Hessian criterion for a flex ($\det \mathrm{Hess}F(p)=0$)

For a plane curve defined by a homogeneous polynomial $F$ of degree $d\ge 2$, assume the field characteristic is different from two and $d-1$ is nonzero in the field. A smooth point of a variety is a flex exactly when the determinant of the homogeneous Hessian matrix vanishes there, provided its tangent line is not a component. Put the point at $[0:0:1]$ with tangent $y=0$ and $F_y(p)\ne 0$. The homogeneous identities give $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, so the determinant is $-(d-1{)}^2{F}_y(p{)}^2{F}_{xx}(p)$. In the completed local ring, write the curve as a formal power series $y=h(x)$. Its linear coefficient is zero and its quadratic coefficient is $-F_{xx}(p)/(2F_y(p))$. Vanishing of that coefficient is exactly intersection multiplicity at least three with the tangent, proving the criterion in these positive characteristics as well as in characteristic zero.

For a cubic homogeneous polynomial $F$, its Hessian matrix has linear entries, so its determinant is a cubic homogeneous polynomial, possibly zero. On a nonsingular plane cubic in characteristic different from two, its zeros are exactly the flexes, by the Hessian criterion for a flex. If the determinant is nonzero, Bézout's theorem supplies intersections; if it is zero, the local criterion itself supplies flexes.