Let $E/K$ be an elliptic curve over a number field, and $\mathfrak{p}$ a prime of good supersingular reduction. Let $p$ be the prime below $\mathfrak{p}$. I believe that the following is true, but I can't prove it, hence my asking here:

$E$ does not possess a $K$-rational $p$-isogeny.

I think this is true because, roughly, supersingular primes are ``rare'', and so too are rational isogenies, so asking for both at once is probably asking for too much. Googling has not helped me.

  • $\begingroup$ This is indeed true over ${\mathbb Q}$, by a theorem of Serre - "Propriétés galoisiennes des points d'ordre fini des courbes elliptiques"), Prop. 12. $\endgroup$ Dec 20, 2012 at 17:30

2 Answers 2


You can't prove it because it is untrue.

Let $E$ be an elliptic curve with CM by $\mathbf{Z}[\sqrt{-p}]$ defined over a number field $K$ which

  • Contains $\mathbf{Q}(\sqrt{-p})$ so that the action of $\mathbf{Z}[\sqrt{-p}]$ is $K$-rational and
  • Over which $E$ has good reduction (in fact, since $E$ has CM, there's a number field over which it has everywhere good reduction)

By Deuring's Criterion, $E$ has supersingular reduction above $p$, and since $K$ contains the CM field, $[\sqrt{-p}]$ is a $K$-rational $p$-isogeny.

In general, you need much finer information about how rare two quantities are in order to make the "both of these are rare, so they should be impossible together!" argument. The more likely occurrence is that if Condition A happens with probability $1/a$ and Condition B happens with probability $1/b$ then they both occur with probability $1/ab$ (and similarly in the probability zero cases).

  • $\begingroup$ Thank you stankewicz! However, my main field of concern is $K = \mathbb{Q}$, and here there is surely a hope that the answer is yes? $\endgroup$
    – Giuseppe
    Dec 20, 2012 at 14:33
  • 1
    $\begingroup$ My previous comment was trivially wrong. Nonetheless, I still doubt that it's true over $\mathbf{Q}$. The basic reason is that the modular curves $X_0(p)$ for $p=2,3,5$ are all isomorphic to $\mathbb{P}^1$ and so have tons of rational points. It seems like there should be a counterexample somewhere in there. $\endgroup$
    – stankewicz
    Dec 20, 2012 at 15:13
  • $\begingroup$ Slick proof, kreck! $\endgroup$
    – stankewicz
    Dec 20, 2012 at 16:26
  • 4
    $\begingroup$ @Giuseppe: You're OK when $e(\mathfrak{p}|p)=1$ (e.g., $K=\mathbf{Q}$, any $p$). Such an isogeny extends to a map between the Neron models, which are abelian schemes over $R=O_{K,\mathfrak{p}}$. This extended map is finite flat (as for abelian schemes over a dvr in general), so its kernel is a finite flat group scheme $G$ over $R$ with order $p$ (by the degree hypothesis) and geometric special fiber $\alpha_p$ (by the ss hypothesis). By Oort-Tate ($e=1$!), no order-$p$ finite flat group scheme over the strict henselization of $R$ has special fiber $\alpha_p$. QED $\endgroup$
    – user29720
    Dec 20, 2012 at 16:52
  • 3
    $\begingroup$ [I had originally posted the above argument mistakenly writing $e \le p-1$, but it only works when $e = 1$, as one sees by taking reduction of the examples in the Oort-Tate list, and cannot be improved: stankewicz gives counterexamples with $e=2$. The comments from stankewicz and Giuseppe appeared while I rewrote my argument to correct the typo.] @Giuseppe: your 2nd comment indicates that you should learn about finite flat group schemes (in the course of which you'll see $\alpha_p$). These are pervasive in most interesting things since the 1970's with the arithmetic of abelian varieties. $\endgroup$
    – user29720
    Dec 20, 2012 at 16:58

Here's another proof in the case $K=\mathbb{Q}$. If $E$ has good supersingular reduction at $p$ then it's well-known that locally at $p$ the Galois representation on the $p$-torsion is irreducible, and indeed induced from (either of) the "niveau 2" character(s) of the abs Galois group of the unram quad extension of $\mathbb{Q}$. Hence even over $\mathbb{Q}_p$ the curve admits no isogenies of degree $p$ (as the kernel would then be a Galois-stable sub).

Although I'm cheating really, because one proof of the "well-known" bit would be via kreck's argument.


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