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I'm reading Silverman's Arithmetic of Elliptic Curves and read that isogenies are group homomorphisms and if $\phi:E_1\longrightarrow E_2$ is a non-zero isogeny between elliptic curves,its kernel $Ker\phi=\phi^{-1}(O)$ where $O$ is the point at infinty of $E_2$.

Now suppose that $\phi\in End(E)$ be a non-zero isogeny for some elliptic curve $E$.Then,applying the first isomorphism theorem we have, $$E/Ker\phi\cong \phi(E)$$ Since every isogeny is a morphism between smooth curves and $\phi\neq O$ it is surjective,which implies,$Ker\phi=O$.That means $\phi$ is injective.But $End(E)-{O}\neq Aut(E)$,so $\phi$ need not be an isomorphism.

1.) Can someone give me an example of some isogeny and elliptic curve where I get to know how it fails to have an inverse,i.e, not be an isomorphism?

Edit 1:
2.) I have another question regarding the argument above.Silverman also defines a multiplication by m-map as follows:

Let $m\in\mathbb{Z}$.Define the map $[m]:E\longrightarrow E, \ P\rightarrow P+...+P$ m times if m>0 and $(-P)+...+(-P)$, $-m$ times if m<0 and $[0]P=O$.Then $[m]$ is a non-constant isogeny for $m\neq0$.

Now this means $[m]\in End(E)$ and by the above argument it implies $Ker[m]=E[m]={O},\forall m\in\mathbb{Z}-\{0\}$ which is not true.So what is actually going wrong in my argument.

Edit 2: Ok,so the argument $E/Ker\phi\cong E \implies Ker\phi={O}$ is not true since $E$ may be an infinite group and for infinite groups this maybe the case:Does $G\cong G/H$ imply that $H$ is trivial?

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  • $\begingroup$ Why do you think that the map being surjective implies that $\ker \phi = 0$? We have $E/\ker \phi \cong E$, but this is still very much possible for this to happen even when $\ker \phi \neq 0$. $\endgroup$ Jul 13, 2023 at 0:47
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    $\begingroup$ Yes, I have mentioned the mistake in my most recent edit(edit 2.). $\endgroup$
    – user631874
    Jul 13, 2023 at 6:30

1 Answer 1

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The Frobenius map from a plane curve to itself is injective and surjective, but its set-theoretic inverse is not algebraic because it involves a "pth root." Silverman will get to this at some point or another.

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    $\begingroup$ Thanks.Just checked that there was an exercise on the frobenius morphism of varieties in chapter 1 not being an isomorphism.I did that but didn't see it for elliptic curves explicitly yet. $\endgroup$
    – user631874
    Jul 11, 2023 at 19:51

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