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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!

the L
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2 Answers2

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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.

Georges Elencwajg
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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
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    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
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    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 Georges, I'm quite amused! Though I actually knew your real name both from having recognized your gravatar from MO and from my days with moderator powers here. I may actually reask the above on MO (if someone doesn't post another answer here that takes care of it) --- for some reason, I anticipated that I was just missing something simple. Regards, – Akhil Mathew Jun 01 '11 at 01:35
  • Dear Akhil, I warmly encourage you to post your excellent question on MO if nobody answers it here. I too would be very happy to see an elementary way of classifying all étale covers of $\mathbb G_m$. – Georges Elencwajg Jun 01 '11 at 07:28
  • Thank you for your answer, although I must admit that I am quite disappointed that there is no easy algebraic proof. – the L Jun 01 '11 at 07:39
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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. – Georges Elencwajg Jun 01 '11 at 08:10
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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.

David E Speyer
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