Let $X$ be a (smooth projective geometrically connected) curve over a number field $K$ of genus $g\geq 2$. Let $d\geq 2$ be an integer.

Does there exist a curve $Y$ over $K$ with a finite etale $K$-morphism $Y\to X$ of degree $d$?

I know how to do this over $\bar{K}$ (and thus over some finite extension of $K$). In fact, it suffices to find a finite degree topological cover $Y_{\mathbf{C}} \to X_{\mathbf{C}}$. This is easy to achieve by embedding $X_{\mathbf{C}}$ into its Jacobian and taking a degree $d$ topological cover of the Jacobian $J = \mathbf{C}^g/\Lambda$ of $X_{\mathbf{C}}$. The latter can be constructed easily by taking a sub-lattice of $\Lambda$ of index $d$.

Two problems arise.

The curve $X$ might not embed into its Jacobian, i.e., it might happen that $X(K)$ is empty. So if it helps, assume $X(K)$ is non-empty.

Also, the etale cover of $J$ constructed over $\mathbf{C}$ might not be defined over $K$ a priori. Can one show that it actually is defined over $K$?

  • 3
    $\begingroup$ Not in general: if $X$ is an elliptic curve with trivial two-torsion subgroup over $K$, then there is no non-trivial étale double cover of $X$. $\endgroup$ – M P Oct 14 '12 at 10:57
  • 3
    $\begingroup$ Unramified double covers of a curve correspond to elements of order 2 in its Jacobian, so MP's observation generalizes to curves whose Jacobian has only the trivial 2-torsion, of which there exist plenty. $\endgroup$ – RP_ Oct 14 '12 at 11:24
  • 4
    $\begingroup$ (What I wrote before was mostly wrong -- so I have deleted it.) Elliptic curves always have étale covers corresponding to mult. by $n$. Generically, these are the only ones. As for higher genus curves, these too always admit non-trivial étale coverings: Choose a non-trivial morphism $f:X \rightarrow J = \mathrm{Jac}(X)$ (no need to assume it injective), then base change $[n]:J \rightarrow J$ along $f$ to obtain a map $Y \rightarrow X$, which is étale since it is the base change of an étale map. I imagine it might well be true that, generically, this is all there is, but I have no proof. $\endgroup$ – RP_ Oct 14 '12 at 12:38
  • 1
    $\begingroup$ @rene. One needs some kind of argument for why the resulting cover is not trivial. E.g. take $f$ constant. $\endgroup$ – Dan Petersen Oct 14 '12 at 15:25
  • 2
    $\begingroup$ @Harry: it would be very surprising if a d-cover of X implies the existence of an element of order d in the Jacobian, because the proof uses the fact that index two subgroups are normal, which fails to hold in general. @Rene: Can't one always chose a morphism $X \to J$ that induces an isomorphism on the first homology? Then it is clear that the covering is nontrivial. $\endgroup$ – Will Sawin Oct 15 '12 at 5:50

This is not a complete answer, but I am trying to translate your issue in something more tractable.

If you assume there exists a rational point $x$, then the image of the corresponding section $s=s_x$ of $\pi_1(X,\overline x)\to \rm{Gal}_k$ is a closed subgroup of $\pi_1(X,\overline x)$ thus corresponds by Galois theory to a pro-cover $X_s\to X$ which is geometrically connected by construction (this works as well, of course, if you assume only the existence of $s$). So in this case you have plenty of covers (all intermediate covers) but it is not obvious to me how to determine their orders.

Going back to the general formulation of your problem, the usual way to solve your first problem is to use the universal albanese torsor. For a curve, this is just the degree-$1$ part of the Picard scheme of $X/k$, the map $X\to {\rm Pic}^1_{X/k}$ corresponds to the data of the isomorphism class of $\mathcal O_{X\times_k X}(\Delta)$, where $\Delta$ is the diagonal of $X\times_k X$. The albanese torsor is a torsor under the jacobian ${\rm Pic}^0_{X/k}$. Moreover, the morphism $X\to {\rm Pic}^1_{X/k}$ induces on $\pi_1$ an isomorphism $\pi_1(X,\overline x)\to \pi_1(X,\overline x)^{[ab]}$, where the last group is the quotient of $\pi_1(X,\overline x)$ by the kernel of $\pi_1(X_{\overline{k}},\overline x)\to \pi_1(X_{\overline{k}},\overline x)^{ab}$, the abelianization of the geometric fundamental group. (One says that $\pi_1({\rm Pic}^1_{X/k},\overline x)$ is the geometric abelianization of $\pi_1(X,\overline x)$). This enables to reduce your problem to the similar problem for a torsor under an abelian variety. Both problems are equivalent if you want to consider abelian covers only.

I am unsure on how to solve this last issue, but I think this is known - let us hope that an expert of abelian varieties will give the answer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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