Supersingular Elliptic Curves with rational isogeny? 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.
 A: 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).
A: 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.
