Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2010 / B / 5

We first prove all three general torsion assertions before treating the lettered cases. If a finite point has $|x{|}_p>1$, the cubic term dominates the integral lower terms, so $v_p(x)=-2m$, $v_p(y)=-3m$ for an integer $m\ge 1$. Its parameter $z=-x/y$ then lies in $p{\mathbb{Z}}_p$, with norm at most $p^{-1}$. The integral Weierstrass model defines its formal group at infinity over ${\mathbb{Z}}_p$: using $w=-1/y$ gives $w=z^3+Az{w}^2+B{w}^3$, an integral power series starting in degree three.

The supplied torsion estimate excludes a nonzero torsion parameter for $p\ge 3$, since $p^{-1}<p^{-1/(p-1)}$, and the lower bound increases for higher $p$-powers. For $p=2$ it excludes orders $2^n$ with $n\ge 2$, because their bound is strictly above $1/2$. The remaining possibility is order two, which would imply the finite point equals its negative and hence $y=0$. A nonintegral $x$ cannot solve the monic cubic with integral coefficients, so this is impossible too. Thus every finite torsion point has integral $x$, and the equation then forces integral $y$.

For good reduction the reduction map is a group homomorphism whose kernel is this formal group at infinity. Its torsion-free kernel proves injectivity on torsion, giving a subgroup of the finite reduced group. Applying local integrality to a rational torsion point at every prime gives $x,y\in \mathbb{Z}$.

If $y\ne 0$, the double of a rational torsion point is finite and integral. The doubling formulatherefore implies $y^2\mid (3x^2+a{)}^2$. Also $y^2=x^3+ax+b$. Evaluating the supplied polynomial identity at the integral $x$ now givesThis establishes the needed Nagell–Lutz theorem conclusions without assuming them in the lettered applications.

Nonsingularity with $b=a^2$ forces $a\ne 0$. TakeIt lies on the curve. Its tangent slope is $a/(2a)=1/2$, so the doubling formula gives $x(2P)=1/4$. If $P$ were torsion, $2P$ would be torsion as well, contradicting the integrality just proved. Therefore $P$ has infinite order.

For $y^2=x^3+x+2$, the point $P=(1,2)$ doubles to $(-1,0)$: its tangent slope is $(3+1)/(2\cdot 2)=1$, giving $x(2P)=-1$ and $y(2P)=0$. Thus $P$ has order four.

The curve has good reduction at $3$, since $4+27\cdot 4=112$ is nonzero modulo three. Its reduced points are the point at infinity, two points above $x=1$, and one above $x=2$, so there are four in total. Injectivity of torsion reduction bounds the rational torsion order by four. The known four points exhaust it:

Modulo five the curve is $y^2=x^3+x$, with good reduction because $4a^3+27b^2\equiv 4(\mathrm{mod}5)$. It has one point above each of $x=0,2,3$ and no others, plus infinity, so its group has order four. Rational torsion therefore has order dividing four.

A nonzero rational point of order two would have $y=0$ and integral $x$, so would give an integer root of $x^3+ax+b$. But modulo two this polynomial is $x^3+x+1$, which is one at both residues. Thus there is no rational 2-torsion. Every nontrivial finite group of order dividing four contains an element of order two, so