Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 29 / 2

The projective completion has identity $O=[0:1:0]$. For this Weierstrass equation of an elliptic curve, negation isFor distinct affine points $P,Q$, draw their chord; for $P=Q$, use the tangent with its intersection multiplicity. If the third intersection is $R$, define $P\oplus Q=-R$. A vertical chord pairs $P$ with $-P$ and gives their sum $O$; also $P\oplus O=P$. A vertical tangent at a point satisfying $2y+1=0$ gives $2P=O$.

For a nonvertical line $y=mx+b$, substitution into the cubic givesThus if $P=(x_1,y_1)$ and $Q=(x_2,y_2)$, the sum has coordinatesHere $m=(y_2-y_1)/(x_2-x_1)$ for a chord, and implicit differentiation givesfor a nonvertical tangent. These formulas include the repeated-root case. Associativity can be seen without a lengthy coordinate calculation: a line cuts the cubic in a divisor equivalent to $3(O)$, and $P\mapsto [(P)-(O)]$ identifies the curve with its degree-zero divisor class group. The chord-and-tangent operation becomes addition of divisor classes, hence is associative.

At $P_1=(0,0)$ the tangent slope is $-2$, so its equation is $y=-2x$. Substitution leaves $x^3=0$; this tangent meets the curve three times at $P_1$. The chord-and-tangent group law therefore givesIn particular $P_1$ has exact order three. The chord through $P_1$ and $P_2=(-2,-4)$ is $y=2x$. Substitution gives $x(x+2)(x-2)=0$, so the third intersection is $(2,4)$. Negating it gives

The primes $2,3,5$ do not divide the elliptic-curve discriminant $-91$, so all three reductions are nonsingular. Count affine solutions and add the point at infinity. At $p=2$ the right side takes values $0,1$ at $x=0,1$, while $y^2+y=0$ for both values of $y$. This gives two affine points, namely $(0,0),(0,1)$.

At $p=3$, the right-side values at $x=0,1,2$ are $0,0,2$. The corresponding numbers of solutions to $y^2+y$ are $2,2,1$, giving five affine points. At $p=5$, the right-side values at $x=0,1,2,3,4$ are $0,3,0,2,0$, with respective solution counts $2,0,2,2,2$, giving eight affine points. Therefore

Use the standard theorem on reduction of torsion points on an elliptic curve: at a prime of good reduction, reduction is injective on torsion whose order is prime to that prime. Reduction at $2$ implies that every odd-primary part of the rational torsion subgroup has order dividing $3$. Thus its only possible odd prime is $3$, and its $3$-primary part has order at most $3$. Reduction at $5$ injects the $2$-primary part into a group of order $9$, so that part is trivial. No other primary part can occur.

Consequently the torsion subgroup has order at most three. Since $P_1$ has exact order three,This argument uses only prime-to-residue-characteristic injectivity, and therefore avoids an unjustified claim of full torsion injectivity at the prime $2$.

We can use reduction instead of expanding a very large rational addition formula. Modulo $2$, both $P_1$ and $P_2$ reduce to $(0,0)$, which generates the three-element reduced group. The reduction homomorphism therefore givesThe rational point $P_2$ is not any of the three torsion points just found, so it has infinite order. In particular $R=P_1+5P_2\ne O$: otherwise multiplication by three would give $15P_2=O$.

If $R$ had integral coordinates, its affine integral representative would reduce to an affine point modulo $2$, not to the point at infinity. This contradicts its reduction to $O$. ThusIndeed, it is not even integral at $2$. This is a reduction certificate for a nonintegral elliptic point, with the crucial nonidentity condition verified by the torsion calculation.