Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 24F / Solution

The degree is the sum of divisor coefficients; principal divisors are $(f)$ for nonzero rational functions. $\mathrm{Cl}(C)=\mathrm{Div}(C)/\mathrm{Prin}(C)$ and ${\mathrm{Cl}}^0$ is its degree-zero subgroup. On ${\mathbb{P}}^1$, $K\sim -2P$; on a hyperelliptic genus-$g$ curve with degree-two fibre $H$, $K\sim (g-1)H$. If ${\mathrm{Cl}}^0=0$, any point divisor has $\ell (P)=2$ by Riemann-Roch, giving a degree-one map to ${\mathbb{P}}^1$; the converse follows because every degree-zero divisor on ${\mathbb{P}}^1$ is principal. For $g\ge 1$, $[p-p_0]=[q-p_0]$ would give a function with divisor $p-q$, hence a degree-one map unless $p=q$, proving injectivity. If $p+q\sim r+s$ and the pairs differ, the resulting degree-two pencil makes $C$ hyperelliptic; therefore on a nonhyperelliptic curve the unordered pairs coincide.

Solved by gpt-5.6-sol high.