Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 25 / 5

The divisors have no common irreducible component, since a common curve would give infinitely many intersection points. The Bézout theorem therefore gives total intersection multiplicity $\mathrm{deg}C\mathrm{deg}D=3\cdot 3=9$. As all intersection is supported at $p$,

Take $D=\{y^{2}z=x^3\}$, $L=\{z=0\}$ and $C=3L$. This projective cuspidal cubic is irreducible and meets $L$ only at $p=[0:1:0]$. At $p$, the derivative with respect to $z$ is $y^2=1$, so $D$ is smooth. In the affine chart $y=1$, it is $z=x^3$, and $L$ is $z=0$, so $I_p(L,D)=3$. Multiplicity is linear in the divisor $C=3L$, giving

Apply a projective transformation to put the cuspidal cubic in the form $D=\{F=y^{2}z-x^3=0\}$. Since $L$ meets $D$ at only one smooth point $p$, the Bézout theorem forces intersection multiplicity three there. Hence $L$ is the tangent line at a flex.

The homogeneous Hessian matrix has determinantBy the Hessian criterion for a flex, every smooth flex lies in $D\cap \{x{y}^2=0\}$. These intersections are $[0:0:1]$, the cusp, and $[0:1:0]$, which is smooth. Thus the unique smooth flex of a cuspidal cubic is $p=[0:1:0]$, with tangent line $L=\{z=0\}$. Conversely the pair in part (b) has exactly the required properties. Therefore there is one projective-equivalence class, represented by