Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2008 / B / 6

For a nonsingular curve $y^2=x^3+Ax+B$ with integers $A,B$, the Nagell–Lutz theorem states that a finite rational torsion point has integer coordinates and satisfies either $y=0$ orThe point at infinity is the identity and is treated separately. Integrality is the result allowed in the source. To prove the remaining divisibility, suppose $y\ne 0$. Then the doubled point is finite and also torsion, so its coordinates are integral. Its tangent slope is $m=(3x^2+A)/(2y)$ and its abscissa is $m^2-2x$. Thus $m^2$ is an integer. A rational number whose square is an integer is itself an integer, so $2y$ divides $3x^2+A$. In particular $y^2$ divides ${\psi }_1(x)=(3x^2+A{)}^2$. It also divides ${\psi }_2(x)=x^3+Ax+B=y^2$. Evaluating the permitted polynomial identity at the integer $x$ therefore proves the divisibility of $4A^3+27B^2$ by $y^2$.

For $y=0$, the point equals its negative and has order two. The theorem gives a finite set of torsion candidates, but the divisibility condition by itself does not assert that every candidate is torsion.

Here $4A^3+27B^2=31$. A nonzero ordinate of a torsion point must have its square dividing 31, so $y=\pm 1$. The curve equation then becomes $x(x^2+1)=0$, whose only integer solution is $x=0$. No point with $y=0$ exists over the rationals: the monic cubic $x^3+x+1$ has no integer root, since its only possible integer roots are $\pm 1$ and neither works.

It remains to test $(0,\pm 1)$. Doubling either gives abscissa $(1/(\pm 2){)}^2=1/4$, which is not an integer. If the original point were torsion its double would satisfy the integrality part of the Nagell–Lutz theorem, a contradiction. ThereforeThe rational point $(0,1)$ consequently has infinite order, so its multiples give infinitely many distinct rational points.

The curve has good reduction at five and seven: its discriminant is $-432D^2$, and the hypotheses make $D$ a unit at those primes. Over ${\mathbb{F}}_5$ the cubing map is a bijection, soThe character sum vanishes after replacing $x^3+D$ by a variable ranging over the whole field. Over ${\mathbb{F}}_7$, where $D=2$, the possible cubes are zero once, one three times, and minus one three times. The corresponding right sides are two, three and one. Two and one are nonzero squares while three is not, so the count is $1+2+0+6=9$.

Rational torsion injects into the reduction group at a good odd prime, a standard reduction theorem. Thus its order divides $\gcd (6,9)=3$. If a point of order three existed, its coordinates would be integral by the Nagell–Lutz theorem. Since $2P=-P$, the duplication formula givesThus either $x=0$ and $y^2=D$, or $x^3=-4D$ and $y^2=-3D$.

At the prime in the hypothesis, $D$ is a nonsquare unit. Since $p\equiv 1(\mathrm{mod}3)$, there is a nontrivial cube root of unity $\zeta \in {\mathbb{F}}_p$, and $(2\zeta +1{)}^2=-3$. Hence $-3$ is a nonzero square, so both $D$ and $-3D$ are nonsquares. Neither ordinate equation is possible even modulo $p$. There is no rational point of order three, proving

Write the equation as $y^2=x^3-5x^2+4x$. Its duplication formula uses slope $m=(3x^2-10x+4)/(2y)$ and $x(2P)=m^2+5-2x$. At the three listed points the slopes are respectively $i$, $-1+i\sqrt{3}$, and $2+\sqrt{3}$. Substitution givesThese are nonidentity points of order two, so each listed point has exactly order four.

For $p\ne 2,3$ the three distinct roots zero, one and four give the full two-torsion subgroup of order four over ${\mathbb{F}}_p$. At least one of $-1,3,-3$ is a square: if neither of the first two is, their product is. If $-1$ is a square, the first listed point exists over the field. If three is a square, the third exists. If $-3$ is a square, the second can be written $(1-\sqrt{-3},3+\sqrt{-3})$ and exists without requiring the two separate square roots. The duplication computations remain valid after reduction; their ordinates do not vanish unless $p=2$ or three. Thus a rational two-torsion point has a half over the field. Together with a different two-torsion point this generates a subgroup of order eight. Consequently $\boxed{8\mid \#E({\mathbb{F}}_p)}$ for every stated prime.

Over the rationals there are already four two-torsion points, including the identity. Counting at five gives eight points: the abscissas zero, one and four give one ordinate each, two and three give two each, and there is the identity. Thus rational torsion has order dividing eight.

There are no rational points of order four. For a curve $y^2={\prod }_j(x-e_j)$, translating $e_i$ to zero and using the duplication formula shows that a half of $(e_i,0)$ must satisfy $(x-e_i{)}^2=(e_i-e_j)(e_i-e_k)$. For the three roots here, the candidate abscissas are $\pm 2$, $1\pm \sqrt{-3}$, and $4\pm 2\sqrt{3}$. The last two pairs are nonrational; at $x=2,-2$ the ordinate squares are $-4,-36$, also impossible over $\mathbb{Q}$. A group of order eight containing this two-torsion subgroup would contain an element of order four, because an elliptic curve has only four points killed by two. Hence