# Division fields of abelian varieties over function fields

Let $k$ be a finitely generated field (for example a finite field or a number field) and $K/k$ a finitely generated regular extension with $trdeg(K/k)=1$. Let $A/K$ be a principally polarized abelian variety. For every prime number $l\neq char(K)$ let $K_l:=K(A[l])$ be the field obtained by adjoining the coordinates of the $l$-torsion points of $A$ to $K$. For such a function field extension $K_l/K$ it is natural to consider the algebraic closure $k_l$ of $k$ in $K_l$, i.e. the field of elements of $K_l$ which are algebraic over $k$. In brief: I am wondering what we can say about $k_l$.

Of course, the existence of the Weil pairing afforded by the principal polarization forces $k(\mu_l)\subset k_l$ for every prime $l\neq char(K)$. This inclusion needs not be an equality. (Consider the case where $A$ has abelian subvarieties defined over $k$.)

Let us call an abelian variety $B/K$ weakly isotrivial, if there is a non-zero abelian variety $B_0/\overline{k}$ and a $\overline{K}$-homomorphism $B_{0, \overline{K}}\to A_{\overline{K}}$ with finite kernel.

Question: Suppose that $A$ is not weakly isotrivial. Is it true that there is a constant $l_0(A/K)$ such that $k_l=k(\mu_l)$ for every prime number $l\ge l_0(A/K)$? (If no, is this true after replacing $K$ by a finite extension and $k$ by its algebraic closure in this extension''?)

Remarks:

a) For example if $char(K)=0$, $End(A)=\mathbb{Z}$ and $\dim(A)=2, 6$ or odd, then the answer to this question is yes''. The proof we have for that case is somewhat special, however, because then we have exceptionally good control over the associated monodromy groups. (We use that $Gal(K_l/K)=GSp(2\dim(A), \mathbb{F}_l)$ for almost all primes $l$, some group theory and the Mordell-Lang theorem.)

b) The special case $\dim(A)=1$ is contained in the book on modular forms by S. Lang.

c) It is clear in the situation of the question that there is a constant $l(A/K)$ such that $[K_l:K]>[k_l:k]$ for all primes $l\ge l(A/K)$. Otherwise we would have $K_l=k_l K$ and hence $K_l\subset \overline{k} K$ for infinitely many primes $l$. This would imply $|A(\overline{k}K)[l]|=l^{2\dim(A)}$ for infinitely many primes $l$, which is not the case by the Mordell-Lang theorem.