# $\pi^{tame}(\mathbb{A}^1_k)$ is trivial

Fixed an algebraically closed field of characteristic $p>0$, it is well known the result of the title: $\pi^{tame}(\mathbb{A}^1_k)\simeq 1$. Where the tame fundamental group, in this situation, classifies all the finite ètale coverings of $\mathbb{A}^1_k$ which are tamely ramified on the infinite point of $\mathbb{P}^1_k$.

How is it proved?

Following Hartshorne Chap IV Par. 2, and using the Riemann-Hurwitz formula, it can be proved that $\widehat{\pi}(\mathbb{P}^1_k)\simeq 1$, hence also $\pi^{tame}(\mathbb{P}^1_k)\simeq 1$. Where $\widehat{\pi}$ is the whole ètale fundamental group.

I observed that it is crucial to look only at tamely ramified coverings on the infinite point, because there exist examples of finite ètale coverings of $\mathbb{A}^1_k$ which are wild ramified on the infinite point.

I thought to approach the problem taking the completion $k[[x]]$ of the Zariski local ring of the infinite point, then cutting out the closed point from the neighborhood $Spec(k[[x]])$ of $\infty$ one would get $Spec(k((x^{-1},x]])$ (where $k((x^{-1},x]]$ is the fraction field of $k[[x]]$) which should be contained in $\mathbb{A}^1_k$ so covered by an ètale morphism. But I didn't get anything nor I'm sure that I didn't write rubbish.

I also tried to control, in the Riemann-Hurwitz formula, the ramification index over $\infty$ with the degree of the covering map. In order to adjust the proof for $\mathbb{P}^1_k$ of Hartshorne. But again it didn't bring me anywhere.

Edit: I want to resume in the edit the progress I made thanks to the generous suggestions of Pete L. Clark.

We are using the notations of corollary 2.4 of chapter IV of Hartshorne. We want to apply the Riemann-Hurwitz formula to a covering $f: X\rightarrow Y$, where $Y=\mathbb{P}^1_k$. Thanks to the fact that the restriction of the covering on $\mathbb{A}^1_k$ is ètale, and applying the same formula to the restricted covering, we can deduce that the degree of $f$ is $1$. Is this true?

So the formula tells us, about the original covering, that $2g(X)=deg(R)$, where $R$ is the ramification divisor. Pete pointed out explicitely that the divisor is trivial everywhere except in one point, namely $\infty$. But I'm confused on how to use this fact for proving $deg(R)=0$.

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If I'm not mistaken, you're there already. There is one version of the Riemann-Hurwitz formula which holds when the ramification is assumed to be tame (so always in characteristic zero) and another version which allows for wild ramification. You're asking about the tame one, so applying that to a cover of $\mathbb{P}^1$ with only one ramification point will show that it has no nontrivial tamely ramified coverings in any characteristic.
Can you use the Riemann-Hurwitz theorem to show that $\mathbb{A}^1$ has no covers of degree $d > 1$ in characteristic zero? If you haven't done that computation already, maybe that's where the problem lies? –  Pete L. Clark Jul 4 '11 at 21:29
In the notations of Corollary 2.4 of chapter IV in Hartshorne, when you refer to an extra term which can be nonzero only if the ramification is wild you are referring to $deg(R)$? If yes I cannot understand why is it true, indeed it should have a contribution from the ramification in $\infty$. I want underscore I'm thinking $Y=\mathbb{P}^1_k$, if on the other hand $Y=\mathbb{A}^1_k$ then $deg(R)$ should be $0$ in any case (tame or wild) because the map is supposed to be ètale on the whole space $\mathbb{A}^1_k$. 'sorry but I'm afraid I'm missing the point... –  Giovanni De Gaetano Jul 4 '11 at 21:34