Let $X$ be a (smooth projective geometrically connected) variety over a field $k$.

Consider the set Et$(X,k)$ of finite etale covers $Y\to X$ over $k$, with $Y$ geometrically connected over $k$.

Assume Et$(X,k)$ is infinite. Consider the following question:

Does $X$ have a $k$-rational point?

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.

What if $X$ is a curve? Is $X(k)$ non-empty?

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.

I'm mainly interested in the characteristic zero case, but comments on the situation in positive characteristic would also be interesting.

  • $\begingroup$ What if $X$ is an elliptic curve ? $\endgroup$ Nov 3, 2012 at 3:02
  • $\begingroup$ @Chandan: Every elliptic curve $E$ has a $k$-rational point, by definition; namely $0$. More general $E(T)$ is a group (hence non-empty) for any $k$-scheme $T$. $\endgroup$
    – jmc
    Nov 3, 2012 at 8:29
  • $\begingroup$ Of course. I got confused as to whether you wanted a rational point or didn't want any... $\endgroup$ Nov 3, 2012 at 8:46
  • $\begingroup$ @Jan Hendrik: Do you allow étale covers $X_{k'}\to X$ obtained by finite separable extensions of $k$ ? $\endgroup$
    – Qing Liu
    Nov 3, 2012 at 10:37
  • $\begingroup$ @Qing Liu. No, I want them to be geometrically connected over $k$. I should have written that in the question. $\endgroup$ Nov 3, 2012 at 11:31

1 Answer 1


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 $J$.

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.

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.

  • $\begingroup$ Are you aware of a nice example of such a curve? It seems very plausible to me but I don't know how you'd construct one. $\endgroup$
    – Will Sawin
    Nov 3, 2012 at 14:48
  • 3
    $\begingroup$ @Will: every curve over a finite field has a divisor of degree 1, but not necessarily à rational point. $\endgroup$ Nov 3, 2012 at 15:39
  • 3
    $\begingroup$ Maybe I also misunderstand the question, but I don't see what the difficulty is over a number field. Take any genus one curve $X $ without a rational point, $3x^3+4y^3+5z^3=0$ over $Q$, for example. Then $X$ will have a finite-to-one map $f:X\rightarrow E$ to its Jacobian $E$, which is an elliptic curve. You can then pull back, say, any isogeny of $E$ of degree prime to that of $f$. You can generalize this easily to other curves without rational points. $\endgroup$ Nov 3, 2012 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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