Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 1

Let $E$ be a smooth plane cubic whose identity $O$ is an inflection point. A line through $P$ and $Q$, using the tangent when $P=Q$, has a third intersection $R$ counted with multiplicity. The chord-and-tangent group law defines $P+Q$ by drawing the line through $R$ and $O$ and taking its third intersection.

The clean verification of the group axioms uses divisors. The line at infinity meets a Weierstrass cubic in $3(O)$, so three collinear points $P,Q,R$ satisfyConsequently the mapsends the chord-and-tangent construction to addition of divisor classes. The principal divisor criterion on an elliptic curve shows that this map is bijective. Associativity and commutativity therefore follow from the abelian group law on ${\mathrm{Pic}}^0(E)$. The tangent convention handles repeated intersections, $O$ represents the zero class, and the third point on the line through $P$ and $O$ represents the inverse of $P$. Hence all group axioms hold.

Forthe negative of $(x,y)$ is $(x,-y-1)$. At $P=(0,1)$ the tangent slope isThe tangent is $y=1-4x/3$, so the addition formulas giveThusThe line through $P=(0,1)$ and $Q=(-2,1)$ is $y=1$. Its third intersection has $x=2$, and reflection under $y\mapsto -y-1$ gives

Direct enumeration of the solutions of $y^2+y=x^3-4x+2$ givesafter adjoining the point at infinity. Therefore

The discriminant is $1909=23\cdot 83$, so $3,5,7$ are primes of good reduction of an elliptic curve. The reduction of torsion points on an elliptic curve injects the prime-to-$p$ torsion into $\tilde{E}({\mathbb{F}}_p)$.

If a prime $\ell$ divided the order of $E(\mathbb{Q}{)}_{\mathrm{tors}}$, its primary subgroup would inject at every one of $3,5,7$ except possibly when $p=\ell$. For $\ell =3$, the counts at $5$ and $7$ have greatest common divisor one; for $\ell =5$ use the counts at $3$ and $7$; for $\ell =7$ use those at $3$ and $5$; every other $\ell$ would divide all three counts. Each possibility is excluded. Hence

Modulo $3$, the affine points areEach equals its own inverse because $-1-1=1$ in ${\mathbb{F}}_3$. Thuswhich is noncyclic. The reductions of $P$ and $Q$ are the distinct nonzero points $(0,1)$ and $(1,1)$, so they form a basis.

If $mP+nQ=O$, reduction modulo $3$ shows that $m$ and $n$ are even. Write $m=2m_1$ and $n=2n_1$. Then $m_{1}P+n_{1}Q$ is a rational point of order dividing two, and part (iii) makes it zero. Repeating the argument shows that $m$ and $n$ are divisible by every power of two, so $m=n=0$. Therefore