Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 24F / Solution

The Riemann-Roch theorem statesOn a genus-one curve $K\sim 0$. If $D$ has degree zero, then $D+P_0$ has degree one, and Riemann--Roch gives $\ell (D+P_0)=1$. Its unique nonzero section has an effective divisor of degree one, say $P$, soIf also $D\sim Q-P_0$, then $P\sim Q$; a nonconstant function with divisor $P-Q$ would define a degree-one map to ${\mathbb{P}}^1$, forcing the genus to be zero. Hence $P=Q$.

This identifies $X$ with ${\mathrm{Pic}}^0(X)$ by $P\mapsto [P-P_0]$ and defines $P+Q=R$ throughFor the stated cubic, the line through $A=(2,1)$ and $B=(-2,-1)$ is $y=x/2$. Its third intersection is $(1/4,1/8)$, so reflection in the $x$-axis gives

Because $P_0$ is an inflection point, a line section is linearly equivalent to $3P_0$. A point $P$ is an inflection point exactly when some line has intersection divisor $3P$, equivalentlyUnder the group-law identification this is precisely $3(P-P_0)=0$, so the inflection points are exactly the three-torsion points.

Solved by gpt-5.6-sol high.