Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In characteristic p there are nontrivial etale covers of the affine line, such as those obtained by adjoining solutions to x^2 + x + f(t) = 0 for f(t) in k[t]. Using an etale cohomology computation with the Artin-Schreier sequence I believe you can show that, at least, the abelianization of the absolute Galois group is terrible.

What is known about the absolute Galois group of the affine line in characteristic p? In addition, can spaces which are not A^1 (or extensions of A^1 to a larger base field) occur as covers?

share|improve this question
add comment

7 Answers

up vote 10 down vote accepted

Indeed, you can get whatever genus you want even with a fixed Galois group G, so long as its order is divisible by p: this is a result of Pries: .pdf here.

In fact, Pries has lots of papers about exactly what can happen; looking at her papers and the ones cited therein should give you a pretty thorough picture.

We don't know the Galois group of the affine line, but we do know which finite groups occur as its quotients; this is a result of Harbater from 1994 ("Abhyankar's conjecture for Galois groups over curves.") Update: As a commenter pointed out, Harbater proved this fact for an arbitrary affine curve; the statement for the affine line was an earlier theorem of Raynaud.

share|improve this answer
1  
I interpreted your answer to mean that you can any attain finite group G as a Galois group as long as p divides #G. This is not true: a necessary condition (proved sufficient by Raynaud) for G to occur is that it be "quasi-p", i.e., generated by its Sylow p-subgroups. For instance an abelian group is quasi-p iff it's a p-group, not iff its order is divisible by p. I'm sure you know this, but others may not. –  Pete L. Clark Dec 28 '09 at 6:17
add comment

As stated by David Speyer and Clark Barwick, the awnser to the second question is the following:

Any smooth projective curve $C$ defined over a field $k$ of positive characteristic $p$ can be realized as a finite cover of the projective line only ramified above one point.

Here is a short constructive proof only based on Riemann-Roch theorem. It can be considered as an illustration of Kedlaya's proof, only dealing with curves.

  • First of all, there exists a generically étale finite cover $C\to\mathbf P^1$, induced by a rational function $f\in k(C)$ (in fact, any element of $k(C)-k(C)^p$ will do the job).
  • Denote by $R\in$ Div$(C)$ the (reduced) ramification divisor of the above cover (i.e. the ramified points are couted without multiplicity). From Riemann-Roch theorem, for large $n$, there exist a rational function $g\in k(C)$ having a pole of order $n$ at each point of the support of $R$. We may take $n$ strictly greater than $\frac{\deg(f)}p$.
  • Then, the rational function $h=f+g^p$ induces a cover $C\to\mathbf P^1$ only ramified above infinity.
share|improve this answer
add comment

As an excuse to talk about one of my favorite results, I thought I'd put this out there (even though I've already mentioned this to Tyler privately).

Abhyankar conjectured that the the collection of finite quotients of the étale fundamental group of the affine line in characteristic $p$ are exactly the quasi-$p$-groups. This was proved by Raynaud (as mentoned above). A slightly more complicated statement (for general curves) was quickly thereafter proved by Harbater.

Here's an even more interesting (to my mind) result:

Suppose $X$ a geometrically connected, projective variety of dimension over any field $K$ of positive characteristic. Suppose $L$ an ample line bundle on $X$, $D$ a closed subscheme of dimension less than $n$, and $S$ a $0$-dimensional subscheme of the regular locus of $X$ not meeting $D$. Then there exists a positive integer $r$ and an $(n+1)$-tuple of linearly independent sections of $L^{\otimes r}$ with no common zero such that the induced finite morphism $f : X \to P^n_K$ of $K$-schemes meets the following conditions.

(1) If $H$ denotes the hyperplane at infinity, then $f$ is étale away from $H$.

(2) The image $f(D)$ is contained in $H$.

(3) The image $f(S)$ does not meet $H$.

This was proved by Abhyankar in dimension $1$, and the general result is due to Kedlaya. The proof is just gorgeous; it's even simpler than his first paper on the subject, which only works for infinite fields $K$.

This says something pretty remarkable: even though, in characteristic $0$, affine spaces are simply connected, in positive characteristic, every variety contains a Zariski open that is an étale cover of affine space! (Katz uses this kind of trick in his notes on Weil II.)

share|improve this answer
add comment

I see that this question is from a while back, but I figured I add this little morsel: Manish Kumar proved for his thesis that the commutator subgroup of the algebraic fundamental group of A^1 (and, in fact, ANY smooth affine curve) over a countable algebraically closed field of characteristic p>0 is profinite free. He later (to people's astonishment, as the original statement was surprising to begin with) continued to prove this for any algebraically closed field of characteristic p>0 in this arxiv pre-print: http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf

share|improve this answer
add comment

Here's a trick to generate a lot of examples over k = algebraic closure of a finite field (used in the Weil conjectures). Take any smooth curve C and a generically etale map f:C -> P^1. Then there is some open U in A^1 such that V = f^{-1}(U) -> U is finite etale. The reduced scheme underlying A^1 - U is a finite set of closed points which generate a finite subgroup G of the additive group A^1(k). Composing with the quotient A^1 -> A^1/G gives a map V -> A^1 - 0 that is finite etale. Composing with the Artin-Schreier isogeny A^1 - 0 -> A^1 sending x to x^p + 1/x gives a finite etale map V -> A^1.

share|improve this answer
add comment

The statement about which finite groups occur as Galois groups of etale covers of the affine line -- any group which does not have a non-trivial quotient of order prime to p -- is due to Raynaud, not Harbater. (Harbater extended Raynaud's theorem to arbitrary affine curves.)

share|improve this answer
add comment

In response to your second question, http://arxiv.org/abs/math/0207150 states "every curve over an infinite field of characteristic p>0 admits a map to P^1 ramified over only one point!" The above paper proves a more general result, and cites the result about curves to a paper of Katz.

share|improve this answer
    
This is an amazing paper. Un/fortunately Kiran writes papers much faster than I can read them, but I will try to make an exception for this one. –  Pete L. Clark Dec 28 '09 at 11:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.