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I'm reading Luca De Feo's slides; on the slide 12 it's written that the multiplication map $[m] : E \rightarrow E$ is an isogeny. I do not understand why this isogeny is non-cyclic. At the first glance, there would be a point of order $m$, which generates all other points in the kernel.

Furthermore, I do not understand the following: if the degree of such an isogeny would be $m^2$, does it mean that such isogeny is inseparable? Let's, for instance, take the curve $E: y^2 = x^3 + x + 4$ over $\mathbb{F}_5$. There are $9$ points on this curve. Let's take the map $P \mapsto 4P$. Obviously, $E[4] = \{\mathcal{O}\}$. What is the degree of an isogeny in this case?

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  • $\begingroup$ Why do you assume a point of order $m$ generates all points in the kernel? $\endgroup$
    – Will Sawin
    Commented Oct 14 at 20:19
  • $\begingroup$ Because $mP = \mathcal{O}$, the subgroup is an additive group and the order of each point in kernel divides $m$. Am I wrong? Can you give an example please? $\endgroup$
    – samething
    Commented Oct 14 at 20:25
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    $\begingroup$ $E[4](\mathbb{F}_5) = \{O\}$ but $E[4]$ as a group scheme has $16$ points over $\mathbb{F}_{5^6}$. This is a basic question about isogenies, please read chapter III.4 in Silverman's "The arithmetic of elliptic curves". $\endgroup$
    – Chris Wuthrich
    Commented Oct 14 at 20:33
  • $\begingroup$ @ChrisWuthrich thanks a lot for the example, I'll read Silverman's chapter. $\endgroup$
    – samething
    Commented Oct 14 at 20:36
  • $\begingroup$ Your argument seems to prove the converse of what you claim: $mP = \mathcal O$, so the order of every multiple of $P$ divides $m$, and a point is in the kernel if and only if its order divides $m$. It follows that every multiple of $P$ is in the kernel but it does not follow that every element of the kernel is a multiple of $P$. If you look at slide 8 of the slides you posted you will see that the kernel has order $m^2$ as long as the characteristic $p$ does not divide $m$. So the isogeny is separable in this case. $\endgroup$
    – Will Sawin
    Commented Oct 14 at 21:23

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The multiplication-by-$m$ map is inseparable if and only if $m$ is equal to $0$ in your field. The kernel of an isogeny generally refers to the kernel over the algebraic closure of your base field. If $\bar K$ is algebraically closed and $[m]:E\to E$ is separable, then $E[m](\bar K)\cong(\mathbb Z/m\mathbb Z)^2$. In particular, it is non-cyclic. If $K$ is not algebraically closed, then the $K$-rational points in $E[m]$, i.e., the subgroup $E[m](K)$, will be a subgroup of $(\mathbb Z/m\mathbb Z)^2$, so may or may not be cyclic. In any case, the multiplication-by-$m$ map always has degree $m^2$, since the degree is defined as the degree of the associated field extension $K(E)/[m]^*K(E)$. The map $[m]$ is separable if and only if the field extension is separable.

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