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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?

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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