Recently, in a study of Rational Points on Elliptic Curves, by Silverman and Tate, I went over a proof of Mordell's Theorem. The group of rational points on an elliptic curve is finitely generated. This was proved using the height function.

My question is: How simply can one prove that the group of complex points on an elliptic curve $E(\mathbb{C})$ is not finitely generated?

My proof starts like this:

Assume towards contradiction that the $E(\mathbb{C})$ is finitely generated. Then there exists a generating subset of $E(\mathbb{C})$. Denote this subset as $H$. Then my idea is that you can add complex points in the group and get the identity before generating the set, and this is the part that I am having trouble doing.

This is all I have so far... I don't know the best approach at deriving a contradiction.


Any assistance would be greatly appreciated.

  • $\begingroup$ You don't know the best approach or you don't know any approach? I don't see any effort in this question. $\endgroup$ – Erick Wong Feb 25 '18 at 19:23
  • $\begingroup$ @ErickWong Sorry, about that. Edited. $\endgroup$ – klorzan Feb 25 '18 at 19:37
  • 1
    $\begingroup$ Thank you, I have retracted my downvote. $\endgroup$ – Erick Wong Feb 25 '18 at 20:00
  • $\begingroup$ Finitely generated abelian groups have finite torsion part. $\endgroup$ – Ferra Feb 26 '18 at 14:08

Finitely generated abelian groups are countable.


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