Let $K$ be a local field of characteristic $0$ with a perfect residue field of characteristic $p$. Fix a uniformizer $\pi$ and an algebraic closure $\overline{K}$. Now fix a choice of a system of $p^n$-th roots of $\pi$ in $\overline{K}$ for all $n\in \mathbb{N}$, noted $\xi=\{ \pi^{(n)} \}_{n\geq 0}$, $\ i.e.\ (\pi^{(n+1)})^p=\pi^{(n)}$ and $\pi^{(0)}=\pi$. Denote $K_{\xi}:=\cup_{n\in \mathbb{N}} K(\pi^{(n)})$.
Now I have some elementary question on the "relation" of $\xi$ with the usual cyclotomic extension $K^{cycl}:=\cup_{n\in \mathbb{N}} K(\zeta_{n})$, where $\zeta_m$ is a primitive $p^m$-th root unity for all $m\in \mathbb{N}$. I feel the situation should be simple at least if we assume $\zeta_n\not\in K, n\geq n_0$ for some fixed $n_0\in \mathbb{N}$, yet I can not describe it precisely. To make things more clear, I start by asking some precise questions as follows, based on the assumption just mentioned (I am also happy to be informed if you feel the assumption is not necessary).
- Is it true that $\pi^{(n+1)}\not \in K^{cycl}(\pi^{(n)})$ when $n\geq n_0$, and why?
- More or less the same question: What is $K^{cycl}\cap K_{\xi}$? Is it just $K$?
Noticing that when we make a different choice of system of $p^n$-th roots of $\pi$, say $\xi^{'}=\{ \pi^{'(n)} \}_{n\geq 0}$, we then know $\pi^{(m)}=\zeta_m^{c(m)}\pi^{'(m)}$ for some $c(m)\in \mathbb{Z}/p^m\mathbb{Z},\ m\in \mathbb{N}$.
- More or less the same question again: Is it true that $K_{\xi}\cap K_{\xi^{'}}=K(\pi^{(m)})$, where $m=\max \{n\in \mathbb{N};\ \pi^{(n)}=\pi^{'(n)} \}$?
Although I feel this question should be an elementary Galois theory exercise, I cannot write down a rigorous proof. Thanks in advance for clarifications or rigorous proofs.
Local field : finite extension of $\Bbb{Q}_p$ or $\Bbb{F}_p((t))$.
$F(\zeta_{p^\infty-1})/F$ is unramified whereas $F(\pi_F^{1/p})/F$ is not.
Thus can replace $F$ by $F(\zeta_{p^\infty-1})$ the problem will stay the same.
It remains to check if $F(\pi_F^{1/p})=F(\zeta_{p^k})$ where $k$ is the least possible, thus $[F(\zeta_{p^k}):F]=p$.
Comparing the degrees and Galoisness we find $char(F)=0,k\ge 2,\zeta_p\in F$. The Galois groups have generators $\sigma(\pi_F^{1/p})=\zeta_p \pi_F^{1/p}$ and $\sigma'(\zeta_{p^k})=\zeta_{p^r}\zeta_{p^k}$, it must be $r=1$ because $[F(\zeta_{p^k}):F]=p$.
If $F(\pi_F^{1/p})=F(\zeta_{p^k})$ then $\sigma=(\sigma')^a,p\nmid a$ so $\sigma(\zeta_{p^k})=\zeta_p^a \zeta_{p^k},\sigma(\pi_F^{-a/p}\zeta_{p^k})=\pi_F^{-a/p}\zeta_{p^k}$ and $\pi_F^{-a/p}\zeta_{p^k}\in F$, a contradiction because $\pi_F^{-a/p}$ would have the same valuation as an element of $F$.