Etale covers of the affine line - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:48:04Z http://mathoverflow.net/feeds/question/868 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line Etale covers of the affine line Tyler Lawson 2009-10-17T12:52:47Z 2011-05-03T07:13:37Z <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line/869#869 Answer by David Speyer for Etale covers of the affine line David Speyer 2009-10-17T13:03:59Z 2009-10-17T13:03:59Z <p>In response to your second question, <a href="http://arxiv.org/abs/math/0207150" rel="nofollow">http://arxiv.org/abs/math/0207150</a> 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.</p> http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line/877#877 Answer by JSE for Etale covers of the affine line JSE 2009-10-17T14:20:58Z 2009-10-19T15:10:35Z <p>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 <a href="http://www.math.colostate.edu/~pries/Preprints/06genus1%5F05.pdf" rel="nofollow">here.</a></p> <p>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.</p> <p>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.") <strong>Update</strong>: 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.</p> http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line/884#884 Answer by ulrich for Etale covers of the affine line ulrich 2009-10-17T15:14:35Z 2009-10-17T15:14:35Z <p>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.)</p> http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line/890#890 Answer by Bhargav for Etale covers of the affine line Bhargav 2009-10-17T16:42:36Z 2009-10-17T16:51:37Z <p>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.</p> http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line/9935#9935 Answer by H. Hasson for Etale covers of the affine line H. Hasson 2009-12-28T04:52:26Z 2009-12-28T04:52:26Z <p>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: <a href="http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf" rel="nofollow">http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf</a></p> http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line/11012#11012 Answer by Clark Barwick for Etale covers of the affine line Clark Barwick 2010-01-07T05:06:12Z 2010-01-07T05:06:12Z <p>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).</p> <p>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.</p> <p>Here's an even more interesting (to my mind) result:</p> <p>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.</p> <p>(1) If $H$ denotes the hyperplane at infinity, then $f$ is étale away from $H$.</p> <p>(2) The image $f(D)$ is contained in $H$.</p> <p>(3) The image $f(S)$ does not meet $H$.</p> <p>This was proved by Abhyankar in dimension $1$, and the general result is due to <a href="http://arxiv.org/abs/math/0303382" rel="nofollow">Kedlaya</a>. The proof is just gorgeous; it's even simpler than his first <a href="http://arxiv.org/abs/math.AG/0207150" rel="nofollow">paper</a> on the subject, which only works for infinite fields $K$.</p> <p>This says something pretty remarkable: even though, in characteristic $0$, affine spaces are simply connected, in positive characteristic, <em>every</em> 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.)</p> http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line/63764#63764 Answer by Leonardo for Etale covers of the affine line Leonardo 2011-05-03T02:33:05Z 2011-05-03T07:13:37Z <p>As stated by David Speyer and Clark Barwick, the awnser to the second question is the following:</p> <p><em>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.</em></p> <p>Here is a short constructive proof only based on Riemann-Roch theorem. It can be considered as an illustration of Kedlaya's <a href="http://arxiv.org/abs/math/0303382" rel="nofollow">proof</a>, only dealing with curves.</p> <ul> <li>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).</li> <li>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 $\frac{\deg(f)}p$.</li> <li>Then, the rational function $h=f+g^p$ induces a cover $C\to\mathbf P^1$ only ramified above infinity.</li> </ul>