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I think that k-isogenous elliptic curves have the same rank as I think rank is an isogeny invariant. However, I am not sure. Does anyone know where could I find a proof? Thanks!

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up vote 6 down vote accepted

An isogeny $A \to B$ is a map $A \to B$ with finite kernel. Choose a splitting of $MW(A)$ into torsion-free and torsion summands. This kernel cannot include any of the torsion-free part of $MW(A)$ and so is injective on the torsion-free part so the rank of $MW(B)$ is at least the rank of $MW(A)$. Since the isogeny goes both ways, this gives you equality.

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Thank you Will! – Patt Geffrey Mar 18 '12 at 23:48
What is the " rank part of $A$ " ? – Chandan Singh Dalawat Mar 19 '12 at 3:28
@Chandan: I think he may be implicitly choosing a splitting of Mordell-Weil into torsion-free and torsion summands. – S. Carnahan Mar 19 '12 at 4:06
I just wanted to point out that it woud be good manners to make the hypotheses explicit. When you write "k-isogenous", you should tell us what k is; when you talk about the rank, you should be over a k for which the group of k-rational points is finitely generated, etc. – Chandan Singh Dalawat Mar 19 '12 at 4:41
It works in any dimension, so yes, isogenies between abelian varieties also preserve rank. – Will Sawin Mar 19 '12 at 19:29

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