I am trying to prove that the following elliptic curve has rank=1: $$y^2=x^3+x^2+x+1$$ From the map $$\delta: E(Q)\rightarrow Q(i)^*/(Q(i)^*)^2$$ $$(x,y)\mapsto (x-i) $$ for $(x,y)\neq O$ and $O\mapsto 1$, I can show that $E(Q)/2E(Q)\cong (Z/2Z)^2$. By Mordell Weil Theorem, I have $r=rank(E)\le 1$ (as I only able to calculate 2-torsion points). And I think to finish, I need to prove that $E$ has a non-torsion point (so that $r\ge 1$) so I pick a point $P = (1,2)$, but I cant prove that $P$ has infinite order. I tried to calculate the canonical height but not succeeded. $$$$ My question is: Is there a way to prove that a point $P$ on the curve has infinite order in this case and in general? (Unfortunately, I am not allowed to use Nagell Lutz THM yet). Thank you very much for your help.
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I am just being stupid. I forgot about the iso of the torsion group T $$T/2T\simeq E(Q)[2]$$ i.e. i got r=1 for free without caring about P. – kimtuan Feb 14 '16 at 11:18
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Yes, one can detect if a point $P$ has infinite order, if one is allowed to apply the theorem of Mazur that classifies all the possible torsion group of elliptic curves defined over $\mathbb{Q}$, see Wikipedia link.
From the theorem one knows that if a point $P$ has order greater than $12$ then the order is infinite. So all you need to check is if $nP = \mathcal{O}$ for any $0 \leq n \leq 12$. Otherwise, the order of the point is infinite.
Jesper Petersen
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Thanks for your comment. Though Mazur THM seems to be an overkill for this question. – kimtuan Feb 12 '16 at 23:50