n-th root of unity in n-th division field of abelian variety? 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
 A: Let $\hat A/K$ be the dual of $A/K$. Then $K(A[n],\hat A[n])$ contains a primitive $n$'th root of unity, since it contains the image of the Weil pairing $e_n:A[n]\times\hat A[n]\to\mu_n$, which is non-degenerate. So if $A/K$ has a principal polarization (i.e., defined over $K$), then $A\cong_K\hat A$, and hence $K(A[n],\hat A[n])=K(A[n])$ contains a primitive $n$'th root of unity. In general, there should be some condition in terms of the smallest degree of a polarization of $A/K$.
A: 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 $\gamma$,
$$\mu_{\ell^m}\subset K.$$
