I'm interested in a specific case of the Mordell equation:
$$E: y^2=x^3+k$$
where $k=p^2$ for some prime $p$.
Most of the literature I've been able to find regarding the Mordell equation either explicitly assumes $k$ to be square-free or avoids the case altogether.
I want to show that the torsion subgroup of $E(\mathbb{Q})$ is (isomorphic to) $\mathbb{Z}/3\mathbb{Z}$.
By the Nagell-Lutz theorem, we can narrow down the possibilities to
$$y \in \{\pm{1},\pm{3},\pm{p},\pm{3p},\pm{p^2},\pm{3p^2}\}$$
By a factoring argument, I can prove that the $y=\pm{1}$ case yields no solutions (except a specific side-case for $p=3$ that yields a non-torsion point).
I can't seem to uncover an analogous argument, or really any argument, for the $y=\pm{3}$ case, which boils down to solving
$$x^3=9-p^2=(3-p)(3+p)$$
Here's what I know:
- $x$ must be an integer if it is a rational solution;
- $9-x^3$ doesn't factor over $\mathbb{Z}$ so we can't use the same argument as in the $p=\pm{1}$ case;
- $p=2$ yields no solutions, so we can assume that $p$ is odd, which leads to $x$ being even, which in turn shows that $8\vert(3-p)(3+p)$.
I've tried to extend out point #3 above but keep on running into dead ends. Are there other techniques that I'm missing here? Any pointers would be helpful.
Thanks!
Best Answer
Here is an alternative view. Let $T(x,y)$ be a torsion point on the given curve $E$ over $\Bbb Q$ with (affine) equation $$ E\ :\qquad Y^2 = X^3 + k\ ,\ k=p^2\ ,\ p>3\text{ prime .}$$
From the OP, we already know that only points $(x,y)$ with $y$ among one of the $12$ divisors in $\Bbb Z$ of $3p^2$ are possible. (Since $y^2$ divides the discriminant $-27k^2=-27p^4$.) And $y$ determines the value of $x$, since $x\to x^3$ is injective on $\Bbb Z$. The values $\pm p$ and $\pm p^2$ are excluded. So there remain only $8$ possibilities.
The $3$-torsion points are simple $(0,\pm p)$. (Easy computations inserted for the convenience of a community reader only at the end.) So the torsion order is either $3$ or $6$. In case of a $6$-torsion, we would have a point of $2$-torsion. This is easily excluded, since the polynomial $X^3-p^2$ has no rational roots. So the torsion order is $3$.
$\square$
Addendum: Computation of the $3$-torsion points.
We compute both sides of the equality $T= -2T$. The point $T$ is $(x,y)$. Let us compute $-2T=(x',y')$. The slope $m$ in $(x,y)$ comes from the formal differential $2Y\; dY=3X^3\; dX$, it is $$m=\frac{3x^2}{2y}\ .$$ The components of $-2T$ are thus given by the intersection of the line $(Y-y)=m(X-x)$ with $E$. plugging in $Y = m(X-x)+y$ into the equation of $E$ we have $(m(X-x)+y)^2=X^3+k$, and the three solutions are $x,x,x'$. Vieta for the coefficient in $X^2$ gives $x'$, so $$ \begin{aligned} x' &= m^2-x-x=\frac{9x^4}{4y^2}-2x\ ,\\ y' &= y + m(x'-x)\ . \end{aligned} $$ From $y'=y$ we get either $x'=x$, so $T=O$, or else $m=0$, so $x=0$. The corresponding points are $T_\pm =(0,\pm A)$, both torsion points. (Opposite to each other.)
Addendum: An other possibility to eliminate a torsion point of the shape $(x,\pm3)=\pm(x,3)$ is as follows. It is enough to consider only the plus sign, we start with $y=3$, and $P(x,3)$ torsion, and compute $2P=(x',y')$, which is also torsion. We expect an "ugly" expression, but $y'$ is also constrained to be one of the divisors of $3p^2$. With the above notations... $$ \begin{aligned} m &=\frac{3x^2}{2y}=\frac{x^2}2\ ,\\ x' &= m^2-2x=\frac {x^4}4-2x\ ,\\ y' &= y+m(x'-x)=3+\frac {x^2}2\left(\frac{x^4}4-3x\right)\\ &=3+\frac{(9-p^2)}2\cdot\frac{(9-p^2)-12}4 =\frac 18(p^4-6p^2-3) \end{aligned} $$ The last number is integer and a divisor of $3p^2$. It is $-3/8\ne 0$ in the field $\Bbb F_p$. So it is relatively prime to $p$, thus a divisor of $3$, and there are four cases ($\pm 1$, $\pm 3$) to be checked.