Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 24I (Algebraic Geometry)

For a divisor on an algebraic curve $D$ on a smooth projective curve $C$ of geometric genus $g$, and a canonical divisor $K_C$, the Riemann-Roch theorem states

Choose a point $Q\in E$. Since $g(E)=1$, a nonzero regular differential has an effective divisor of degree $2g-2=0$, so $K_E\sim 0$. The Riemann-Roch theorem therefore givesChooseThe rational functions $x$ and $y$ have their only poles at $Q$, of orders two and three respectively.

The seven functionsbelong to the six-dimensional Riemann-Roch space $L(6Q)$, so they satisfy a nontrivial linear relation. Comparing pole orders shows that the coefficients of both $x^3$ and $y^2$ are nonzero. After rescaling, the relation has the Weierstrass formThe divisor $3Q$ is very ample, so its complete linear system embeds $E$ in ${\mathbb{P}}^2$ and the displayed relation cuts out its image; this is the Weierstrass construction from Riemann-Roch in genus one.

Because $\mathrm{char}k\ne 2$, completing the square removes the terms linear in $y$, and because $\mathrm{char}k\ne 3$, translating $x$ removes the quadratic term. Thus the equation becomesSince $k$ is algebraically closed, the cubic factors asSmoothness of $E$ says that the elliptic-curve discriminant is nonzero, equivalently the three roots are distinct. Hence

DefineIf $D$ is a divisor class of degree zero, Riemann--Roch gives $\ell (D+Q)=1$, because $K_E\sim 0$ and the divisor $-D-Q$ has negative degree. A nonzero section therefore yieldsfor some point $P$, proving that $\alpha$ is surjective. If $\alpha (P)=\alpha (P^{\prime })$, then $P-P^{\prime }$ is a principal divisor. For $P\ne P^{\prime }$ its defining function would have one simple pole and give a degree-one map $E\to {\mathbb{P}}^1$, impossible for a genus-one curve. Thus $P=P^{\prime }$. The map and its inverse are regular, so this proves the Abel-Jacobi map of a genus-one curve is an isomorphism.

Transport the addition of the divisor class group through $\alpha$: define $P⊞R$ byThen $Q$ is the identity, and this is the chord-and-tangent group law on the plane cubic.

For the displayed equation, inversion is $(x,y)\mapsto (x,-y)$. Thus $P⊞P=Q$ exactly when $P=-P$. Besides the point at infinity $Q$, this requires $y=0$. Therefore the complete set of two-torsion points is