Compute the torsion subgroup of the elliptic curve $y^2=x^3+5x^2+3x+7$.

I am only used to computing torsion groups when our equation is in 'short Weirstrass form'; i.e. $y^2=x^3+Ax+B$ for integer $A,B$. In that case, we can reduce over primes $p$ of good reduction, and use the fact that this $\mathcal{E}(\mathbb{Q})$ injects into $\mathcal{E}(\mathbb{F}_p)$. Here is the solution given by my instructor:

Reduce mod $3$ to get $y^2=x^3+2x^2+7$. The RHS has no solutions over $\mathbb{F}_3$ so must be squarefree, and this reduction is therefore smooth. One can check to see there are $5$ points over $\mathbb{F}_3$, so the torsion group has order dividing $5$. Therefore we only need to check that there is a point in the torsion group with order exactly $5$ instead of infinite order. $(1,4)$ is obviously a point on the curve, and by point duplication we can how it has order $5$.

This solution confuses me for the following reasons:

  • Why didn't we just write $y^2=x^3+2x^2+1$ modulo $3$? This gives a different answer?

  • What exactly is meant by 'smooth' reduction? Is this just a synonym for good reduction? If so how have they deduced this from the fact there are no solutions on the RHS?

  • In general, is the strategy behind computing the torsion subgroup (save for using Nagell-Lutz) to find the primes of good reduction, compute the sizes of the groups manually for small enough primes $p$, and then try to spot points on the curve; then, showing they have certain order? i.e. if we know the torsion group has order dividing, say, $4$, then we ought to find a point of order $4$ to show the torsion group has order exactly $4$?

  • $\begingroup$ For your third question, yes. For example, Magma's implementation over number fields first finds a bound on the order of the torsion group by reducing modulo a bunch of primes, then uses the division polynomial in a clever way to check the finitely many cases left. Note that using Schoof's algorithm (and extensions) point counting is extremely efficient over finite fields. $\endgroup$ Commented May 15 at 21:03

1 Answer 1

  1. $x^3+2x^2+1$ is a perfectly good way to write this and everything still works.
  2. Smooth reduction is a synonym for good reduction (good reduction means the reduction is smooth). By the Jacobian criteria, one can check that $[0:1:0]$ is nonsingular on the projective closure of $V(y^2=x^3+ax^2+bx+c)$, so the singular points are $(x_0,y_0)$ satisfying $2y_0=0$ and $3x_0^2+2ax_0+b=0$. But $x_0$ is a root of $x^3+ax^2+bx+c$ and its derivative iff $x^3+ax^2+bx+c$ has a multiple root, which for cubics means it must factor as $(x-\alpha)^2(x-\beta)$ for $\alpha,\beta\in\Bbb F_3$, or $x^3+ax^2+bx+c$ must have a root in $\Bbb F_3$. It doesn't, so the cubic is smooth.
  3. I'm not an expert on this and I don't do research with it, but frequently this is a worthwhile technique for basic problems: if you can find a small enough prime of good reduction, you get a very effective bound on the size of the torsion group since there simply cannot be that many points. The "i.e." question is just first-year abstract algebra - if you have an abelian group with order dividing $n$ and you find an element of order $n$, your group must actually be $\Bbb Z/n\Bbb Z$.

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