Let $K$ be a field and $G_K$ be its absolute Galois group. Let $E_1,E_2$ be two elliptic curves over $K$. Assume that there exists an isogeny $f:E_1\rightarrow E_2$. Let $p$ be a prime number. Then $f$ induces an isomorphism of rational Tate module $V_p(E_1)\cong V_p(E_2)$ as representations of $G_K$: let $f^{\vee}:E_2\rightarrow E_1$ be the dual isogeny, then $f\circ f^{\vee}:E_1\rightarrow E_1$ is $[\mathrm{deg}(f)]$ and on the Tate module $T_p(E_1)$ $f\circ f^{\vee}$ is just multiplicated by $\mathrm{deg}(f)$. So after tensoring $\mathbb{Q}_p$, $f$ induces an isomorphism and the isomorphism is $G_K$-equivariant.

My question is whether we can always get an isomorphism of integral Tate module $T_p(E_1)\cong T_p(E_2)$ as $G_K$-modules over $\mathbb{Z}_p$? Notice that the isomorphism doesn't need to be induced by $f$.

There is a possible way to find a counterexample. Tate's isogeny theorem tells that the isogeny classes of elliptic curves over a finite field $\mathbb{F}_q$ is determined by the rational representations. If we could construct elliptic curves over a finite field for any integral representations, then we just need to find two non-isomorphic integral models for a rational representations then we get a counterexample for the original question.

If the counterexample exists for general fields, can it be true for some special fields?

  • $\begingroup$ You mean $T_p(E_i) =\{ Q \in E_i, \exists n, p^n Q = O\} $ which is isomorphic to a subgroup of $\mathbb{Q}_p^2 / \mathbb{Z}_p^2$ and $V_p(E_i) = T_p(E_i) \otimes_{\mathbb{Z}_{p} } \mathbb{Q}_{p}$ ? Then it depends if $T_p(E_i) \cong \deg(f) T_p(E_i) $. Did you try with $K $ a finite field ? $\endgroup$ – reuns Dec 16 '18 at 20:32
  • $\begingroup$ $T_p(E_i)=\lim_n E[p^{n}]$ and $V_p(E_i)=T_p(E_i)\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$. The isomorphism of underlying $\mathbb{Z}_p$-modules is automatic. The problem is about Galois representations. We don't need to restrict to $f$ since if $E_1=E_2$ and $f=[p]$, then the isomorphic of Tate module as Galois representation is trivial and has no relationship to the isogeny. $\endgroup$ – wuzx Dec 16 '18 at 20:34
  • $\begingroup$ Can you put clearly what you think you know about $V_p(E_i)$. ? $E_1/\ker(f)$ and $E_2$ are isomorphic as groups, $\mathbb{Z}_p$ and Galois modules. Then what ? $\endgroup$ – reuns Dec 16 '18 at 20:48
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    $\begingroup$ The answer is no (i.e. there are counterexamples). Explicit examples can be found in this question. $\endgroup$ – Brandon Carter Dec 21 '18 at 3:36
  • $\begingroup$ @BrandonCarter Thank you! It is very explicit! $\endgroup$ – wuzx Dec 22 '18 at 19:51

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