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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.

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What if $X$ is an elliptic curve ? –  Chandan Singh Dalawat Nov 3 '12 at 3:02
    
@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 $T$. –  jmc Nov 3 '12 at 8:29
    
Of course. I got confused as to whether you wanted a rational point or didn't want any... –  Chandan Singh Dalawat Nov 3 '12 at 8:46
    
@Jan Hendrik: Do you allow étale covers $X_{k'}\to X$ obtained by finite separable extensions of $k$ ? –  Qing Liu Nov 3 '12 at 10:37
    
@Qing Liu. No, I want them to be geometrically connected over $k$. I should have written that in the question. –  Jan Hendrik Nov 3 '12 at 11:31
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1 Answer

up vote 6 down vote accepted

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.

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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. –  Will Sawin Nov 3 '12 at 14:48
3  
@Will: every curve over a finite field has a divisor of degree 1, but not necessarily à rational point. –  Laurent Moret-Bailly Nov 3 '12 at 15:39
3  
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. –  Minhyong Kim Nov 3 '12 at 16:06
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