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.