Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 125 / 2

A one-dimensional commutative formal group law over ${\mathcal{O}}_K$ is a power series $F(X,Y)\in {\mathcal{O}}_K[[X,Y]]$ satisfyingFor $r\ge 1$, the ideal ${\pi }^r{\mathcal{O}}_K$ becomes a group, denoted $F({\pi }^r{\mathcal{O}}_K)$, under $x{+}_{F}y=F(x,y)$; convergence follows because both inputs lie in the maximal ideal.

Over the characteristic-zero field $K$, there is a unique formal logarithmsatisfying ${\log }_F(F(X,Y))={\log }_F(X)+{\log }_F(Y)$. Its coefficients have bounded denominator growth, so for sufficiently large $r$ both ${\log }_F$ and its inverse formal group exponential converge on ${\pi }^r{\mathcal{O}}_K$ and preserve that ideal. They givewhere the final isomorphism is multiplication by ${\pi }^{-r}$.

The curve has good reduction at $2$ and $3$ because its discriminant $-53$ is a unit at both primes. Direct counting givesIndeed, at $p=2$ the affine solution counts for $x=0,1$ are $2,1$, and at $p=3$ the counts for $x=0,1,2$ are $2,2,2$; in each case one then adds the point at infinity.

For good reduction at $p$, the reduction of an elliptic curve gives a mapwhose kernel is its formal group of an elliptic curve, and the torsion in that kernel is $p$-primary. Hence the prime-to-$p$ part of any torsion subgroup injects into $\tilde{E}({\mathbb{F}}_p)$.

At $p=2$, the odd part of $E(\mathbb{Q}{)}_{\mathrm{tors}}$ must divide $4$, so it is trivial. The torsion group is therefore a $2$-group. At $p=3$, all of this group has order prime to $3$ and injects into a group of order $7$, so it too is trivial. Thus

Let $E_r({\mathbb{Q}}_2)$ be the $r$th term in the filtration of elliptic-curve points over a local field. Reduction givesof order four, whilehas order two. The formal logarithm is injective on $E_2({\mathbb{Q}}_2)$ and identifies it with an additive subgroup of ${\mathbb{Q}}_2$, so $E_2({\mathbb{Q}}_2)$ is torsion-free. A finite subgroup of $E({\mathbb{Q}}_2)$ therefore injects into $E({\mathbb{Q}}_2)/E_2({\mathbb{Q}}_2)$, whose order is $4\cdot 2=8$. Thus $|E({\mathbb{Q}}_2{)}_{\mathrm{tors}}|$ divides $8$.

The reduction of $P=(0,0)$ is a nonidentity point of the group $\tilde{E}({\mathbb{F}}_3)$ of prime order seven. Hence $7P$ reduces to the identity and lies in $E_1({\mathbb{Q}}_3)$. Part i shows that $P$ has infinite order, so $7P\ne O$. Every nonidentity point in $E_1({\mathbb{Q}}_3)$ has formal parameter $z=-x/y$ of positive valuation and thereforehas negative valuation. Thus $7P$ does not have integral coordinates.