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I am trying to determine the etale fundamental group of $V = A^1 - \{0\}$ over an algebraically closed field $k$. I am trying to stay in the comfortable zone of non-singular varieties.

To do this, I wonder if there is an easy way to determine all finite etale maps $f:W\to V$ where $W$ is a non-singular variety over $k$.

Any hints how to find all these maps? can I compute the etale fundamental group without finding these?

By finiteness, I guess $W$ must be a finite union of space curves and points.

As a start, I checked what are the finite etale automorphisms of $V$, these are simply given by $a \mapsto a^n$.

Thanks!

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    as a start I will be happy to hear about the characteristic zero case.2011-05-31

2 Answers 2

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Suppose $X$ is a scheme locally of finite type over $\mathbb C$. Then the category of finite étale covers of $X$ is equivalent to the catgory of finite analytic étale covers over $X^{an}$, where $X^{an}$ is the analytic space canonically associated to $X$. The equivalence associates to the étale cover X'\to X its analytification (X')^{an}\to X^{an}.

In your case this implies that the only étale covers of $\mathbb G_m=V=\mathbb A_k^1 \setminus \{0\}$ are the morphisms you mentioned $\mathbb G_m\to \mathbb G_m:z\mapsto z^n$.

The same result is true over any algebraically closed field of characteristic $0$, and implies that the algebraic fundamental group of $ \mathbb G_m $ is the profinite completion $\pi_1^{alg}(\mathbb G_m )=\hat{\mathbb Z}$ of the topological fundamental group $\pi_1^{top}(\mathbb G_m^{an})=\mathbb Z $ . I recommend extreme prudence in characteristic $p$, since as far as I know even the structure of the algebraic fundamental group of $\mathbb A^1_k$ is not known in characteristic $p$ !

Bibliography The equivalence of categories mentioned above is due to Grauert-Remmert. There is a shorter proof in Grothendieck's SGA 1, Théorème 5.1, which however uses Hironaka's resolution of singularities in characteristic zero.

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    Dear anonymous, I share your frustration but once again I only stated *my* ignorance of an easy proof . Maybe a more competent user will come and show us one such proof.2011-06-01
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I'm not sure how elementary/algebraic you will consider this. Let $k$ be an algebraically closed field of characteristic $0$.

Let $W \to V$ be an etale map. Assume $W$ is connected; a general etale map will then be the disjoint union of several examples of this sort. Since $W$ is etale over $V$, we know that $W$ is smooth and one dimensional. Let $\overline{W}$ be the complete curve containing $W$, so we have a map $\overline{W} \to \mathbb{P}^1$. Let the degree of this map be $n$; let $g$ be the genus of $\overline{W}$; let $e^0_1$, ..., $e^0_r$ be the ramification degrees of the points over $0$ and let $e^{\infty}_1$, ..., $e^{\infty}_s$ be the ramification degrees of the points over $\infty$.

The Riemann-Hurwitz formula gives $2g-2 = -2n + \sum (e^0_i-1) + \sum (e^{\infty}_i-1).$ (This is the step which is invalid in positive characteristic.) The right hand side is $-2n + \sum e^0_i + \sum e^{\infty}_i - r -s = -2n+n+n-r-s=-r-s.$ So $2g+r+s=2.$

But $g \geq 0$ and $r$ and $s \geq 1$. So this can only hold if $g=0$ and $r=s=1$. The fact that $g=0$ means that $\overline{W} \cong \mathbb{P}^1_k$. The fact that $r=s=1$ means that there is one point of $\overline{W}$ lying over $0$, and one point lying over $\infty$; without loss of generality, let those points be $0$ and $\infty$.

So our map is of the form $t \mapsto p(t)/q(t)$ for some relatively prime polynomials $p$ and $q$, and the preimages of $0$ and $\infty$ are $0$ and $\infty$. So the only root of $p$ can be $0$, and $q$ can have no roots at all. We conclude that our map is of the form $t \mapsto a t^n$, as desired.


You definitely can give purely algebraic proofs that every curve embeds in a complete curve, and of Riemann-Hurwitz. I feel like one should be able to give pretty elementary ones, but I don't know a reference which does it in an elementary way.