p^×)^p$ is isomorphic to $\Bbb Z/p\Bbb Z×\Bbb Z/p\Bbb Z$

2. Does the result differ when $p=2$?

There is an explanation here What is the index of the $p$-th power of $\mathbb Q_p^\times$ in $\mathbb Q_p^\times$ but it feels like there should be an easier way to see this, and this explanation is a bit more complicated than I can follow.

Answer 1

We will use the multiplicative decompositions $$ \mathbf Q_p^\times = p^\mathbf Z \times \mu_{p-1} \times (1+p\mathbf Z_p) $$ for $p > 2$ and $$ \mathbf Q_2^\times = 2^\mathbf Z \times \mu_2 \times (1+4\mathbf Z_2). $$ It tells us for $k \geq 1$ that $$ {\mathbf Q_p^\times}^k = p^{k\mathbf Z} \times \mu_{p-1}^k \times (1+p\mathbf Z_p)^k $$ for $p > 2$ and $$ {\mathbf Q_2^\times}^k = 2^{k\mathbf Z} \times \mu_2^k \times (1+4\mathbf Z_2)^k. $$ Therefore $$ \mathbf Q_p^\times/{\mathbf Q_p^\times}^k \cong \mathbf Z/{k\mathbf Z} \times \mu_{p-1}/\mu_{p-1}^k \times (1+p\mathbf Z_p)/(1+p\mathbf Z_p)^k $$ for $p > 2$ and $$ \mathbf Q_2^\times/{\mathbf Q_2^\times}^k \cong \mathbf Z/{k\mathbf Z} \times \mu_{2}/\mu_{2}^k \times (1+4\mathbf Z_2)/(1+4\mathbf Z_2)^k $$ Since $\mu_{p-1}$ is cyclic of order $p-1$ and $\mu_2$ is cyclic of order $2$, $\mu_{p-1}/\mu_{p-1}^k = \mu_{p-1}/\mu_{(p-1)/(p-1,k)}$, so its order is $(p-1)/((p-1)/(p-1,k)) = (p-1,k)$. Similarly, $\mu_{2}/\mu_{2}^k$ has order $(2,k)$, which of course is also easy to check directly (taking cases if $k$ is even or odd).

Therefore $$ [\mathbf Q_p^\times:{\mathbf Q_p^\times}^k] = k(p-1,k)[(1+p\mathbf Z_p):(1+p\mathbf Z_p)^k] $$ for $p > 2$ and $$ [\mathbf Q_2^\times:{\mathbf Q_2^\times}^k] = k(2,k)[(1+4\mathbf Z_2):(1+4\mathbf Z_2)^k]. $$

To work out the indices of the $1$-units, we use the $p$-adic logarithm to make things additive: for $p > 2$, the $p$-adic log is a topological group isomorphism (even an isometry) $1+p\mathbf Z_p \to p\mathbf Z_p$. Composition with division by $p$ shows $1+p\mathbf Z_p \to \mathbf Z_p$ as groups, so $$ [(1+p\mathbf Z_p):(1+p\mathbf Z_p)^k] = [\mathbf Z_p:k\mathbf Z_p]. $$ Writing $k = p^{v_p(k)}u$ with $u \in \mathbf Z_p^\times$, so $k\mathbf Z_p = p^{v_p(k)}\mathbf Z_p$, we get $$ [\mathbf Z_p:k\mathbf Z_p] = [\mathbf Z_p:p^{v_p(k)}\mathbf Z_p] = p^{v_p(k)} $$ when $p > 2$. Similarly, since the $2$-adic logarithm is a group isomorphism $1+4\mathbf Z_2 \to 4\mathbf Z_2$, $$ [(1+4\mathbf Z_2):(1+4\mathbf Z_2)^k] = [\mathbf Z_2:k\mathbf Z_2] = 2^{v_2(k)} $$ where $k = 2^{v_2(k)}u$ where $u \in \mathbf Z_2^\times$.

Thus we finally get the index formulas $$ [\mathbf Q_p^\times:{\mathbf Q_p^\times}^k] = k(p-1,k)p^{v_p(k)} $$ for $p > 2$ and $$ [\mathbf Q_2^\times:{\mathbf Q_2^\times}^k] = k(2,k)2^{v_2(k)}. $$ The difference between these formulas for $p > 2$ and $p = 2$ is that when $p > 2$ the middle factor is $(p-1,k)$, while when $p = 2$ the middle factor is $(2,k)$, not $(2-1,k) = 1$.

For a finite extension $F$ of $\mathbf Q_p$, Lang's Algebraic Number Theory has a formula for $[F^\times:{F^\times}^k]$ in the chapter on completions. In the second edition it is Proposition $6$ of Chapter II, Section 3: $$ [F^\times:{F^\times}^k] = k|\mu_k(F)|p^{dv_p(k)}, $$ where $d = [F:\mathbf Q_p]$ and $\mu_m(F)$ is the group of $m$th roots of unity in $F$. This is consistent with the above formulas when $F = \mathbf Q_p$, in which case $d = 1$.

Answer 2

For $p$ odd, $(1+p)^{m p^d} \equiv 1+m p^{d+1} \bmod p^{d+2}$, so starting with $a_0\in 1+p\Bbb{Z}_p$ there is a unique sequence $a_d,m_d$ such that $a_{d+1}=a_d (1+p)^{m_d p^d}$, $a_d \equiv 1\bmod p^{d+1}$, which gives that $$a_0^{-1}=\prod_{d\ge 0} (1+p)^{m_d p^d}= (1+p)^{\sum_{d\ge 0} m_d p^d}$$ ie. $$1+p\Bbb{Z}_p = (1+p)^{\Bbb{Z}_p}$$ Next, Hensel lemma gives that $\zeta_{p-1}\in \Bbb{Z}_p$. From there $$\Bbb{Q}_p^\times = p^\Bbb{Z} \langle \zeta_{p-1} \rangle (1+p\Bbb{Z}_p)=p^\Bbb{Z} \langle \zeta_{p-1} \rangle (1+p)^{\Bbb{Z}_p}$$ $$\Bbb{Q}_p^\times/\Bbb{Q}_p^{\times k} = \frac{p^\Bbb{Z}}{p^{k\Bbb{Z}}} \frac{\langle \zeta_{p-1} \rangle }{\langle \zeta_{p-1}^k \rangle } \frac{(1+p)^{\Bbb{Z}_p}}{(1+p)^{k\Bbb{Z}_p}}, \qquad[\Bbb{Q}_p^\times:\Bbb{Q}_p^{\times k}]= k \gcd(k,p-1) p^{v_p(k)}$$ If $p=2$ then it works the same way from $\Bbb{Q}_p^\times = 2^\Bbb{Z} \langle \zeta_2 \rangle (1+4)^{\Bbb{Z}_2}$.