Relation between fields generated by torsion points of two isogenous abelian varieties Let $K$ be a finite unramified extension of $\mathbb{Q}_p$ and assume that $A$ and $A'$ are two isogenous abelian varieties with good reduction defined over K. What can be said about the fields K(A[p]) and K(A'[p]) ?
For example, if $A$ and $A'$ are elliptic curves then possibly one can prove $A[p]$ and $A'[p]$ generate same fields over a finite unramified extension of $K$.
Further, if $A'$ is the dual abelian variety of $A$, then is there any nice relation between the fields $K(A[p])$ and $K(A'[p])$ ?
 A: All my experience and intuition come from the one-dimensional case, and even there, my ideas are a bit fuzzy. Perhaps a real expert can jump in and amend and improve what I say here.
It seems to me that everything depends on the nature of the particular isogeny $\varphi:A\to A'$. Surely, when the degree is prime to $p$, the restriction of $\varphi$ to $A[p]$ is now an isomorphism, and the two fields are the same.
Let us then concentrate on the case that $\deg\varphi$ is a power of $p$. By factoring out powers of the endomorphism $[p]_A$, we may assume (in the one-dimensional case) that $\ker\varphi$ is cyclic. The interesting case now seems to me to be when $A$ is ordinary. Let’s just look at the simplest case $\deg\varphi=p$; then the kernel may be the local subgroup of $A[p]$, which in characteristic $p$ is the kernel of Frobenius. I’m unsure of myself here, but this seems to suggest that $A'$ is the frobeniized $A$, i.e. what you get by applying Frobenius to the defining equation(s) of $A$. Surely then, again, the fields are the same?
It’s when $A$ is ordinary and the kernel of $\varphi$ is another cyclic subgroup of $\ker[p]$ that I feel uncertain. I guess the complementary isogeny $\varphi':A'\to A$ falls into the case already discussed. If so, this covers the cases when $A$ is ordinary and $\deg\varphi=p$.
The supersingular one-dimensional case is the one where I am most comfortable. For here, $A[p]$ is $F[p]$, where $F$ is the formal group. Easy considerations of ramification show you that there are no isogenies of degree $p^m$ defined over the (unramified) base and with cyclic kernel. In other words, this case does not occur.
Maybe this will help you to reach a full answer to your question.
A: $\newcommand\Z{\mathbf{Z}}$
"What can be said" and "is there a nice relation" are not actual questions.
It's certainly false that $K(A[p])$ and $K(A'[p])$ generate the same field over an unramified extension of $K$. Indeed this is false even for elliptic curves. One can certainly find $E$ and $E'$ isogenous over $\mathbf{Q}_2$ such that, as Galois modules, $E[2] = (\Z/2\Z)^2$, and $E'[2]$ is the extension of $(\Z/2\Z)$ by $(\Z/2\Z)$ which splits over $\mathbf{Q}_2(i)$. (The existence of such curves follows from Serre-Tate theory.) More generally, one can find $E/\mathbf{Q}_p$ so that $E[p] = (\Z/p\Z) \oplus \mu_p$ and $E'[p]$ is a non-split extension of $(\Z/p\Z)$ by $\mu_p$ defined by taking the $p$th root of a unit.
One can say that the semi-simplification of the Galois representations associated to $A[p]$ and $A'[p]$ will be the same. To see this, let $V$ be an irreducible representation that occurs in the semisimplification. Suppose that it occurs to multiplicity $m$. Equivalently, $A[p]$ has a filtration with simple filtered pieces where $V$ occurs $m$ times, and $A'[p]$ has a filtration where $V$ occurs $m'$ times. Then $A[p^k]$ has a filtration with simple filtered pieces where $V$ occurs $mk$ times, and $A'[p^k]$ has a filtration where it occurs $m'k$ times. However, there is a map
$$\phi: A[p^k] \rightarrow A'[p^k]$$
induced by the isogeny which has kernel and cokernel of order bounded by $k$. It follows by comparing the filtered pieces that
$$mk = m'k + O(1),$$
and so $m = m'$.
