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The following argument is, I believe, based on the premise that an isogeny (or a morphism of curves that is a group homomorphism) doesn't ramify:

Considering the multiplication by $n$ map $[n]$ on a elliptic curve $E/K$, where char$(K)\not|\,\,n$ so that $[n]$ is separable:

$|[n]^{−1}(Q)| ≤ deg[n] = n^2$ with equality for all but finitely many $Q ∈ E.$ But $[n]$ is a group homomorphism, so $|E[n]| = n^2$

$E[n]$ is by definition $|[n]^{−1}(\mathcal O_E)|$ so this is at least saying that $\mathcal O_E$ is not one of those points where the map ramifies (but I believe this implies that no other point ramifies). Why is this so?

Rodrigo
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1 Answers1

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In positive characteristics, isogenies may ramify. In general, see Silverman's "The Arithmetic of Elliptic Curves", Theorem 4.10 in Chapter III, where he shows that if $\phi:E_1\to E_2$ is a non-constant isogeny then

(1) For every $Q\in E_2$, we have $\# \phi^{-1}(Q)$ equals the separable degree of $\phi$.

(2) For every $P\in E_1$, the ramification index of $\phi$ at $P$ equals the inseparable degree of $\phi$.

(3) If $\phi$ is separable, then $\phi$ is unramified and $\# \ker(\phi)$ is equal to the (separable) degree of $\phi$.

The proof of (1) uses the fact that if you have a non-constant map of curves, then $\# \phi^{-1}(Q)$ equals the separable degree for all but finitely many $Q$, together with the fact that $\phi$ is a group homomorphism, to show that in fact $\# \phi^{-1}(Q)=\deg_s(\phi)$ holds for all $Q$.

Álvaro Lozano-Robledo
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