Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For this curve $y^2=x^3+b^2x^2-a^2b^2x$ where $a \neq b$ and $a,b$ are rational. I can prove that if $b^2+4a^2$ is square then torsion group of curve is $\mathbb Z2 \times \mathbb Z2$, and when $b^2+4a^2$ is not square I want prove that torsion group is $\mathbb Z 2$. So for second part I can prove there is not point of order 3,4 but really I don't know how I can prove there is not point of order 5 . How I can prove that there is not point of order 5 when $b^2+4a^2$ is not square?

share|improve this question
add comment

1 Answer

First of all, your question depends on only one and not on two parameters: Upon setting $b=ta$ for a new variable $t$, and replacing $x$ and $y$ with $a^2x$ and $a^3y$, your elliptic curve assumes the simpler form $y^2=x^3+z(x^2-x)$ with $z=t^2$. The $j$-invariant of this elliptic curve is $\frac{256(z+3)^3}{z+4}$. If the curve has a point of order $5$, then it has an isogeny of degree $5$. By an old result (for instance in Fricke's 1922 book on elliptic functions), the $j$-invariant of an elliptic curve with an isogeny of degree $5$ has the form $\frac{(s^2+10s+5)^3}{s}$. So one first has to solve \begin{equation} \frac{256(z+3)^3}{z+4}=\frac{(s^2+10s+5)^3}{s}, \end{equation} and then to check when $z=t^2$ for a rational $t$. This equation simplifies upon setting $(z+3)w=s^2+10s+5$ for a new variable $w$, and eliminating $z$. The resulting curve in $s$ and $w$ has genus $0$, and Maple's function parametrization (do with(algcurves) before that) gives a rational parametrization, and hence $z=R(u)$ for a rational function $R$ and a variable $u$. The equation $t^2=R(u)$ is easily transformed to an elliptic curve. One can compute (with Sage) that this elliptic curve has torsion order $2$ and rank $0$.

So there are at most finitely many candidates for $t$ to check. It seems (but I didn't check very carefully) that no $t$ gives an isogeny of degree $5$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.