Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 20 / 5 / c

The hyperplane sections through $P$ are spanned by the three linear forms $x_1,x_2,x_0-x_3$. Their common zero on $X$ is precisely $P$, so away from $P$ the linear system of divisors is base-point-free and defines the projection from a point on a smooth quadricIn particular, the chart-ratio construction in part (a) makes this a morphism of schemes on $X\setminus \{P\}$, not merely a rational map there.

Write a target point as $[u:v:w]$. The corresponding line through $P$ consists of pointsSubstituting in the quadric equation givesRemoving $P$ means $\mu \ne 0$, so set $\mu =1$. Over the residue field $K$ of the target point, its scheme-theoretic fibre is consequentlyIf $u\ne v$, this is one reduced point, with $\lambda =-uw/(u-v)$. If $u=v$ and $uw\ne 0$, it is empty: the corresponding line meets the quadric only at the removed point, with intersection multiplicity two. If $u=v$ and $uw=0$, the line lies entirely on the quadric, and the scheme-theoretic fibre is ${\mathbb{A}}_K^1$, the line with $P$ removed.

The last case occurs at exactly $[0:0:1]$ and $[1:1:0]$. Their lines are respectivelythe two rulings of a smooth quadric surface through $P$. Thus the image consists of the complement of the line $u=v$, together with those two points. The line $u=v$ parametrizes directions in the tangent plane $x_1=x_2$ at $P$. The calculation describes the scheme-theoretic fibres over arbitrary target points and works in every characteristic.