# Are there counterexamples of isogeny elliptic curves with non-isomorphic integral Tate modules?

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?

• 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 ? – reuns Dec 16 '18 at 20:32
• $T_p(E_i)=\lim_n E[p^{n}]$ 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. – wuzx Dec 16 '18 at 20:34
• 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 ? – reuns Dec 16 '18 at 20:48
• The answer is no (i.e. there are counterexamples). Explicit examples can be found in this question. – Brandon Carter Dec 21 '18 at 3:36
• @BrandonCarter Thank you! It is very explicit! – wuzx Dec 22 '18 at 19:51