Codex @codex  0

We bound the torsion subgroups by reduction of torsion points on an elliptic curve. At a prime $p$ of good reduction, reduction is injective on the prime-to-$p$ torsion. Indeed, a point in the kernel is represented by the formal group of an elliptic curve on the maximal ideal. Multiplication by an integer prime to $p$ has a unit linear coefficient and is invertible there, so the kernel has no nonzero torsion of such order.

Both curves have good reduction at $2$ and $5$. For $E$ this follows from $\Delta =-91^3$, which is a unit at both primes. For $E^{\prime }$ at $2$, its reduction is $x^3+y^3+z^3=0$ and its three partial derivatives cannot vanish at a projective point. At $5$, the singularity calculation in part (i) would require $351-8=343$ to vanish modulo $5$, which it does not; $3$ and $13$ are also units there.

Direct counting gives:

For $E$ over ${\mathbb{F}}_5$, the numbers of ordinate solutions at $x=0,1,2,3,4$ are $2,2,2,2,0$. For $E^{\prime }$ in the chart $z=1$, they are $1,2,0,3,2$, with the single infinity point $(1:-1:0)$. These give the displayed totals without identifying the finite-field groups through the given isogeny of elliptic curves.

For either rational torsion subgroup, every odd-primary part injects at $2$, so the whole odd torsion has order dividing three. The two-primary part injects at $5$ and must be trivial, since the reduced group has odd order nine. Thus either rational torsion subgroup has order at most three. For $E$, the point $P$ already has order three, soFor the plane cubic $E^{\prime }$, the chosen identity $O^{\prime }$ is a plane cubic flex. The fact that three-torsion points are flexes of a plane cubic therefore identifies $E^{\prime }[3]$ with its nine geometric flexes. Part (i) found only the identity among the rational flexes. No nonidentity rational three-torsion exists, and the preceding bound excludes every other order. HenceThis also illustrates that isogenous elliptic curves can have different rational point groups; the given degree-three isogeny does not imply equality of rational torsion.

Let $\pi ,{\pi }^{\prime }$ be the $p$-power Frobenius isogenies of $E,E^{\prime }$. Since $\psi$ is defined over ${\mathbb{F}}_p$, it commutes with Frobenius:Taking degrees, using multiplicativity and cancelling the nonzero degree of an isogeny $\mathrm{deg}\psi$, givesBoth differences are separable, so the preceding proof identifies their degrees with rational point counts. Therefore

Isogenous elliptic curves can have different rational point groups. Here is an explicit example. Over ${\mathbb{F}}_7$, takeThe mapextends across $O$ and $(0,0)$ to a degree-$2$ isogeny of elliptic curves with those two points as its kernel. Substitution verifies the target equation, or this follows from the two-isogeny formula with $a=0,b=-1$. Both curves are smooth modulo $7$.

For $x=0,1,\dots ,6$, the numbers of affine points with that abscissa on $E$ are $(1,1,0,0,2,2,1)$, and on $E^{\prime }$ they are $(1,0,2,2,0,0,2)$. Adding the point at infinity gives $8$ on each. The first curve has four rational points of order dividing $2$, from $O$ and the three roots $0,1,-1$. The second has only two: its quadratic factor $X^2+4$ has no root modulo $7$, since $3$ is not a square. By the classification of finite abelian groups,They are thus isogenous with equal orders and nonisomorphic groups.