Past exam of the geometry course of the University of Oxford / 2009 / B / 5 / i / Solution

An inflection point of an algebraic plane curve is a smooth point where the tangent has intersection multiplicity at least three. For a homogeneous defining polynomial $P$ of degree $d$, its Hessian curve is defined by the determinant of the matrix of second partial derivatives, of degree $3(d-2)$. In characteristic zero, the smooth inflection points are exactly the smooth points of $C\cap \mathrm{Hess}(C)$.

For a local verification, move a smooth point to $[0:0:1]$ and its tangent to $y=0$, and scale so $P(x,y,1)=y+a{x}^2+bxy+c{y}^2+\cdots$. The tangent contact is at least three exactly when $a=0$. The homogeneous Euler identities give the Hessian matrix at that point as $\left(\begin{array}{ccc}2a& b& 0\\ b& 2c& d-1\\ 0& d-1& 0\end{array}\right)$, whose determinant is $-2a(d-1{)}^2$. This proves the criterion for $d>1$.

For the given cubic, the singularity equations are $x^2+\lambda yz=y^2+\lambda xz=z^2+\lambda xy=0$. No coordinate can vanish at a projective solution, since that would force all three to vanish. Multiplying the equations therefore gives $1=(-\lambda {)}^3$, so singularity requires ${\lambda }^3=-1$. Conversely, under that condition the point $[1:1:-1/\lambda ]$ satisfies all three equations, and the Euler identity places it on the cubic. Thus