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


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 $\gamma$, $$\mu_{\ell^m}\subset K.$$

  • $\begingroup$ It seems that one can also combine Joe Silverman's answer with your observation that $A^4$ has a principal polarization. $\endgroup$ – ACL Jul 1 '15 at 8:42
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    $\begingroup$ Actually, $A^4$ does not have to be principally polarized. In fact, it does not have to be isomorphic to its dual. (As an example, you may take an abelian surface $A$ over an algebraically closed field $K$ with $\End(A)=\Z$ and such that $\Hom(A,A^t)$ is generated by the polarization $\lambda: A \to A^t$ with $\ker(\lambda)$ being a product of two cyclic groups of prime order $\ell \ne char(K)$.) It is $(A \times A^t)^4$, which is always principally polarized. $\endgroup$ – Yuri Zarhin Jul 1 '15 at 12:39
  • $\begingroup$ My mistake! Sorry for the confusion and thank you for your answer and your reply to my comment. $\endgroup$ – ACL Jul 2 '15 at 9:15

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 $A/K$.

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    $\begingroup$ But one knows (“Zarhin's trick”) that $A^4$ has a principal polarization. Since $K(A[n])=K(A^4[n])$, this implies that the result holds in general! $\endgroup$ – ACL Jul 1 '15 at 8:41
  • $\begingroup$ @ACL Very nice! $\endgroup$ – Joe Silverman Jul 1 '15 at 10:58
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    $\begingroup$ @JoeSilverman: Unfortunately, I made a confusion and (see Zarhin's comment above), it is $(A\times A^{\vee })^4$ which is principally polarized. $\endgroup$ – ACL Jul 2 '15 at 9:17

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