Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 32 / 1

For the Weierstrass equation of an elliptic curve in this part, $a_1=a_2=a_6=0$, $a_3=1$, $a_4=-1$. HenceThe elliptic-curve discriminant is thereforeEvery prime other than $37$ has unit discriminant, so this integral equation gives good reduction. At $37$ the valuation is one. An Admissible change of Weierstrass coordinates changes discriminant valuations by a multiple of twelve; consequently no integral model can lower this value to zero. Thus the equation is minimal there and the reduction is genuinely bad, rather than an artifact of a nonminimal model. Explicitly $(5,18)$ is singular on the reduced affine cubic: both partial derivatives vanish there. The set of bad reduction primes is

Over ${\mathbb{F}}_2$, both $x=0,1$ give $x^3-x=0$, and both $y=0,1$ give $y^2+y=0$. There are four affine rational points, together with the point $O$ at infinity. Over ${\mathbb{F}}_3$, each $x=0,1,2$ gives $x^3-x=0$, while precisely $y=0,2$ solve $y^2+y=0$. ThusBoth reductions are nonsingular by the discriminant calculation, so these are the point counts of the reduced elliptic curves, with their group laws.

Use reduction of torsion points on an elliptic curve, which is injective on torsion of order prime to the residue characteristic. Any two-primary torsion point of an elliptic curve must inject into the group of order seven at three, so is zero. Any three-primary torsion point of an elliptic curve injects into the group of order five at two, so is zero. For a prime $q\ne 2,3$, its primary torsion injects into both reduced groups and would have order dividing both five and seven, so is also zero. Decomposing the cyclic group generated by any torsion point of an elliptic curve into its primary parts shows that $E(\mathbb{Q})$ has no nonidentity torsion at all. This is the two-prime bound on rational elliptic torsion, with the residue-characteristic exceptions handled explicitly. Since $P=(0,0)$ is not $O$,

The inverse on this Weierstrass equation of an elliptic curve is $(x,y)\mapsto (x,-y-1)$. A line $y=\lambda x+\nu$ through the points being added has third intersection $R$ with $x(R)={\lambda }^2-x_1-x_2$, and the sum is $-R$.

At $P=(0,0)$, implicit differentiation gives $\lambda =(3x^2-1)/(2y+1)=-1$. The tangent is $y=-x$. Substitution into the equation gives $x^2=x^3$, so its third intersection is $(1,-1)$, whose inverse is $(1,0)$. Next the line through $(0,0)$ and $(1,0)$ is $y=0$; its remaining intersection is $(-1,0)$, whose inverse is $(-1,-1)$. Consequently

The good-reduction map respects elliptic addition. Since the finite groups at two and three have orders five and seven, respectively, $5P$ reduces to $O$ at two and $7P$ reduces to $O$ at three. These multiples are not $O$, since $P$ has infinite order. The formal kernel of a minimal Weierstrass equation gives, for every nonidentity point in the reduction kernel at $p$,Thus both coordinates of $5P$ have negative two-adic valuation, and both coordinates of $7P$ have negative three-adic valuation. In particularThere is also an elementary way to see that both coordinates must fail integrality here. An integral affine pair cannot reduce to the point at infinity. If one coordinate had negative valuation and the other nonnegative valuation, the equation would give different valuations on its two sides. When both are negative the lowest-valuation terms give $2v_p(y)=3v_p(x)$, so both are indeed negative.

For a direct numerical confirmation, continuing the chord-and-tangent group law calculation givesThe original printed part repeats its request about $7P$ at the end; the two distinct requested multiples are $5P$ and $7P$.