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Let $r^3=d\ne 0$ and let $\zeta$ be a primitive cube root of unity. The nine inflection points of a plane cubic on $X^3+Y^3+d{Z}^3=0$ areThe Hessian criterion for flexes of a plane cubic gives the Hessian determinant $216dXYZ$, so every flex lies on one of the three coordinate lines. Their intersections give precisely the displayed points. With $(1:-1:0)$ as identity, these are exactly the three-torsion points are flexes of a plane cubic configurations: the tangent divisor at a flex is $3P$, giving $3(P-O)=0$.

Over $\mathbb{Q}$ only the first point with $j=0$ is rational on $Z=0$. The other two coordinate lines each give one rational point precisely when $d$ is a rational cube. Consequently the rational three-torsion subgroup is $\mathbb{Z}/3\mathbb{Z}$ when $d$ is a cube and is trivial otherwise.

The Hessian criterion for flexes of a plane cubic applied to the defining polynomial givesConsequently the inflection points of a plane cubic on this curve are exactly its intersections with the three coordinate lines. Choose $r\in \overline{\mathbb{Q}}$ with $r^3=d$, and a primitive cube root of unity $\zeta$. The nine distinct flexes of a diagonal cubic areWith the chosen flex as identity, these points are precisely $E_d[3]$. A tangent cutting the Weil divisor $3P$ gives $3(P-O)=0$; conversely $3(P-O)=0$ makes the hyperplane Weil divisor equivalent to $3P$, which supplies a tangent with triple contact. This is the three-torsion points are flexes of a plane cubic correspondence.

On $Z=0$ only $O$ is rational, since the other cube roots of unity are not rational. On each other coordinate line a rational point exists exactly when $d$ is a rational cube, and then exactly one does. ThereforeIn particular the rational torsion subgroup cannot contain two independent points of order $3$.

One can also determine the whole rational torsion of a diagonal cubic. First it is killed by $6$. To see this, use the Weierstrass equation of an elliptic curve from (iii). At primes $\ell \equiv 2(\mathrm{mod}3)$ outside the finite set containing $2,3$ and the prime divisors of the numerator and denominator of $d$, the cube map permutes ${\mathbb{F}}_{\ell }$, sowhere the quadratic character is extended by $\chi (0)=0$, and the character sum vanishes after substituting $t=x^3-432d^2$. Reduction of torsion points on an elliptic curve injects the prime-to-$\ell$ part of rational torsion into this finite group. This injection follows because multiplication by an integer prime to $\ell$ is invertible on the formal group of an elliptic curve, and hence its reduction kernel has no such torsion.

For any prime $q>3$, choose a good $\ell \equiv 2(\mathrm{mod}3)$ with $\ell \equiv 1(\mathrm{mod}q)$; the Dirichlet theorem on primes in arithmetic progressions and the Chinese remainder theorem supply such primes outside the finite bad set. Then $q\nmid \ell +1$, excluding $q$-torsion. Similarly good $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point has order dividing $6$.

The inversion formula in (i) shows that the only possible nonidentity rational point of order $2$ has $X=Y$ and $Z\ne 0$. It exists precisely when $(X/Z{)}^3=-d/2$, that is, when $d/2$ is a rational cube. There is at most one such point. This condition and $d$ being a cube cannot hold together, since $2$ is not a rational cube. Combining it with the three-torsion calculation gives the full answer:

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.

For a smooth projective curve of genus $g$, the Riemann-Roch theorem stateswhere $D$ is a divisor, $K$ a canonical divisor, and $\ell (D)=\dim H^0(V,{\mathcal{O}}_V(D))$. In particular, applying it to $D=0$ and $D=K$ gives $\ell (K)=g$ and $\mathrm{deg}K=2g-2$.

For a nonconstant morphism $f:C\to D$ of degree $d$ in characteristic zero, choose local parameters with $t\circ f=u^e$ up to a unit at a point of ramification index $e$. Pulling back a differential multiplies its local order by $e$ and adds $e-1$, because $d(u^e)=e{u}^{e-1}du$ and $e\ne 0$ in the field. ThusThe degree of a divisor of a pulled-back divisor is $d$ times its degree. Taking degrees and using the Riemann-Roch theorem gives the Riemann-Hurwitz formula

A smooth plane cubic has genus one by the genus-degree formula. In the displayed normal form, $F(0,X_1)=c{X}_1^3$ with $c\ne 0$: otherwise $X_0$ divides $F$ and the cubic is reducible. Therefore the line $X_0=0$ meets the cubic in the divisor $3P_0$, and ${\mathcal{O}}_V(1)\cong {\mathcal{O}}_V(3P_0)$. The elliptic curve group law identifies $P$ with the degree-zero divisor class $[P-P_0]$. Order three means $3P\sim 3P_0$. Consequently ${\mathcal{O}}_V(3P)\cong {\mathcal{O}}_V(1)$ and has a section with zero divisor $3P$.

Every section of ${\mathcal{O}}_V(1)$ is the restriction of a linear form: ambient linear forms form a three-dimensional space and restrict injectively, while Riemann-Roch theorem gives $h^0({\mathcal{O}}_V(1))=3$, since $\mathrm{deg}(K-3P_0)=-3$ forces the dual term to vanish. Thus this section is some nonzero $H$, proving $V\cap V(H)=\{P\}$, with intersection multiplicity three. This is the three-torsion points are flexes of a plane cubic property.

Projection forgetting $X_3$ defines a morphism $W\to V$: its only possible indeterminacy point $(0:0:0:1)$ does not satisfy the second equation. It is a double cover. Away from the zeros of $H_1{H}_2$, its two square roots are distinct. Each line intersects $V$ in three simple points, since it is nowhere tangent; the two triples are disjoint because the lines do not meet on $V$. In a local trivialization near any of these six points the covering equation is $s^2=ua(u)$ with $a(0)\ne 0$, so the ramification index is two. There is no other ramification. The Riemann-Hurwitz formula therefore gives