Computing torsion subgroup of elliptic curve

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 $$4$$?

• 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. Commented May 15 at 21:03

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 $$\Bbb Z/n\Bbb Z$$.