Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 89 / 4 / i

Use the polynomial graded ring $R[x_0,x_1,x_2,x_3]$, as in the PDF; the doubled brackets in the converted TeX are an OCR error. Let $I$ denote the homogeneous defining ideal. The four cubic coordinates in the proposed map never vanish simultaneously on ${\mathbb{P}}_F^1$, since $y_0^3$ and $y_1^3$ already have that property. Substitution makes each generator of $I$ zero, so they define a morphism into the generic fibre.

The two standard opens $D_{+}(x_0)$ and $D_{+}(x_3)$ cover $X$. Indeed, a homogeneous prime containing $x_0,x_3$ contains $x_2$ by $x_0{x}_3^2=x_2^3$, and then contains $x_1$ by $x_1^2={\pi }^2{x}_0{x}_2$. Such a prime contains the irrelevant ideal and gives no point of Proj.

On $D_{+}(x_0)$ put $a=x_1/x_0$, $b=x_2/x_0$, $c=x_3/x_0$. Over $F$, where $\pi$ is invertible, the equations giveThus the generic-fibre chart is exactly $\mathrm{Spec}F[t]$. The proposed map restricts to this isomorphism with $t=y_1/y_0$. On $D_{+}(x_3)$ put $s=x_2/x_3$. The equations giveso this chart is $\mathrm{Spec}F[s]$, and the map has $s=y_0/y_1$. On the overlap $s=t^{-1}$. These are the standard two charts of the projective line, proving the scheme-level isomorphismEquivalently, rescaling $x_1$ by ${\pi }^{-1}$ gives the usual twisted cubic. The chart proof also shows that the extra cubic relation introduces no additional generic-fibre scheme structure.