why the division field of an abelian variety contains a cyclotomic field?

Question

Given an abelian variety $A$ defined over $\mathbb{Q}$, for a positive integer (we can suppose prime) $\ell$, let $A[\ell]$ denote the group of points of $A$ that are annihilated by $\ell$, the division field $\mathbb{Q}(A[\ell])$ is obtained by adjoining to $\mathbb{Q}$ the coordinates of the points of $A[k]$.

Why the $\ell$-th cyclotomic field is contained in $\mathbb{Q}(A[\ell])$?

I saw somewhere that this is because of the existenece of the Weil pairing (for simplicity, we can suppose that $A$ has a principal polarization)

$$ e_\ell: A[\ell]\times A[\ell] \rightarrow \mu_\ell $$

From here I know that there exist points $P,Q\in A[\ell]$ such that $e_\ell(P,Q)$ is a primitive $\ell$-th root of unity, but I dont understand how this is related to the coordinates of the points in $A[\ell]$. Thanks in advance.

Answer 1

The key fact, that you can find in any standard text, is that the Weil pairing is Galois-equivariant. That is, $$\sigma(e_\ell(P,Q)) = e_\ell(\sigma(P),\sigma(Q))$$ for all $\sigma \in \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, where $\sigma(P)$ means the point given by applying the Galois action to coordinates of $P$. In particular, if we consider any $\sigma \in \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ such that $\sigma(P) = P$ for all $P \in A[\ell]$, we find that $\sigma$ fixes all elements in the image of $e_\ell$. By the result that you linked, any such $\sigma$ therefore fixes all $\ell$-th roots of unity. However, the set of such $\sigma$ is precisely the group $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}(A[\ell]))$, so the $\ell$-th roots of unity lie in the fixed field $\mathbb{Q}(A[\ell])$.