Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 125 / 3

Apply the Riemann-Roch theorem to the divisors $nO$. Since the genus is one and the canonical divisor is trivial, $\ell (nO)=n$ for $n\ge 1$. ChooseThen $x$ and $y$ have exact pole orders two and three at $O$. The seven functionslie in the six-dimensional space $L(6O)$, so they satisfy one relation. Comparing pole orders and completing squares and cubes gives a nonsingular Weierstrass equation of an elliptic curveThe functions $1,x,y$ define the morphism $P\mapsto (x(P):y(P):1)$ away from $O$. Their pole orders show that it extends with $O\mapsto (0:1:0)$. It has degree one and is therefore an isomorphism of smooth projective curves.

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The Hessian determinant of $u^3+v^3+w^3$ is a nonzero scalar multiple of $uvw$. Hence the inflection points areThere are nine of them. Since $O_C=(1:-1:0)$ is an inflection point, these are exactly $C[3]$. They are all defined overso $C(K)[3]\cong (\mathbb{Z}/3\mathbb{Z}{)}^2$.

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The Weil pairing $e_3:C[3]\times C[3]\to {\mu }_3$ is nondegenerate and Galois equivariant. If all of $C[3]$ is rational over a field, pairing a basis produces a primitive cube root of unity in that field. Thus a quadratic field with full 3-torsion must contain $\mathbb{Q}({\zeta }_3)$ and must equal it.

More generally, if $E({\mathbb{F}}_p)[\ell ]\cong (\mathbb{Z}/\ell \mathbb{Z}{)}^2$, Frobenius acts as the identity on $E[\ell ]$. Its characteristic polynomial $X^2-aX+p$ is therefore congruent to $(X-1{)}^2$ modulo $\ell$. Comparing coefficients givesIn particular $\#E({\mathbb{F}}_p)=1+p-a$ with $a\equiv 2(\mathrm{mod}\ell )$.

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Writing $X=u$, $Y=v$, and $Z=-w$ turns the cubic into $X^3+Y^3=Z^3$. The birational coordinatesgive the Weierstrass equationThe original projective cubic has no common zero of its three partial derivatives modulo any $p\ne 3$, so it is already a smooth proper model and $C$ has good reduction at every such prime.

If $p\equiv 1(\mathrm{mod}3)$, then ${\mu }_3\subset {\mathbb{F}}_p$, all nine inflection points are rational, and $C({\mathbb{F}}_p)$ contains $(\mathbb{Z}/3\mathbb{Z}{)}^2$, so it is not cyclic. If $p\equiv 2(\mathrm{mod}3)$, cubing is a bijection on ${\mathbb{F}}_p$. Counting on the Fermat model gives $\#C({\mathbb{F}}_p)=p+1$. A finite elliptic-curve group has the form $\mathbb{Z}/d\mathbb{Z}\times \mathbb{Z}/e\mathbb{Z}$ with $d\mid e$ and $d\mid p-1$, so $d\mid \gcd (p-1,p+1)=2$. For odd $p$, the equation $x^3=432$ has exactly one root because cubing is bijective, so there is only one nonzero rational 2-torsion point and $d\ne 2$. Hence the group is cyclic. For $p=2$ it has order three and is cyclic as well.

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