Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2007 / B / 6

If $p=0$ as an element of $K$, the stated lower bound is zero and is immediate. Otherwise set ${\rho }_1=|p{|}^{1/(p-1)}$. The supplied formal group expansion has $g(0)=0$, because $[p](0)=0$. For $z\in \mathcal{M}$, the non-Archimedean estimates giveIf $0<|z|<{\rho }_1$, then $|z{|}^p<|p||z|$, so the two terms cannot cancel and $|[p](z)|=|p||z|>0$. The same strict inequality persists under iteration, so no such $z$ can have finite $p$-power order. Every nonzero $p$-power torsion point therefore satisfies $|z|\ge {\rho }_1$.

For a point of order $p$, $[p](z)=0$ itself forces $|p||z|\le |z{|}^p$, proving the bound for $n=1$. For $n\ge 2$, put $w=[p](z)$, of order $p^{n-1}$. Since $|z|\ge {\rho }_1$, the preceding maximum is $|z{|}^p$. By induction,This proves the formal torsion radius for prime power order. Taking the positive $p$th root gives

For a short Weierstrass equation of an elliptic curve with coefficients in the p-adic integers, a point with nonintegral $x$ has valuations $v_p(x)=-2m$ and $v_p(y)=-3m$ for some integer $m\ge 1$: the term $x^3$ strictly dominates $Ax+B$ in the equation, so $2v_p(y)=3v_p(x)$. Its formal parameter $T=-x/y$ therefore has valuation $m\ge 1$. These points, together with the identity, form the formal neighborhood $E_1({\mathbb{Q}}_p)$ of the identity, with parameter in $p{\mathbb{Z}}_p$ and the elliptic formal group law over the p-adic integers. For a point with $x\in {\mathbb{Z}}_p$, the equation gives $y\in {\mathbb{Z}}_p$ too.

The formal neighborhood has no nontrivial torsion. Multiplication by an integer coprime to $p$ has series $[m](T)=mT+O(T^2)$ and preserves $|T{|}_p$, so any torsion there must have $p$-power order. If $p\ge 3$, part (i) gives $|T{|}_p\ge p^{-1/(p-1)}>p^{-1}$, contradicting $T\in p{\mathbb{Z}}_p$.

The printed hypotheses also allow $p=2$. In that case, a nontrivial torsion point in the formal neighborhood would have order a power of two and hence yield a point of order two in that neighborhood. But every nonidentity point of order two on a short Weierstrass equation of an elliptic curve has $y=0$ and $x$ a root of the monic cubic over the p-adic integers $X^3+AX+B$. Such a root lies in the p-adic integers, so it cannot belong to the formal neighborhood, whose nonidentity points have nonintegral coordinates. This excludes the remaining case. This is the torsion-free formal subgroup for a short Weierstrass equation. Thus every affine torsion point satisfiesWith good reduction, the reduction homomorphism $E({\mathbb{Q}}_p)\to \tilde{E}({\mathbb{F}}_p)$ has kernel exactly this torsion-free formal neighborhood. Its restriction to the torsion subgroup is therefore injective, proving the required subgroup isomorphism. The short model over the p-adic integers is essential to the prime-two argument; no blanket torsion-injectivity assertion for arbitrary models is needed.

The discriminant is $-64D^3$, which is nonzero modulo three because $D\equiv 2(\mathrm{mod}3)$. Thus there is good reduction at three. For every $x\in {\mathbb{F}}_3$, the reduced right side is $x^3-x=0$, so its only affine points are $(0,0),(1,0),(2,0)$. Together with the identity they form $C_2\times C_2$.

By part (ii), the rational torsion subgroup injects into this four-element group and hence has exponent two. Over $\mathbb{Q}$, the two-torsion points have $Y=0$, requiring $X(X^2+D)=0$. Since $D>0$, the only rational root is zero. Therefore