Galois group of maximally tamily ramified extension over the maximally unramified extension of a global function field F

Question

Suppose $F$ were a local nonarchimedean field of characteristic zero of residue characteristic $p$. Let $F^{un}$ the maximally unramified extension. Let $F^{un}\subset F^{tame}$ be the maximally tame extension. If I understand correctly $F^{tame}$ is obtained from $F^{un}$ by adjoining $\sqrt[n]{p}$ for all $n\in \mathbb{N}$ where $p$ does not divide $n$. Moreover these extensions are separable and have cyclic Galois group.

Let $F$ be a global function field over a finite field $\mathbb{F}_q$ of characteristic $p$. By analogy I suppose that $F^{tame}$ is obtained by adjoining n-th roots of T, where T is a trancendental element. That is we take the union of all splitting fields $E_n$ of $X^n-T$ where $p$ does not divide $n$. Let $\sqrt[n]{T}$ be a root of this polynomial. I believe the roots are $\gamma \sqrt[n]{T}$ with $\gamma \in F^{un}$ a n-th root of unity.

The extension is separable and normal so Galois. I expect Gal$(E|F^{un})\cong \mathbb{Z}/n\mathbb{Z}$ given by the Frobenius automorphisms $m: \gamma \sqrt[n]{T}\mapsto \gamma^{m}\sqrt[n]{T}$ where $m\in \mathbb{Z}/n\mathbb{Z}$. Finally this implies Gal$(F^{tame}|F^{un}) \cong \varprojlim \mathbb{Z}/n\mathbb{Z} \cong \prod_{p \not= l} \mathbb{Z}_l$.

Is this correct?

Answer 1

No, the analogy is not right. It would be right if you were looking at the maximal tame extension of a local function field like $\mathbb{F}_q((T))$, in which case your proof goes through perfectly.

But a general curve will have covers more complicated than the ones you described (indeed, every cover you have described is abelian, so it suffices just to construct a non-abelian cover to see this).