I have already asked this question on stackexchange without much luck. I apologize if the question is too trivial to be asked here.


Let $A$ be an abelian variety defined over a number field $K$, $P \in A(K)$ a point of infinite order. Let $K_{\ell^n} = K(A[\ell^n])$ be the field of definition of the points in $A[\ell^n]$ and $K_{\ell^n, P} = K_{\ell^n}(Q)$ be the field obtained by extending $K_{\ell^n}$ such that the point $Q \in A(\overline{K})$ satisfying $\ell^nQ = P$ is defined over $K_{\ell^n, P}$.

Let $\nu$ be a prime of $K$ that splits completely in $K_{\ell^n}$ but not in $K_{\ell^n, P}$. Moreover assume that $A$ has good reduction at the prime $\nu$. Let $k_{\nu}$ be the residue field of $K$ at the prime $\nu$ and $A_{\nu}$ be the reduction of the abelian variety $A$ at the prime $\nu$. Then there is a reduction homomorphism
$$
A(K) \xrightarrow{sp_{\nu}} A_{\nu}(k_{\nu})
$$
I have read without proof the following statement :

*order of $sp_{\nu}(P)$ is divisible by $\ell$.*

Any reference containing a proof or complete proofs are much appreciated. Thanks!