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I have been having trouble with this question.

Let $m\in \mathbb{Z}$ with $m > 0$ and define $E_m : y^2 = x^3 −x+m^2$ Then $E_m$ is an elliptic curve Determine the group sturcture of $E_m(\mathbb{Z}_5)$ and hence show that when $m\nmid 5$, $P=(0,m) \not\in 2E_m(\mathbb{Q})$

I have no trouble with the first part. If $m\equiv0$ $(mod$ $5)$ then $E_m(\mathbb{Z}_5) = \{\infty,(0,0),(1,0),(2,1),(2,4),(3,3),(3,2),(4,0\}$ which has three points of order 2 (second coordinate zero) so this group is isomorphic to $\mathbb{Z}_4\oplus\mathbb{Z}_2$. Similarly if $m\not\equiv0$ $(mod$ $5)$ then $E_m(\mathbb{Z}_5) \cong \mathbb{Z}_8$

But I have no idea on how this should be applied to the next part.

Any hints?

Shevak
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1 Answers1

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There is a group morphism $\rho : E_m(\Bbb Q) \to E_m(\Bbb Z_5)$ obtained by reduction modulo $5$.

If $\rho(P) \notin 2E_m(\Bbb Z_5)$, then you can deduce that $P \notin 2E_m(\Bbb Q)$. You have to check this for all $4$ possible choices of $m$ mod $5$.

mercio
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