several weeks ago I asked about a statement form Silverman's "The Arithmetic of Elliptic Curves", Cap. VII.4 (The Action of Inertia) recall, we consider a local complete separable field $K$ with residue field $k$. Furthermore, we denote by $K^{nr}$ the maximal unramified extension of $K$.
I asked about the claim why the Galois groups $G_{K^{nr}/K}$ and $G_{\overline{k}/k}$ are equal. Recall, that $\overline{k}$ is the algebraic closure. as correctly pointed out by Ferra in his answer the argument can be looked up at wiki page.
now I would like to know if the isomorphism between $G_{K^{nr}/K}$ an be constructed $G_{\overline{k}/k}$ more directly in a way I would like to explain now:
assume that $K$ if finite extension of $\mathbb{Q}_p$; thus the residue field $k$ is finite extension of $\mathbb{F}_p$ and thus of characteristic $p$.
as
$$G_{K^{nr}/K}= \varprojlim_{L/K \text{ unramified finite ext }}G_{L/K}$$ and $$G_{\overline{k}/k}= \varprojlim_{l/k \text{ finite ext }}G_{L/K}$$
are inverse limits we can reduce our claim to finite case, i.e. we want to see that for any unramified finite extension $L/K$ and induced extension $l/k$ of residue fields we want construct the isomorphism $Gal_{L/K}=Gal_{l/k}$ explicitly.
Q: can the isomorphisms $Gal_{L/K}=Gal_{l/k}$ be constructed 'directly'? what I mean by directly? well, $l/k$ is an extension between finite fields with characteristic $p$ and thus there exist $s <n$ with $k= \mathbb{F}_{p^s}, l= \mathbb{F}_{p^n}$ and the Galois group $Gal_{l/k}$ is concretely generated by relative Frobenius $\sigma^s$ where $\sigma$ is the Frobenius automorphism given by $a \mapsto \sigma(a)=a^p$ defined for all fields of characteristic $p$.
back to my Question: from above we know that $Gal_{L/K}=Gal_{l/k}$ and my question is what is the 'canonical' generator $\alpha \in Gal_{L/K}$ that is the unique lift of relative Frobenius $\sigma^s$. can the constrution of $\alpha$ from $\sigma^s$ be described explicitely?
Best Answer
If $f\in \Bbb{Z}_p[x]_{monic}$ is separable $\bmod p$ then $f\bmod p$'s splitting field is $\Bbb{F}_{p^f} = \Bbb{F}_p(\xi_{p^f-1})$ and (by Hensel lemma) $f$ splits completely in $\Bbb{Z}_p[\zeta_{p^f-1}]$
Thus $$\Bbb{Z}_p^{unr}=\Bbb{Z}_p[\zeta_{p^\infty-1}]=\bigcup_{p\ \nmid\ n} \Bbb{Z}_p[\zeta_n]=\bigcup_N \{\sum_{j\ge 0} c_j p^j, c_j \in S_N\}$$ where $$S_N=\{0\} \cup \{ \zeta_n,n\le N, p\nmid n\}\}$$
The lift of $Frob \in Gal(\overline{\Bbb{F}}_p/\Bbb{F}_p)$ to $\phi \in Gal(\Bbb{Z}_p^{unr}/\Bbb{Z}_p)$ is $$\phi(\sum_{j\ge 0} c_j p^j) = \sum_{j\ge 0} c_j^p p^j$$ and $$Gal(\Bbb{Z}_p^{unr}/\Bbb{Z}_p)= \varprojlim \phi^{\Bbb{Z/nZ}}$$
It works the same way in any complete (or Henselian) DVR $R$ with residue field $R/(\pi)\subset \overline{\Bbb{F}}_p$, replacing $p^j$ by $\pi^j$.