Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 28 / 2

Put $F=x^3+y^3+13z^3-2xyz$. The Hessian criterion for flexes of a plane cubic applies because the projective cubic is smooth. To check smoothness, a singular point with all coordinates nonzero would satisfy $3x^2=2yz$, $3y^2=2xz$, $39z^2=2xy$. Multiplication would force $351=8$, a contradiction. If one coordinate is zero, the three equations force all coordinates to vanish, which is not a projective point.

For completeness, the Hessian criterion for flexes of a plane cubic follows locally by putting a smooth point at $(0:0:1)$ and its tangent at $y=0$. After normalization the homogeneous cubic has terms $y{z}^2+a{x}^{2}z+bxyz+c{y}^{2}z$ and terms cubic in $x,y$. Its tangent has contact order three precisely when $a=0$. Its Hessian matrix at the point has determinant $-8a$, giving exactly the same condition.

For the present cubic the Hessian determinant isOn $F=0$ this is $2744xyz=8\cdot 343xyz$. Thus the plane cubic flexes are exactly the points on the cubic with $xyz=0$. Let $c=\sqrt[3]{13}$ and let $\omega$ be a primitive cube root of unity. The complete list over the algebraic closure isThese nine points are distinct, and the Hessian calculation proves there are no others. They are the flexes of a Hesse cubic. The only rational one is $O^{\prime }=(1:-1:0)$: the other two points with $z=0$ require a nonrational cube root of unity, and a rational point in either remaining triple would give a rational cube root of $-13$. By prime valuations, $13$ is not a rational cube.

For this Weierstrass equation of an elliptic curve, the inverse is $(x,y)\mapsto (x,13-y)$. At a point with $2y-13\ne 0$, implicit differentiation gives the tangent slopeIf the tangent line is $y=mx+b$, substitution into the equation shows that the three intersection abscissae sum to $m^2-1$. Two of them equal the tangency abscissa $x$, so the third is $x_3=m^2-1-2x$. Reflecting the third intersection by the inverse formula gives the elliptic curve group law from Riemann-Roch doubling formulaFor $P$, the slope is $13/(-13)=-1$, giving $x(2P)=0$ and $y(2P)=13$. For $Q$, the slope is $(12-4+13)/(6-13)=-3$, giving $x(2Q)=9-1+4=12$ and $y(2Q)=13-3+(-3)(-2-12)=52$. HenceIn particular $P$ has order exactly three: it is nonidentity and $2P=-P$.

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.

Double $2Q=(12,52)$ using the tangent formula. The slope issoThis finite point is neither $P$ nor $-P$, since its abscissa is nonzero. Therefore $P\pm 4Q$ are both finite nonidentity points.

For a direct exact-coordinate proof, the line from $P$ to $4Q$ has slope $985/104$, while the line from $P$ to $-4Q=(264/169,-3944/2197)$ has slope $-493/429$. For a secant through $P=(0,0)$ and $R=(u,v)$, its slope is $m=v/u$, and the group law gives $x(P+R)=m^2-1-u$, $y(P+R)=13-mx(P+R)$. Simplification yieldsThe first abscissa has odd numerator and denominator $64$. The second has denominator $1089=3^2\cdot 11^2$ and numerator not divisible by either $3$ or $11$. Thus neither point has integral coordinates.

Reduction also explains the obstruction without requiring the full secant arithmetic. At the good prime $2$, $\overline{Q}=(0,1)=-\overline{P}$, and $\overline{P}$ has order three, so $\overline{P+4Q}=O$. At the good prime $11$, the displayed coordinates of $4Q$ reduce to $(0,0)=\overline{P}$, so $\overline{P-4Q}=O$. A finite point with integral affine coordinates reduces to an affine point, not to the identity at infinity. Membership in the kernel of reduction of an elliptic curve therefore certifies the same two nonintegrality conclusions, at primes $2$ and $11$ respectively.