Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 14 / 3

Every projective line on the smooth algebraic surface through $P$ lies in the projective tangent plane $T_{P}S$, since its tangent direction is contained in the surface's Zariski tangent space. The plane section $S\cap T_{P}S$ is a nonzero plane cubic. It cannot be the entire plane, since a smooth cubic surface is irreducible and has no plane component.

Each distinct projective line through $P$ is therefore a distinct degree-one factor of the cubic restricted to $T_{P}S$. A degree-three polynomial has at most three such factors. Hence there are at most three lines through any point. This is the lines through a point of a smooth cubic surface bound; equality is allowed when the tangent section consists of three concurrent lines.

Use coordinates in which $l=\{x_2=x_3=0\}$ and write the defining cubic aswith homogeneous polynomials $Q_2,Q_3$ of degree two. Their restrictions $A=Q_2{|}_l$, $B=Q_3{|}_l$ have no common zero: at such a zero all first partial derivatives of $F$ would vanish. The rational projection $[x_2:x_3]$ therefore extends to a morphism of algebraic varietiesIndeed, wherever $Q_2\ne 0$, the surface equation gives $[x_2:x_3]=[-Q_3:Q_2]$, which remains regular on $l$; the analogous formula works wherever $Q_3\ne 0$.

For the plane with $x_2=sw$, $x_3=tw$, its plane section is $l$ together with the residual plane conicThese residual plane conics are exactly the fibres of a morphism of $f$, giving the conic bundle from a line on a smooth cubic surface. The family itself is smooth: off $l$ it is the graph of projection, and over $l$ the condition $sA+tB=0$ chooses the unique value $[-B:A]$.

Write $q_{s,t}=z^{\mathsf{T}}M(s,t)z$ with $z=(u,v,w{)}^{\mathsf{T}}$ and $M$ symmetric. The degrees of its entries in $(s,t)$ areThus the conic discriminant $\Delta (s,t)=\det M(s,t)$ is a homogeneous polynomial of degree five.

Smoothness also shows that every zero of $\Delta$ is simple and has matrix rank two. To see the first point, at a rank-two singular fibre let $z_0$ span the kernel and use a local base coordinate $\tau$. All fibre-coordinate derivatives vanish at $z_0$, so smoothness of the total family requires $z_0^{\mathsf{T}}{M}^{\prime }(\tau )z_0\ne 0$. Since the adjugate matrix of a rank-two symmetric matrix is a nonzero scalar multiple of $z_0{z}_0^{\mathsf{T}}$, the determinant derivative is nonzero. If the rank were at most one, the projective kernel would contain a line, on which the quadratic $z^{\mathsf{T}}{M}^{\prime }z$ has a zero over the algebraically closed field; that point would be singular in the total family. This proves the reduced singular fibres of a smooth conic bundle assertion and also rules out $\Delta$ vanishing identically.

The discriminant quintic of a cubic surface conic bundle consequently has five distinct zeros in ${\mathbb{P}}^1$. At each, a rank-two plane conic splits into two distinct projective lines $l_i,l_i^{\prime }$. Neither equals $l$: that would require $sA+tB$ to vanish identically, making $A,B$ proportional, contrary to their having no common zero. The original plane section is therefore $l\cup l_i\cup l_i^{\prime }$, proving the required coplanarity and giving five distinct pairs.

The two projective lines in the $i$th pair form the fibre of a morphism $f^{-1}([s_i:t_i])$ of the extended conic bundle from a line on a smooth cubic surface constructed above. Distinct points of the projective line have disjoint inverse images, soThe extension of $f$ across $l$ matters: arguing only that distinct planes through $l$ meet along $l$ would not by itself exclude intersections there. The formula $f{|}_l=[-B:A]$ excludes them as well. Lines within a single pair do intersect; the assertion concerns different pairs.

Choose one projective line from each of two different pairs. They are skew lines by the preceding disjointness result. After a projective linear coordinate change call themFor $A\in L$, $B\in M$, their joining line meets the cubic surface at $A,B$ and a third point. This defines a rational map $L\times M⇢S$ on the Zariski-open set where that third intersection is distinct and the joining line is not contained in $S$.

More explicitly, because $F$ vanishes on both projective lines,The third point is $[VA-UB]$. Both polynomials $U,V$ are nonzero: if either vanished identically, one of the two projective lines would lie in the singular locus. Thus $UV\ne 0$ supplies a nonempty Zariski-open set.

Conversely, a point $P=[x_0:x_1:x_2:x_3]$ outside $L\cup M$ lies on the unique joining line whose endpoints have coordinates $[x_0:x_1]$ on $L$ and $[x_2:x_3]$ on $M$. This is a rational inverse on the same generic locus. The rational parametrization of a cubic surface from two skew lines givesThe latter contains the affine plane as a dense open subset, as does ${\mathbb{P}}^2$. Hence $S$ is birational to ${\mathbb{P}}^2$.