Let $K$ be a number field and $A/K$ an abelian variety over it. 



> 
> Can it be that $K(A[n])$ does not contain a primitive $n$-th rooth of unity? 
> 
> 
> If the answer is yes is it always possible to bound the bad $n$ uniformly in the degree of the number field and the dimension of $A$? 
> 
> 
> 


Both references and arguments are well appreciated! Thanks in advance



n-th root of unity in n-th division field of abelian variety?
Actually, this is an exercise in Serre's Lectures on Mordell--Weil Theorem:


$K(A[n])$ always contains $\mu\_n$ if $char(K)$ does not divide $n$ and $A$ is an abelian variety of positive dimension over $K$. (You don't need to assume that $K$ is a number field) 


Here is a solution. First, it suffices to check the case when $n=\ell^m$ is a power of a prime $\ell$.


Second, if $A^t$ is the dual of $A$ then let us take a $K$-polarization $\lambda: A \to A^{t}$ of smallest possible degree. Then $\lambda$ is not divisible by $\ell$, i.e., $\ker(\lambda)$ does not contain the whole $A[\ell]$. (Otherwise, dividing $\lambda$ by $\ell$ we get a $K$-polarization of lesser degree.) 


Then the image $\lambda(A[\ell^m])\subset A^t[\ell^m]$ contains a point of exact order $\ell^m$, say $Q$. Otherwise, 
$$\lambda(A[\ell^m])\subset A^t[\ell^{m-1}]$$
 and therefore $A[\ell]=\ell^{m-1}A[\ell^m]$ lies in the kernel of $\lambda$, which is not the case. 


Since $A[\ell^m]\subset A[K]$ and $\lambda$ is defined over $K$, the image $\lambda(A[\ell^m])$ lies in $A^t(K)$. In particular, $Q$ is a $K$-rational point on $A^t$.


Third, there is a nondegenerate Galois-equivariant Weil pairing
$$e\_n: A[\ell^m] \times A^t[\ell^m] \to \mu\_{\ell^m}.$$
I claim that there is a point $P \in A[\ell^m]$ such that $e\_n(P,Q)$ is a primitive $\ell^m$th root of unity. Indeed, otherwise
$$e\_n(A[\ell^m],Q) \subset \mu\_{\ell^{m-1}}$$
and therefore nonzero $\ell^{m-1}Q$ is orthogonal to the whole $A[\ell^m]$ with respect to $e\_n$, which contradicts the nondegeneracy of $e\_n$.


So,
$$\gamma:=e\_n(P,Q)$$
is a primitive $\ell^m$th root of unity that lies in $K$, because both $P$ and $Q$ are $K$-points. Since cyclic $\mu\_{\ell^m}$ is generated by $\gamma$,
$$\mu\_{\ell^m}\subset K.$$


