Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 23I / b / Solution

The cubic equation is $x_0{x}_2^2-x_1^3+x_0^2{x}_1=0$. On $x_0\ne 0$, put $u=x_1/x_0$, $v=x_2/x_0$, giving $v^2=u^3-u$. The projection is $u$, so its function-field extension is generated by a square root of $u^3-u$. This is not a square in $k(u)$, since its zeros at $0,1,-1$ have odd order. In characteristic different from two the extension is separable of degree two.

The only apparent basepoint is $P_{\infty }=(0:0:1)$. In the chart $x_2=1$, put $s=x_0$, $t=x_1$. The equation is $s-t^3+s^{2}t=0$, so $t$ is a local parameter and $s=t^3/(1+st)$. The ratio $s/t=t^2/(1+st)$ is regular and the projection extends there as $(s/t:1)$, with value $(0:1)$. Thus the displayed rational projection does define a morphism after this unique extension; it is not literally evaluable by the raw pair $(x_0,x_1)$ at that point.

At a finite point with $v\ne 0$, the derivative with respect to $v$ is $2v\ne 0$, so $u-u(P)$ is a parameter and $e_P=1$. At $v=0$, the roots $u=0,\pm 1$ are simple (also in characteristic three), and $v$ is a parameter with $u-u(P)$ of order two. At infinity the target parameter $s/t$ likewise has order two. ThereforeThe affine gradient and the nonzero $\partial f/\partial x_0$ at infinity also verify nonsingularity for every characteristic other than two. The ramification description is on geometric points; it descends to these stated rational points over $k$.