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Jan 15, 2022 at 21:56 comment added Maarten Derickx Hi Ariyan, Thanks for the reference. I think Thm 1.3 should indeed be enough to answer this question in an affirmative way. There is some work to be done to go from that statement to the statement that I am after. But it shouldn't be to hard.
Jan 15, 2022 at 21:41 history edited Maarten Derickx CC BY-SA 4.0
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Jan 15, 2022 at 21:39 comment added Ariyan Javanpeykar 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).
Jan 15, 2022 at 21:28 comment added Maarten Derickx @SashaP thanks for your comment. There was indeed a mistake in my formulation. I do not necessarily require the subgroups to be distinct, however I do require that the union of all the cosets is not equal to $A(K)$. I guess an equivalent formulation would be that I want there to be a non empty coset to be contained in $A(K) \setminus f(X(K))$.
Jan 15, 2022 at 21:24 history edited Maarten Derickx CC BY-SA 4.0
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Jan 15, 2022 at 21:23 comment added Maarten Derickx 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.
Jan 15, 2022 at 20:59 comment added SashaP 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 $H\subset A(K)$
Jan 15, 2022 at 20:52 comment added Ariyan Javanpeykar 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?
Jan 15, 2022 at 20:48 comment added Ariyan Javanpeykar 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.)
Jan 15, 2022 at 20:23 history asked Maarten Derickx CC BY-SA 4.0