What is the index of the $p$-th power of $\mathbb Q_p^\times$ in $\mathbb Q_p^\times$

Question

In this book it is listed as an exercise to compute the index $[\mathbb Q_p^\times:(\mathbb Q_p^\times)^p]$.

This exercise is appended to a section concerning the structure of unit-group filters, investigating some properties of the homomorphism raising to the $p$-th power. The result is that, when $i<e/p-1$ (resp. when $i>e/p-1$), then $U_i/U_{i+1}$ is mapped isomorphically to $U_{i+e}/U_{i+e+1}$ (resp. $U_{pi}/U_{pi+1}$). Furthermore, when $i=e/{p-1}$ then the kernel of raising to $p$-th power is of order $p$.
Therefore, if $p \neq 2$ then the index $[\mathbb Q_p^\times:(\mathbb Q_p^\times)^p]$ is $p$; while if $p=2$, then the index is $2\cdot 2=4$. Are there any errors in the above reasoning? Thanks for the attention in any case in advance.

Answer 1

This exercise is a fun way to write down some stuff I've been doing for my $p$-adic class.

Teichmuller characters. Suppose $a\in{\bf Z}_p$. Then using the binomial theorem,

$$\left|\frac{p\cdot p\cdot p\cdots p}{1\cdot 2\cdot 3\cdots k}p^n\cdot(p^n-1)\cdots(p^n-(k-1))\right|_p\le\frac{1}{p^n}, $$

$$\left|(1+pa)^{p^n}-1\right|_p\le \max_{1\le k\le p^n}\left|{p^n\choose k}p^k\right|_p\le\frac{1}{p^n}\to0,$$

and hence $(1+pa)^{p^n}\to1$ (compare with $t^\epsilon\to1$ as $\epsilon\to0$ in $\bf R$ for $t$ in a nbhd of $1$). If $x\in{\bf Z}_p$, then $x^{p-1}\in1+p{\bf Z}_p$ (reduce mod $p$ and invoke Euler's theorem), hence $(x^{p^n})_{n=0}^\infty$ is Cauchy and has a limit in the $p$-adic integers, since $x^{p^{n+1}}-x^{p^n}=x^{p^n}\left((x^{p-1})^{p^n}-1\right)$. In an ultrametric space, the distance between successive terms tending to $0$ is sufficient to be Cauchy. Since multiplication and hence fixed powers are by definition continuous in a topological ring, $x\mapsto x^p$ is continuous, so

$$\omega(x):=\lim\limits_{n\to\infty}x^{p^n}\implies \omega(x)^p=\left(\lim_{n\to\infty} x^{p^n}\right)^p=\lim_{n\to\infty}x^{p^{n+1}}=\omega(x),$$

in which case we have $\omega(x)^{p-1}=1$ when $\omega(x)$ is a unit. Since $x^p\equiv x\bmod p$, and ${\bf Z}_p\to{\bf Z}/p{\bf Z}$ is a continuous topological ring homomorphism (the latter having discrete topology), $\omega(x)\equiv x\bmod p$.

This tells us the image of units under $\omega$ consists of the $(p-1)$th roots of unity. Choosing any representatives of ${\bf F}_p$ inside ${\bf Z}_p$, this also gives us a homomorphism ${\bf F}_p^\times\to{\bf Z}_p^\times$ whose image consists of the Teichmuller representatives.

Unit groups. $U(n):=({\bf Z}/n{\bf Z})^\times$ (not to be confused with unitary groups). The CRT allows us to classify these in terms of $U(p^r)$ for prime powers $p^r$; we have $U(p^r)\cong C(p-1)\oplus C(p^{r-1})$ for $p>2$ and $U(2^r)\cong C(2)\oplus C(2^{r-2})$ for $r>1$, where by $C(n)$ we mean the cyclic group of order $n$.

Primitive roots. Let $p$ be an odd prime. Since ${\bf F}_p^\times$ is cyclic, there is a generator (in fact there are $\phi(p-1)$ generators). Such primitive roots will always lift to primitive roots modulo arbitrarily high powers of $p$ (see Lemma 14 here).

Inverse limits. There is a definition ${\bf Z}_p:=\varprojlim\,{\bf Z}/p^n{\bf Z}$ (see here for a good exposition on projective limits in the category of topological rings). Taking inverse limits commutes with direct sums and taking unit groups, so by CRT we obtain

$$\begin{array}{cl} {\bf Z}_p^\times & \cong\left(\varprojlim\frac{\bf Z}{p^n\bf Z}\right)^\times \\ & \cong\varprojlim\left(\frac{\bf Z}{p^n\bf Z}\right)^\times \\ & \cong\varprojlim\left(\frac{\bf Z}{(p-1)}\oplus\frac{\bf Z}{p^{n-1}\bf Z}\right) \\ & \cong\left(\varprojlim\frac{\bf Z}{(p-1)}\right)\oplus\left(\varprojlim\frac{\bf Z}{p^{n-1}\bf Z}\right) \\ & \cong\frac{\bf Z}{(p-1)}\oplus ({\bf Z}_p,+),\end{array}$$ and similarly ${\bf Z}_2\cong{\bf Z}/(2)\oplus({\bf Z}_2,+)$. Observe also ${\bf Q}_l^\times\cong{\bf Z}\oplus{\bf Z}_l^\times$ for any $l$.

Given a sequence of primitive roots $g_n\in{\bf Z}/p^n{\bf Z}$ such that $g_{n+1}\equiv g_n\bmod p^n$, $g:=\lim\limits_{n\to\infty}g_n$ exists as a $p$-adic integer. Then $\tau=g^{p-1}$ has maximal $p$-power order mod every power of $p$, and is in particular an element of $1+p{\bf Z}_p$. There is an explicit isomorphism, then, given by

$${\bf F}_p^\times\oplus{\bf Z}_p\to {\bf Z}_p^\times:(a,x)\mapsto \omega(a)\lim_{n\to\infty}g^{(x~\bmod~p^n)}.$$

Note further that multiplication on the second coordinate of the domain above induces a fairly accessible ${\bf Z}_p$-module structure on $1+p{\bf Z}_p$, and $g$ is a topological generator.

Therefore, in conclusion,

$$\frac{{\bf Z}_p^\times}{({\bf Z}_p^\times)^n}\cong\frac{\bf Z}{r(n,p)}\oplus\frac{\bf Z}{(p^{v_p(n)})},\quad [{\bf Z}_p^\times:({\bf Z}_p^\times)^n]=r(n,p)p^{v_p(n)}$$

$$\frac{{\bf Q}_p^\times}{({\bf Q}_p^\times)^n}\cong\frac{\bf Z}{(n)}\oplus \frac{{\bf Z}_p^\times}{({\bf Z}_p^\times)^n},\quad [{\bf Q}_p^\times:({\bf Q}_p^\times)^n]=n[{\bf Z}_p^\times:({\bf Z}_p^\times)^n],$$

where $\displaystyle r(n,p):=\frac{p-1}{(n,p-1)}$ if $p>2$ and $\displaystyle\frac{2}{(n,2)}$ if $p=2$.