**Setup.** Let $K$ be an algebraically closed field of characteristic zero, and let $A/K$ be a simple abelian variety of dimension $n$. Let $\{ x\_1,x\_2,\dots,x\_{m^{2n}}\}$ denote the $m$-torsion points of $A$.


**Question.** Does there exist an irreducible divisor $D$ on $A$ such that the collection of divisors $\{ D + x\_i\}\_{1\leq i \leq m^{2n}}$ are in general position i.e., for any $I \subset \{ 1,\dots,m^{2n}\}$, $\dim(\bigcap\_{i\in I}\text{Supp}(D + x\_i)) = n - \#I$ if $\#I \leq n$ and $\bigcap\_{i\in I} \text{Supp}(D + x\_i) = \emptyset$ if $\#I > n$?


**Motivation.** My motivation from this comes from a [paper of Corvaja--Zannier](https://arxiv.org/pdf/math/0206100.pdf) concerning integral points on surfaces. They work over a number field and in this setting, they show in Example 1.4 that for such a divisor $D$ on an abelian surface with $m = 4$, the complement $A\setminus[4](D)$ has finitely many integral points where $[4]\colon A\to A$ is the multplication by $4$ map. I was wondering if any such divisor actually existed as they do not give a construction for such a divisor.


**Remarks.** In the setting of an abelian surface, the first condition that any two have zero dimensional intersection is not restrictive, however I do not have any good ideas of how to study this latter condition that any three do not intersection. In [this](https://mathoverflow.net/questions/151965/intersection-of-translates-of-theta-divisor) MO post, it is shown that $D$ cannot be a theta divisor in the setting where $A$ is an abelian surface and $m = 2$.


Any answers, references, or suggestions on how to approach such a problem are greatly appreciated! Also, if it makes things easier, I am happy to assume that $A$ is an abelian surface.



Intersection of translate of divisors on abelian variety
I think one can perhaps make a dimension count argument to show that a generic divisor and its torsion translates, as in your question, are in general position. I'll assume $A$ is an abelian surface.


Suppose $L$ is a very ample line bundle on $A$ and let $\mathbb{P}=\mathbb{P}(H^0(A,L))\cong \mathbb{P}^N$ be its complete linear system of divisors, with $N>>0$. Let $\iota: A\rightarrow \mathbb{P}^N$ denote the embedding induced by $L$. Suppose there are non-zero, distinct torsion points $P,Q$ such that, for any $D\in \mathbb{P}$, we have $D\cap (D-P)\cap (D-Q)\neq \emptyset$.


Let $\pi: C\rightarrow \mathbb{P}$ be the universal divisor, i.e. the fiber of $C$ above $D\in \mathbb{P}$ is given by the divisor $D\subset A$ itself. Define $S\subset C$ to be
$$
S=\{x\in C|x+P, x+Q\in \pi(x)\}.
$$
Here, for a point $x\in C$, $\pi(x)\in \mathbb{P}$ is the divisor that $x$ lives on.


By our assumption $S$ surjects onto $\mathbb{P}$, and I believe that over an open dense subset of $\mathbb{P}$ the map is finite, so that we have $\dim S=N$. Now we have a natural map $S\rightarrow A$, which we may as well assume hits the identity $e\in A$, and let $S\_e$ be the fiber above $e$. Then $\dim S\_e\geq N-2$, and the same for the image $\pi(S\_e)$. But then each divisor $D\in \pi(S\_e)\subset \mathbb{P}$ is given by intersecting $\iota(A)$ with a hyperplane containing the three distinct points $\iota(e), \iota(P), \iota(Q)$, and the dimension of such hyperplanes is $N-3$, which is a contradiction.


