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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?

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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).

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    $\begingroup$ Thank you, I already suspected that it was not true. In the local case one knows these tamily ramified extensions are unique but in the global case this is not true. Might there be some version that is true? For instance is the abelianization of $Gal(F^{tame}|F^{un})$ isomorphic to $\prod_{p \not= l} \mathbb{Z}_l$? $\endgroup$ – Lee Wang Aug 2 '15 at 13:25

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