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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 $a \mapsto a^n$.


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Is $k$ characteristic zero or not? The positive characteristic case will be much harder. – David Speyer May 31 '11 at 20:25
as a start I will be happy to hear about the characteristic zero case. – the L May 31 '11 at 20:38
up vote 9 down vote accepted

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 elgeorges, is there a purely algebraic proof? The statement reduces to the classification of finite algebraic extensions of $k(t)$ (for $k$ alg. closed, char. 0) where all primes except $(t)$ are unramified. But I don't know how to get this easily. – Akhil Mathew May 31 '11 at 21:52
Dear Akhil, I don't know about the existence of a purely algebraic proof . What I know is that until a few years ago there was no purely algebraic computation of the algebraic fundamental group of the projective complex line minus three points (namely $\pi_1^{alg}(\mathbb P^1_{\mathbb C} \setminus \{0,1, \infty\})=\widehat{\mathbb Z\ast \mathbb Z}$). – Georges Elencwajg May 31 '11 at 23:25
Dear elgeorges, Very interesting! Thanks for the response. – Akhil Mathew May 31 '11 at 23:27
Dear Akhil, your comments have made me abandon my pseudonym! I realize that it is ridiculous to force a friendly young man like you to call me "dear elgeorges" while I call you "dear Akhil". Anyway I never wanted to really hide behind this transparent handle. I just liked the fact that since the first two letters of my name are "El", like the Spanish definite article, my username sounded like some secondary character in an old "Zorro" movie... – Georges Elencwajg May 31 '11 at 23:49
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. – Georges Elencwajg Jun 1 '11 at 8:10

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.

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Fantastic! thank you very much! – the L Jun 1 '11 at 22:02
Very interesting, David. I suppose there is a typo on line 5: you can't map $\bar W$ to $V$. – Georges Elencwajg Jun 1 '11 at 22:52
Typo fixed, thanks! – David Speyer Jun 2 '11 at 2:35

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