Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 22I (Algebraic Geometry)

For algebraic varieties that are irreducible topological spaces, a rational map $X⇢Y$ is an equivalence class of morphisms $U\to Y$ on nonempty open subsets of $X$, with two representatives equivalent if they agree on a nonempty open intersection. Such subsets are dense. A birational map has a rational inverse; equivalently it restricts to an isomorphism between dense open subsets. In function field language it induces an isomorphism $k(Y)\cong k(X)$ over the ground field.

Work over the usual algebraically closed ground field of characteristic zero. Away from $x=0$, a line of slope $t=y/x$ through the singular point meets the curve with $t^2=x-1$. ThusSubstitution verifies the defining equation. On $x\ne 0$ and $t^2+1\ne 0$, the two maps are inverse morphisms, so this is a birational map to ${\mathbb{A}}^1$. The two parameters $t=\pm i$ both map to the node $(0,0)$, explaining why the globally defined parametrization is not a global isomorphism.

Let $F(y)=y_1^2{y}_2+y_2^2{y}_3+y_3^2{y}_1$. The proper transform isIndeed, in the blowup chart $y_i\ne 0$, write $x=ty$ with $y_i=1$. Homogeneity gives $F(x)=t^{3}F(y)$. Away from the exceptional divisor $t=0$, the equation is $F(y)=0$, and its closure retains that equation, removing the exceptional factor $t^3$. Every point with $t=0,F(y)=0$ is approached algebraically along the line $(ty,[y])$, so no component in this displayed set is extraneous.

For $x\ne 0$, the fibre is the single point $[y]=[x]$, and the projection is an isomorphism over $Y\setminus \{0\}$. Over $0$ the fibre is the projective plane cubic $F(y)=0$. In characteristic zero this cubic is smooth: a coordinate zero in the three derivative equations would force all coordinates zero; with all nonzero, multiplying $2y_1{y}_2=-y_3^2$, $y_1^2=-2y_2{y}_3$, $y_2^2=-2y_3{y}_1$ gives $1=-8$, impossible. Thus the exceptional fibre is a smooth genus-one plane cubic. The fibre description itself does not require this optional smoothness conclusion; characteristic $3$ requires separate treatment.