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In Silverman, The Arithmetic of Elliptic Curves, GTM 106, Corollary 4.11 says: If we have three elliptic curves $E_1$, $E_2$, $E_3$ with non-constant isogenies $f:E_1 \to E_2$ and $g:E_1 \to E_3$ and $f$ separable and $\ker f \subseteq \ker g$, then an isogeny $h:E_2 \to E_3$ with $g = h f$ exists.

Now I wonder if "$f$ separable" is really necessary and if the galois theoretic proof there is the "best" possible. Could one not simply say: $h$ is defined by its graph $\Gamma_h \subseteq E_2 \times E_3$ and $\Gamma_h$ is the scheme-theoretic image of $\theta:E_1 \to E_2 \times E_3$ with $\theta:P \mapsto (f(P), g(P))$. As $E_1$ and $E_2 \times E_3$ are reduced and everything is noetherian and $\theta$ is proper we should have that $\Gamma_h$ is simply the set-theoretic image $\theta(E_1) \subseteq E_2 \times E_3$ with its reduced-induced scheme-structure.

Is this too good to be true? Where is the catch?

Jürgen Böhm
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1 Answers1

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This formulation probably has a lot to do with Silverman’s foundations, which are closer to classical algebraic geometry than to scheme theory.

When $f$ is not separable, it’s harder to make the inclusion “$\ker{f} \subset \ker{g}$” meaningful (if $f$ is purely inseparable, then $\ker{f}$ is a point — but one with multiplicity, which is a bit more difficult to work with).

Regardless, it’s indeed pretty clear that once you work with schemes, only the inclusion $\ker{f} \subset \ker{g}$ is needed.

Indeed, in the category of fppf sheaves, $E_2$ is the quotient of $E_1$ by $\ker{f}$, so, by the universal property of the quotient, $g: E_1 \rightarrow E_3$ factors through $E_2$. But by Yoneda, morphisms between the fppf-sheaves attached to the $E_i$ are the same as morphisms of schemes between the $E_i$.

Regarding your proof, to produce $h$, you still need to show that $\Gamma_h$ is a genuine graph, i.e. that the first projection $\Gamma_h \rightarrow E_2$ is an isomorphism. How do you do that? And where do you use the fact that $\ker{f}$ contains $\ker{g}$?

Aphelli
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  • In fact this $w:\Gamma_h \to E_2$ being an isomorphism seems now a bit problematic to me - as $w = p_1 \circ i$ it is proper, surjective and because of $\ker f \subseteq \ker g$ also injective. So by Zariski's main theorem finite with one point per fiber. Is this enough in the light of possible inseparability? – Jürgen Böhm Dec 21 '25 at 10:27
  • The map $E_1 \to \Gamma_h \xrightarrow{w} E_2$ is $f$, so if $f$ is separable then $w$ is too - but the advantage of my idea is gone. – Jürgen Böhm Dec 21 '25 at 11:35