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

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.

• The field generated by the coordinates is the one corresponding the the elements of the Galois group that stabilizes the $\ell$-torsion point-wise. So you can fraze everything in a coordinate free way. – S. carmeli Nov 20 '18 at 13:59

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 $$\mathbb{Q}(A[\ell])$$.