Finitely many elliptic curves isogenous to a given one (over number fields)

abelian-varietiesarithmetic-geometryelliptic-curves

Let $K$ be a number field and $E/K$ be an elliptic curve (or an abelian variety).

Is there an "easy" proof that there are only finitely many isomorphism classes of elliptic curves $E' / K$ that are isogenous to $E$ over $K$? At least when $K = \Bbb Q$, it follows from works of Mazur and Kenku, but there is probably easier arguments (available over number fields).

It should essentially amount to showing that $E(\overline K)$ has only finitely many Galois-invariant finite subgroups (e.g. finite subgroups of $E(K)$ are in finite number since $E(K)$ is finitely generated).

Best Answer

Regarding Mindlack's answer, I'll point out that Serre's proof (if I remember correctly) uses the fact that there are only finitely many $K$-isomorphism classes of elliptic curves with good reduction outside of a given finite set of primes $S$, i.e., Shafarevich's theorem for elliptic curves, which was extended to abelian varieties by Faltings. And the fact about isogenies is also an immediate consequence of this fact, since $K$-isogenous curves have the same set of primes of bad reduction. The most direct way to prove the fact about primes of good reduction outside $S$ is to first expand $S$ so that the ring of $S$-integers is a PID and so that $S$ includes all primes over 2 and 3. Then every curve $E/K$ with good reduction outside $S$ has a model $y^2=x^3+Ax+B$ with $A,B\in R_S$ and with $4A^3+27B^2\in R_S^*$. Now using the fact that $R_S^*/{R_S^*}^6$ is finite, we end up needing to show that for a given $c\in R_S^*$, the equation $u^3+v^2=c$ has only finitely many solutions $u,v\in R_S$. So now we're reduced to the fact that an elliptic curve has only finitely many $R_S$-integral points. This is the generalized version of Siegel's theorem due (I think) to Mahler. It's proof depends on something like the Thue-Siegel-Roth theorem on diophantine approximation, although one doesn't need the full strength of Roth's result. However, one does need a result that I would classify as a deep theorem.