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Let $K$ be a number field $A$ an abelian variety over $K$ and $f: X \to A$ a possibly ramified covering of degree $d$ with $X$ a proper variety. My question is:

Suppose that $f(X(K)) \neq A(K)$, can ond then always find a finite set of strict subgroups $H_1,\ldots, H_n \subsetneq A(K)$ and a sequence of points $x_1, \ldots, x_n$ such that $f(X(K)) \subseteq \bigcup x_i + H_i \neq A(K)$?

This question is only interesting when $\# A(K) = \infty$. And the general result also easily follows from the case where $X$ is irreducible.

Now there are two cases in which I know the answer to be positive:

  1. When $A$ is an elliptic curve. Because in the case $X$ is normal if $X(K) \neq \emptyset$ then either $X$ is an elliptic curve as well in which case $f(X(K))$ is already a coset of a finite index subgroup, or $X$ has genus > 1 in which case $X(K)$ is finite and the statement is trivially true. The general case follows from the normal case, cause $X$ has only finitely many singular points.
  2. $X$ is a subvariety of an abelian variety $B$. Because in this case one can use a result of Faltings theorem to cover the rational points of $X$ by translates of abelian subvarieties of $B$, and the result follows as well.

I personally even have no idea what I should expect the answer to be. Although I am hoping for a positive answer because of some applications I have for this statement.

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  • $\begingroup$ Are you assuming $X$ to be normal? (When $A$ is an elliptic curve, $X$ could be a singular curve whose normalization is an elliptic curve.) $\endgroup$
    – Ariyan Javanpeykar
    Commented Jan 15, 2022 at 20:48
  • $\begingroup$ If $X$ is normal, and $X\to A$ is a finite surjective not-unramified (i.e., not etale) morphism, then we expect $X(K)$ to be non-dense. (Indeed, such a variety $X$ dominates a positive-dimensional variety of general type by a theorem of Kawamata proven in his PhD thesis, and Lang conjectured non-density of rational points on varieties of general type.) Does this non-density imply what you want? $\endgroup$
    – Ariyan Javanpeykar
    Commented Jan 15, 2022 at 20:52
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    $\begingroup$ Are you assuming that the subgroups $H_1,\dots H_n$ are distinct? Otherwise the conclusion seems to be vacuous: we can take $H_1=\dots =H_n=H$ for $H$ of finite index and let $x_1,\dots, x_n$ be a set of coset representatives of $H\subset A(K)$ $\endgroup$
    – SashaP
    Commented Jan 15, 2022 at 20:59
  • $\begingroup$ Hi @AriyanJavanpeykar, Indeed I was a bit sloppy with the elliptic curve argument. However, when $X$ is a curve it will have finitely many singular points anyway, so one could cover the rational singular points by cosets of the subgroup {0}. Anyway I would happily accept any answer that gives an answer in the case that $X$ is normal. $\endgroup$
    – Maarten Derickx
    Commented Jan 15, 2022 at 21:23
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    $\begingroup$ If $A(K)$ is dense, then $A(K)\setminus f(X(K))$ contains a finite index coset of $A(K)$ by Thm 1.3 in arxiv.org/abs/2011.12840 To deal with the case that $A(K)$ infinite, you can maybe take the closure of $A(K$) and restrict your covering to some well-chosen positive-dimensional component of $\overline{A(K)}$ (which, by Faltings, is the translate of an abelian subvariety). $\endgroup$
    – Ariyan Javanpeykar
    Commented Jan 15, 2022 at 21:39

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