# Multiplication by $m$ isogenies of elliptic curves in characteristic $p$

I've been attempting to prove some comments I've read on MO by myself for my undergrad thesis regarding étale morphisms of elliptic curves. My definition of an étale morphism is taken from Milne's notes (page 19) which roughly says that for a field $K$ with algebraic closure $\bar{K}$, a morphism $\phi: X\rightarrow Y$ of varieties (all defined over $K$) is étale at $P\in X(K)$ if it is étale at $P$ when we consider everything as being defined over $\bar{K}$. In this case being étale means the induced linear map $d_P\phi: T_P X \rightarrow T_{\phi(P)}Y$ (as $\bar{K}$-vector spaces) on tangent spaces is an isomorphism. We say $\phi$ is étale if it is étale at every point.

Assuming I've done things correctly I've managed to prove that for an elliptic curve $E$ over a field $K$,

1. If $\operatorname{char}(K)=0$ then every multiplication-by-$m$ isogeny $[m]:E\rightarrow E$ is étale;
2. If $\operatorname{char}(K) = p>0$ then $[m]:E\rightarrow E$ is étale as long as $m$ is not a multiple of $p$.

The way I proved this (using Theorem III.4.10(c) in Silverman's Arithmetic of Elliptic Curves) broke down in characteristic $p$ when I looked at isogenies of the form $[np]$ because I could not show these isogenies had to be separable. Initially I had just supposed they had to be zero, but I don't think this is right. Indeed I've seen that an object called a "$p$-divisible group" apparently captures $p$-torsion in characteristic $p$ better than Tate $p$-modules, and this suggests that in fact these morphisms aren't always zero. So isogenies of the form $[np]$ can be finite, and I guess could potentially even be étale. I really have no intuition in characteristic $p$ though.

If anyone could provide me with an answer to whether multiplying by (a multiple of) $p$ on an elliptic curve in characteristic $p$ is étale or not - or some pointers on how to prove this - I would really appreciate it.

Edit: I've been reminded of theorem III.4.2(a) in AEC in a comment below which says that for nonzero $m$ the multiplication-by-$m$ isogeny is never zero, no matter what $m$ or the characteristic of the field are. Thus the isogenies $[np]$ are nonzero even in characteristic $p$. Whether or not they are étale, however, is still open.

Further edit: it has been pointed out in the comments that the multiplication by $p$ isogeny in characteristic $p$ can be purely inseparable (in the case that the elliptic curve is "supersingular"). Hence in some cases we can't apply theorem III.4.10c to deduce they are unramified, which is what I needed for the other parts of this proof. So I think the question boils down to whether or not these morphisms are unramified and how to show this in a different way.

• Indeed the isogeny $[m]:E\to E$ is nonzero as long as $m\neq 0$, even if $E$ is defined over a field of characteristic dividing $m$ (this is proposition III.4.2a in Silverman). $[p]:E\to E$ where $E$ lives over a field of characteristic $p$ is purely inseparable, see Silverman theorem V.3.1a. – Stahl Jan 17 '16 at 23:04
• Thanks for both of these references - I had completely forgotten about III.4.2a! But does pure inseparability have any bearing on being étale? – Alex Saad Jan 17 '16 at 23:15
• I don't believe inseparability is incompatible with \'{e}tale-ness of the morphism, I simply wanted to include that fact as well since you stated that you couldn't show that the isogenies needed to be separable in the question - and they are not (perhaps you meant some other sort of separability?). – Stahl Jan 17 '16 at 23:18
• @Stahl Right, this makes sense - and I meant the usual definition of separability. I suppose I wasn't sure whether or not they were separable and didn't know of this theorem in chapter V. Now I know they are purely inseparable it follows that any proof of these isogenies being étale has to go along different lines to my attempt. – Alex Saad Jan 17 '16 at 23:28
• The comments at the top of page 6 here might be useful to you. – Stahl Jan 18 '16 at 2:02

Lemma (38.9.9): Let $k$ be a field. Let $A$ be an abelian variety over $k$. Then $[d]:A\to A$ is étale if and only if $d$ is invertible in $k$.
Proof: Observe that $[d](x+y)=[d](x)+[d](y)$. Since translation by a point is an automorphism of $A$, we see that the set of points where $[d]:A\to A$ is étale is either empty or equal to $A$ (some details omitted). Thus it suffices to check whether $[d]$ is étale at the unit $e\in A(k)$. Since we know that $[d]$ is finite locally free (Lemma 38.9.8) to see that it is étale at $e$ is equivalent to proving that $d[d]:T_{A/k,e}\to T_{A/k,e}$ is injective. See Varieties, Lemma 32.14.8 and Morphisms, Lemma 28.36.16. By Lemma 38.6.4 we see that $d[d]$ is given by multiplication by $d$ on $T_{A/k,e}$.