Fano scheme of lines ($F_1(X)$)

The Fano scheme for projective lines. For $X\subseteq {\mathbb{P}}^3$, it is a closed subscheme of $\mathrm{Gr}(2,4)$ and hence is projective. Requiring the defining homogeneous polynomials of $X$ to vanish along a line gives its closed defining conditions.

Cubic forms in four variables form ${\mathbb{P}}^{19}$. Lines form the four-dimensional Grassmannian $G=\mathrm{Gr}(2,4)$. The incidence pairs $(\ell ,[F])$ with $F{|}_{\ell }=0$ form a projective-space bundle with fibre ${\mathbb{P}}^{15}$ over $G$, since restriction to a line imposes four independent linear conditions. Hence the incidence variety is irreducible of dimension 19. For $F=x_0^2{x}_2+x_1^2{x}_3$, a nearby line written $x_2=a_0{x}_0+a_1{x}_1$, $x_3=b_0{x}_0+b_1{x}_1$ lies on $F$ precisely when all four parameters vanish. This is an isolated point of that fibre. The fiber dimension theorem forces the image to have dimension 19, and projectivity makes the image closed. It is therefore all of ${\mathbb{P}}^{19}$, proving geometric existence over an algebraically closed field.

A projective bundle with fibre ${\mathbb{P}}^{15}$ over the four-dimensional Grassmannian $\mathrm{Gr}(2,4)$. The condition of containing a fixed line imposes four independent linear conditions on the twenty cubic coefficients. The projection to ${\mathbb{P}}^{19}$ is proper. The line $\{x_2=x_3=0\}$ is an isolated point over $x_0^2{x}_2+x_1^2{x}_3$; isolated fibre point forces dominance in equal dimensions then proves that every cubic surface contains a line.

For a line $L$ on a degree-$d$ smooth surface in ${\mathbb{P}}^3$, the adjunction formula gives $L^2=2-d$. If $d\ge 3$, all lines are negative curves, hence there are only countably many by countability of negative curves. Their projective Fano scheme of lines cannot have a positive-dimensional complex component, since such a component has uncountably many points. A zero-dimensional projective scheme of finite type has finitely many points, proving the assertion.