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Let $E$ be an elliptic curve over $\mathbb{Q}$ and $p$ be a prime. Then what does it mean by "$E$ has a $\mathbb{Q}$-isogeny of degree $p$"?

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  • $\begingroup$ If you're in the context of arithmetic geometry, it most likely means an isogeny defined over $\mathbb{Q}$ whose kernel has $p$ elements. In some cases a $\mathbb{Q}$-isogeny is meant as an isogeny tensored with a rational number, but I think the first interpretation is more likely. $\endgroup$ – rfauffar Mar 7 '14 at 15:45
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As Robert Auffarth said in a comment, a $\mathbb{Q}$-isogeny of degree $p$ is a non-constant morphism of elliptic curves $E\to E'$ defined over $\mathbb{Q}$, that sends zero to zero, i.e., $\mathcal{O}_E\mapsto \mathcal{O}_{E'}$, and has degree $p$. Since the degree of the map equals the size of the kernel, the degree $p$ condition means that the kernel has size exactly $p$.

As it turns out (this is hard), there are only $\mathbb{Q}$-rational isogenies of elliptic curves of degree $p$ for finitely many primes, namely $2,3,5,7,11,13,17,19,37,43,67$, and $163$.

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  • $\begingroup$ Of course, in your last statement, you are excluding the isogenies given by multiplication by $[n]$ on an elliptic curve. So you should say "$\mathbf Q$-rational isogenies of non-isomorphic elliptic curves", I think. $\endgroup$ – Ariyan Javanpeykar Mar 9 '14 at 21:25
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    $\begingroup$ @Ari, the multiplication by n map has degree n^2, which is not a prime. I am only speaking about isogenies of prime degree. $\endgroup$ – Álvaro Lozano-Robledo Mar 10 '14 at 0:04
  • $\begingroup$ You are absolutely right! $\endgroup$ – Ariyan Javanpeykar Mar 10 '14 at 13:59

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