Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 25H / Solution

The Riemann-Hurwitz formula for a nonconstant degree-$n$ morphism
$f:C\to C^{\prime }$ of smooth projective connected curves in characteristic zero iswhere $R_f$ is the effective ramification divisor. If $g(C^{\prime })\ge 2$ and
$g(C)<g(C^{\prime })$, the right side is at least $2g(C^{\prime })-2>2g(C)-2$, impossible. If $g(C^{\prime })=1$, then $g(C)=0$ and the left side is $-2$ while the right side is nonnegative. Thus every morphism is constant.

Choose $p\in C_d$ and project from $p$ to the pencil of lines through it. Bézout's theorem says that a general such line meets $C_d$ in $p$ plus $d-1$ further points, so the resolved projectionhas degree $d-1$. Since the genus of a smooth plane curve is
$g=(d-1)(d-2)/2$, Riemann--Hurwitz givesEvery branch point receives at least one ramification point, hence

The adjunction formula givesSince $\mathrm{deg}D=d$ and $\mathrm{deg}{K}_{C_d}=d(d-3)$, linear equivalence
$D\sim K_{C_d}$ would force $d=d(d-3)$, hence $d=4$. It is therefore impossible for $d\ge 5$.

For a smooth plane quartic, projection from a point of the curve gives a degree-three map, so its gonality is at most three. Its genus is three and
$K_C\sim D$, so its plane embedding is the canonical map. A genus-three curve with a degree-two map to ${\mathbb{P}}^1$ would be hyperelliptic, and its canonical map would factor through that double cover and map onto a conic rather than embed the curve. Thus no degree-two map exists, and a degree-one map is excluded by the genus. The quartic's gonality isFor a genus-one curve, Riemann--Roch supplies a degree-two map to ${\mathbb{P}}^1$, while degree one would make it isomorphic to ${\mathbb{P}}^1$. Its gonality is therefore

Solved by gpt-5.6-sol high.