most recent 30 from http://mathoverflow.net 2013-05-19T02:08:22Z http://mathoverflow.net/feeds/question/111329 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111329/varieties-with-infinitely-many-etale-covers-and-rational-points Jan Hendrik 2012-11-03T01:19:50Z 2012-11-03T13:38:09Z

<p>Let $X$ be a (smooth projective geometrically connected) variety over a field $k$. </p> <p>Consider the set Et$(X,k)$ of finite etale covers $Y\to X$ over $k$, with $Y$ geometrically connected over $k$.</p> <p>Assume Et$(X,k)$ is infinite. Consider the following question:</p> <p>Does $X$ have a $k$-rational point?</p> <p>The answer should be negative in general. In fact, I think one can construct a surface with infinitely many etale covers but no rational points by taking the product of two curves $C$ and $D$ over $k$, where $C$ has infinitely many etale covers and a rational point, but $D$ doesn't have any rational points. Then $C\times D$ has no rational points, but infinitely many covers.</p> <p>What if $X$ is a curve? Is $X(k)$ non-empty?</p> <p>Note that the converse is true if we consider curves of positive genus. That is, if $X$ is a curve of positive genus over $k$ with a $k$-rational point, then it has infinitely many etale covers. </p> <p>I'm mainly interested in the characteristic zero case, but comments on the situation in positive characteristic would also be interesting.</p>

http://mathoverflow.net/questions/111329/varieties-with-infinitely-many-etale-covers-and-rational-points/111331#111331 Felipe Voloch 2012-11-03T01:54:57Z 2012-11-03T13:38:09Z

<p>Maybe I misunderstand something, but don't all curves have etale covers? Embed $X$ in $J^1$ (divisors of degree $1$ modulo linear equivalence). Then $J^1$ is a torsor for the Jacobian $J$ and since $J$ has etale covers, e.g. coming from multiplication by an arbitrary $n$, $J^1$ does too. Certainly, for those curves with a rational divisor of degree one, they have covers, as $J^1$ is isomorphic to $J$.</p> <p>EDIT: Upon further reflection, I guess it's not true that $J^1$ always has covers, as it may not be in the divisible part of the Weil-Chatelet group of $J$. But there definitely exist curves with no points having divisors of degree one, and therefore covers of arbitrarily large degree. </p> <p>However, you question is a good one and you might be heading in the direction of Grothendieck's section conjecture: For finitely generated fields $k$, $X(k)$ is non-empty if and only if there is a section $G_k \to \pi_1(X)$ of the canonical projection $\pi_1(X) \to G_k$, where $G_k$ is the absolute Galois group of $k$ and $\pi_1$ is the etale fundamental group.</p>