I'm studying elliptic curves and I have a question

Take two $k$-isogenous elliptic curves defined over a number field $k$ and fix a place $v$ of good reduction.

Are the reduced curves $\mathrm{mod} \:v$ isogenous?


Yes, they are. In fact, note that if $E,E'$ are $k$-isogenous, then their $l$-adic Tate modules are isomorphic as $Gal(\overline{k}/k)$-modules, where $l$ is any prime number not lying below $v$. Since $v$ is a place of good reduction, the Tate modules of the reduced curves are isomorphic as $Gal(\overline{\mathbb F_v}/\mathbb F_v)$-modules. Now a theorem of Tate (see for example Silverman, III.7.7) tells you that the map $$\hom (E_1,E_2)\otimes\mathbb Z_l\to \hom (T_l(E_1),T_l(E_2))$$ is an isomorphism when $E_1$, $E_2$ are elliptic curves over a finite field, and this proves your claim.

  • $\begingroup$ Thanks a lot! I have some questions about your answer. I was thinking the way of isogeny theorem but I had a problem to show that k-isogenous curve have isomoprhic tate module. the second isomorphism you use come from the Neron ogg shafarevich criterion that is the isomorphism pass to the quotient $Gal(\bar K/K)/I_v$ am I right? $\endgroup$ – user262440 Sep 10 '15 at 9:23
  • 1
    $\begingroup$ You can see it in this easy way: suppose your isogeny $\varphi$ has degree $n$, and take a prime $l$ not lying below $v$ and such that $l\nmid n$. Now compose $\varphi$ with its dual to get an isogeny $E\to E$ wich coincides with multiplication by $n$. This induces multiplication by $n$ on the Tate module, which is an isomorphism since $n\in \mathbb Z_l^*$. Therefore $\varphi$ must induce an isomorphism of Tate modules, which is $G_k$-equivariant because the isogeny is defined over $k$. About your second question, the answer is yes. $\endgroup$ – Ferra Sep 10 '15 at 9:36
  • $\begingroup$ Ok thanks! only one thing in wich way do I obtain that $\varphi$ induces an isomorphism between the tate modules using $[n]:T_{\ell} (E) \longrightarrow T_{\ell}(E)$? $\endgroup$ – user262440 Sep 10 '15 at 15:37
  • 1
    $\begingroup$ because both $\varphi\circ \widehat{\varphi}$ and $\widehat{\varphi}\circ\varphi$ coincide with multiplication by $n$, so they are both bijective (on the Tate modules of course), so from the first you get that $\varphi$ is injective and from the second that it is surjective. Hence it is an iso. $\endgroup$ – Ferra Sep 10 '15 at 16:37
  • $\begingroup$ Yes you are right i was thinking complicated thing! Thanks for your help! $\endgroup$ – user262440 Sep 10 '15 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.