Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 19 / 3

Let $L/{\mathbb{Q}}_p$ be finite, with ring of integers ${\mathcal{O}}_L$ and maximal ideal ${\mathfrak{m}}_L$. The formal group of an elliptic curve on an integral Weierstrass model is a formal group law $F(X,Y)\in {\mathcal{O}}_L[[X,Y]]$, and $\hat{E}(L)$ denotes its points with parameter in ${\mathfrak{m}}_L$. Its multiplication series isSince $p\nmid m$, the linear coefficient $m$ is a unit. Construct a compositional inverse $g(T)=m^{-1}T+{\sum }_{j\ge 2}{b}_j{T}^j$ recursively: at degree $j$, the coefficient of $b_j$ in $[m{]}_F(g(T))$ is $m$, so the equation making that degree vanish uniquely determines $b_j\in {\mathcal{O}}_L$. The inverse on the other side is the same series, by uniqueness of formal composition inverses.

Both series converge on ${\mathfrak{m}}_L$, since their coefficients are integral and powers of an element of ${\mathfrak{m}}_L$ tend to zero. Their formal identities therefore hold as identities of convergent values. Thus for every $P$ there is a unique $Q$ with $[m]Q=P$, andThis proves unique divisibility for every finite extension, including ramified extensions and negative $m$. It uses prime-to-residue-characteristic multiplication on a formal group, not an unjustified logarithm isomorphism on the entire maximal ideal.

Now assume good reduction of an elliptic curve. For a finite local extension $M$, the reduction sequence isProperness defines reduction for all points, smoothness and the Hensel lemma make it surjective, and its kernel is precisely the formal group. Let $K$ be the maximal unramified extension and take $P\in E(K)$. Its coordinates lie in a finite unramified extension $L/{\mathbb{Q}}_p$. Over the algebraic closure of the residue field, the isogeny $[m]$ is surjective, so choose $\overline{Q}$ with $[m]\overline{Q}=\tilde{P}$. This residue point is defined over a finite residue extension. Choose the corresponding finite unramified extension $M/L$ inside $K$, and lift it to $Q_0\in E(M)$ by smooth reduction.

Then $P-[m]Q_0$ lies in the reduction kernel. The first argument gives a unique kernel point $R$ with $[m]R=P-[m]Q_0$. Hence $Q=Q_0+R\in E(M)\subseteq E(K)$ satisfies $[m]Q=P$. ConsequentlyEvery lifting and convergence argument was made in a finite complete extension; $K$ itself need not be complete. Unique divisibility is asserted only for the formal group. For $|m|>1$, good reduction also lifts the prime-to-$p$ torsion, so $E(K)$ has nonzero $m$-torsion and division there is not unique. The construction is unramified division points at good reduction.

We prove that the only finite-order rational point is the origin: $\boxed{E(\mathbb{Q}{)}_{\mathrm{tors}}=\{O\}}$. The identity always has order one, so “no elements of finite order” must be understood as no nonidentity torsion points.

The given discriminant is a unit at $5$ and $7$, so the integral Weierstrass equation of an elliptic curve has good reduction at both primes. For an odd prime $p$, completing the square in $y^2+xy=f(x)$ shows that the number of $y$ above $x$ is determined byA nonzero quadratic residue gives two points, zero gives one, and a nonresidue gives none. Direct evaluation givesIncluding the origin gives $\#\tilde{E}({\mathbb{F}}_5)=10$ and $\#\tilde{E}({\mathbb{F}}_7)=13$.

At a good prime $p$, reduction is injective on torsion of order prime to $p$: a point in its kernel belongs to the formal group, where multiplication by that order is bijective by part(i), so a point killed by it must be zero. Suppose a nonidentity rational point has finite order. Taking a suitable multiple gives a point of prime order $\ell$. If $\ell \notin \{5,7\}$, reduction at both primes forces $\ell$ to divide both $10$ and $13$, impossible. If $\ell =5$, reduction at $7$ forces $5\mid 13$, impossible. If $\ell =7$, reduction at $5$ forces $7\mid 10$, also impossible. This exhausts all primes and proves the claim. The torsion exclusion from good reduction counts explicitly handles the two residual characteristics instead of incorrectly assuming prime-to-$p$ injectivity also controls $p$-torsion.