Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 24F / Solution

A morphism $\varphi :{\mathbb{P}}^2\to {\mathbb{P}}^1$ is represented by homogeneous forms $F,G$ of the same degree with no common projective zero. If that degree were positive, the two plane curves $F=0$ and $G=0$ would intersect, producing a base point. Thus the degree is zero and $\varphi$ is constant.

If a closed subvariety of $({\mathbb{P}}^1{)}^3$ were isomorphic to ${\mathbb{P}}^2$, composing its embedding with each coordinate projection would give three constant maps. Their product would be constant, contradicting that it is an embedding.

The Riemann--Hurwitz theorem statesfor a degree-$d$ nonconstant map of smooth projective curves.

Project $X$ to the second ${\mathbb{P}}^1$. This is a degree-two map. Its quadratic discriminant in the $x$ variables is homogeneous of degree six in $y$, so smoothness gives six branch points counted with multiplicity. Henceand $g_X=2$. A smooth plane curve has genus $(d-1)(d-2)/2$, which is never two for an integer $d$, so $X$ is not isomorphic to a smooth plane curve.

Solved by gpt-5.6-sol high.