Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 24 / 3

For the affine polynomial $f(x,y)=y^2+y-x^3+x$, a singular point of an algebraic variety must satisfy $f=f_x=f_y=0$. Modulo $37$, the derivative equations are $2y+1=0$ and $3x^2=1$. Thus $y=18$ and $x^2=25$, giving candidates $x=5$ and $x=32$. At $(5,18)$, the two sides of the curve equation are both $9$ modulo $37$. At $(32,18)$, the right side is $28$, so that candidate is not on the curve. The point at infinity is nonsingular: for the homogeneous equation $Y^{2}Z+Y{Z}^2-X^3+X{Z}^2=0$, the derivative with respect to $Z$ is one at $(0:1:0)$. Therefore the unique singular point of an algebraic variety is

Count affine solutions and then add the point at infinity. Over the finite field ${\mathbb{F}}_2$, both values of $x$ give $x^3-x=0$, and both values of $y$ solve $y^2+y=0$. Hence there are four affine points. Over ${\mathbb{F}}_3$, all three values of $x$ give zero, and each has the two ordinates $0,2$, giving six affine points.

Over the finite field ${\mathbb{F}}_{11}$, completing the square gives $(2y+1{)}^2=1+4(x^3-x)$. The square residues are $0,1,3,4,5,9$. For $x=0,1,\dots ,10$, the right sides and the corresponding numbers of ordinates areTheir sum is sixteen. The resulting elliptic-curve point counts over a finite field are

The elliptic-curve discriminant is $37$, so the elliptic curve has good reduction at both $2$ and $3$. At a prime $\ell$ of good reduction, the reduction of torsion points on an elliptic curve is injective on torsion of order prime to $\ell$. Here is the formal reason: the kernel of reduction of an elliptic curve is described by an integral formal group law on $\ell {\mathbb{Z}}_{\ell }$, and its multiplication series is $[n](t)=nt+O(t^2)$. If $\ell \nmid n$, the first term has strictly smaller valuation than every higher term for $t\ne 0$. Thus this kernel has no nonidentity point killed by $n$.

Apply this to each prime-primary component of the rational torsion subgroup. A component at $2$ injects into ${\tilde{E}}_3({\mathbb{F}}_3)$, of order $7$, and is therefore trivial. A component at $3$ injects into ${\tilde{E}}_2({\mathbb{F}}_2)$, of order $5$, and is trivial. Every other prime-primary component injects into both groups, so its order divides $\gcd (5,7)=1$. This proves the two-prime bound on rational elliptic torsion in the present case:

At $2$ there is good reduction, and the surjectivity of good reduction over a local field gives an exact sequenceThe quotient has order $5$ and is consequently a cyclic group. For every odd integer $n$, multiplication by $n$ is an automorphism of $E_1({\mathbb{Q}}_2)$: its formal group law has multiplication series $nt+O(t^2)$ with unit linear coefficient, whose inverse formal power series has integral coefficients and converges on $2{\mathbb{Z}}_2$. This is the prime-to-residue-characteristic multiplication on a formal group.

Thus the odd-primary torsion subgroup injects into the order-five quotient. It also realizes that quotient: choose a lift $P$ of any reduced point, so $[5]P\in E_1$. There is a unique $Q\in E_1$ with $[5]Q=[5]P$. Then $P-Q$ is killed by five and has the prescribed reduction. This proves the prime-to-p torsion lifts uniquely at good reduction directly here. Therefore, for an odd prime $p$,The five-primary subgroup consists of the unique five-torsion lifts of the five reduced points. These lifts need not be rational over $\mathbb{Q}$; the preceding answer about rational torsion points of an elliptic curve does not contradict their existence over the p-adic field.

For every prime $p\ne 37$, the elliptic-curve discriminant is a unit modulo $p$, giving good reduction. Hasse's theorem on elliptic curves givesEquivalently, the elliptic-curve point count over a finite field lies between $p+1-2\sqrt{p}$ and $p+1+2\sqrt{p}$, including for $p=2$.