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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!

flag	asked Mar 18 2012 at 23:31 Patt Geffrey 131●2

Will Sawin · Accepted Answer · 2012-03-19 05:11:28Z UTC

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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.

edited Mar 19 2012 at 5:11

answered Mar 18 2012 at 23:45

Will Sawin
19.3k●1●24●48

Thank you Will! – Patt Geffrey Mar 18 2012 at 23:48

What is the " rank part of $A$ " ? – Chandan Singh Dalawat Mar 19 2012 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 2012 at 4:06

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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 2012 at 4:41

@Chandan and S.Carnahan - yes, that is correct. Sorry for the lack of clarity. – Will Sawin Mar 19 2012 at 5:10

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Rank of isogenous elliptic curves

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