Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 25J / Solution

A rational map $X⇢Y$ between irreducible projective varieties is an equivalence class of morphisms from nonempty open subsets of $X$ to $Y$, with representatives identified when they agree on a nonempty open subset. It is regular at $p$ if some representative is defined on a neighborhood of $p$.

Forthe coordinates vanish together exactly at $(0:1:0)$ and $(1:0:0)$. Near the first point, the paths $z=0$ and $x=0$ have limiting images $(1:0:0)$ and $(0:0:1)$; near the second, the paths $z=0$ and $y=0$ have limiting images $(1:0:0)$ and $(0:1:0)$. No continuous extension exists there. Elsewhere the coordinates do not vanish together, so the map is regular.

DefineWhere all coordinates are nonzero,and similarly $\varphi \psi$ is the identity. Thus $\varphi$ is birational and is an isomorphism on ${\mathbb{P}}^2\setminus Z(xyz)$.

For irreducibility of $P=x^2{z}^4-x^3{y}^3+z^6$, dehomogenize at $z=1$. Over $k(x)$ this is, up to a unit,The rational function on the right is not a cube, since its valuation at either root of $x^2+1$ is one. The cubic is therefore irreducible over $k(x)$; Gauss's lemma gives irreducibility in $k[x,y]$. Since $z$ does not divide $P$, homogenization preserves irreducibility.

Writing $(u:v:w)=(xy:xz:z^2)$ transforms the equation intoIts partial derivatives are $-3u^2$, $2vw$, and $v^2+3w^2$ up to common signs. Simultaneous vanishing forces $u=v=w=0$, impossible projectively, so $C$ is nonsingular. The inverse $\psi$ restricts on dense open subsets, proving that $V$ and $C$ are birational.

Solved by gpt-5.6-sol high.