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Suppose $A$ is an abelian variety over an algebraically closed field $k$. I'm confused about the notion of translates of abelian subvarieties. I looked over a few related papers, but I couldn't find a precise algebro-geometric definition. People simply used $x+B$, and I assume they are interested in functor of points. In my opinion, given a point $x:\mathrm{Spec}k'\rightarrow A$, a translate should live in the base change $A_{k'}$. Namely there is an abelian subvariety $B$ of $A_{k'}$ such that $x+B$ is our translate.

If $x$ is a closed point of $A$ and $B\subset A$, then $x+B$ would be a closed subvariety of $A$.

If $x$ is not closed, a translate through $x$ is really something lives in a base change of $A$ (so that $x$ becomes a closed point in the base change).

I'm wondering about the following question:

Mordell exceptional locus of a closed subvariety $X\subset A$ is defined as the union of images of positive-dimensional translates inside $X$. Do we need to consider the translates through non closed points? Namely, if we take the union of translates through closed points, will that give us the same thing?

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    $\begingroup$ Usually, when one speaks of "translate of an abelian subvariety" it really is in the sense "translate of an abelian subvariety by a closed point". To answer your question: Let $X\subset A$ be a closed subvariety over an abelian variety $A$ over $k$. Assume $k$ is algebraically closed (of characteristic zero?). Let $Sp(X)$ be the union of positive-dimensional translates (by closed points of $A$) of abelian subvarieties contained in $X$. Let $L/k$ be an extension of algebraically closed fields. Then $Sp(X)_L = Sp(X_L)$, where $Sp(X_L)$ is the union of ... $\endgroup$ Commented Mar 31, 2020 at 17:09
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    $\begingroup$ ...the positive-dimensional translates of abelian subvarieties by closed points of $A_L$ contained in $X_L$. Thus, the Mordell exceptional locus behaves well with respect to field extensions. You can prove this using spreading out and specialization arguments; see also Proposition 3.7 in arxiv.org/abs/1909.12187 $\endgroup$ Commented Mar 31, 2020 at 17:09
  • $\begingroup$ @AriyanJavanpeykar: That's really helpful! Thanks a lot! So I assume if we want an argument as Proposition 3.7, we would need to know $Sp(X_L)$ is closed?Do you think there is a way to do it without assuming closedness of the set (so $Sp(X)_L$ just means the inverse image of $Sp(X)$ in $X_L$)? $\endgroup$ Commented Mar 31, 2020 at 23:48
  • $\begingroup$ @AriyanJavanpeykar: Also, if you have a nontrivial map from a group variety to $X$, would it factor through $Sp(X)$? $\endgroup$ Commented Apr 1, 2020 at 3:48
  • $\begingroup$ The fact that $Sp(X)$ is closed in $X$ was proven first by Kawamata; see Thm. 4 in Y. Kawamata, On Bloch’s conjecture, Invent. Math. 57 (1980), 97-100. Ueno subsequently proved that $Sp(X)\neq X$ if and only if $X$ is of general type. Similar statements are true for closed subvarieties of semi-abelian varieties; see Abramovich numdam.org/item/CM_1994__90_1_37_0 ...cont'd below... $\endgroup$ Commented Apr 1, 2020 at 8:16

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Usually, when one speaks of "translate of an abelian subvariety" it really is in the sense "translate of an abelian subvariety by a closed point".

To answer your question: Let $X\subset A$ be a closed subvariety of an abelian variety $A$ over $k$. Assume that $k$ is algebraically closed of characteristic zero. Let $Sp(X)$ be the union of positive-dimensional translates (by closed points of $A$) of abelian subvarieties contained in $X$. Kawamata proved that $Sp(X)$ is closed in $X$; see Thm. 4 in Y. Kawamata, On Bloch’s conjecture, Invent. Math. 57 (1980), 97-100.

Side remark. Ueno proved that $Sp(X) \neq X$ if and only if $X$ is of general type. Similar statements are true for closed subvarieties of semi-abelian varieties in positive characteristic; see Abramovich numdam.org/item/CM_1994__90_1_37_0

Let $L/k$ be an extension of algebraically closed fields. Then $Sp(X)_L = Sp(X_L)$, where $Sp(X_L)$ is the union of the positive-dimensional translates of abelian subvarieties of $A_L$ contained in $X_L$. Thus, the "special" locus of $X$ behaves well with respect to field extensions. Let me explain how to prove this in a more general context.

Let $\Delta^{gr}_X$ be the groupless-exceptional locus. That is, $\Delta^{gr}_X$ is the Zariski closure of the union of the images of non-constant morphisms $U\to X$, where $U$ is a dense open subset of a connected finite type group scheme $G$ over $k$ such that $\mathrm{codim}_G(G\setminus U)\geq 2$. Then, $\Delta^{gr}_X =Sp(X)$; see Theorem 13.1 in https://arxiv.org/pdf/2002.11981.pdf . To prove this equality of sets, use the following three facts:

1) every rational map $B\dashrightarrow X$ with $B$ an abelian variety extends to a morphism $B\to X$ (use that $X$ has no rational curves).

2) if $G$ is a connected linear algebraic group, then every rational map $G\dashrightarrow X $ is constant. This is because linear algebraic groups are covered by (non-compact) rational curves.

3) The image of a morphism of abelian varieties $B\to A$ is the translate of a abelian subvariety of $A$.

In the absence of an ambient abelian variety, the general statement you are looking for is the following:

Proposition 3.7 in https://arxiv.org/abs/1909.12187

Let $L/k$ be an extension of algebraically closed fields of characteristic zero. Let X be a proper variety over $k$. Then $(\Delta_X^{gr})_L = \Delta^{gr}_{X_L}$.

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  • $\begingroup$ I'm thinking about the following example: if $X=E\times C$ where $E$ is an elliptic curve and $C$ is a curve of genus $2$. Then $Sp(X)$ is supposed to be $X$. But the union of translates by closed points seems not including the codim $2$ generic point? I'm probably missing a point here. Could you help me out? $\endgroup$ Commented Apr 1, 2020 at 15:48
  • $\begingroup$ @ggttttll Sp(X) is scheme-theoretically really the closure of the closed subset (of the variety $X(k)$) given by the union of translates of positive-dimensional abelian subvarieties. It is then clear that $Sp(X) = X$ as a scheme if $X = E \times C$. $\endgroup$ Commented Apr 1, 2020 at 16:12
  • $\begingroup$ Thanks for all the help! Take care! $\endgroup$ Commented Apr 1, 2020 at 16:54

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