= Solution
Let $F$ define the <nonsingular plane cubic> and let $H_F=\det\operatorname{Hess}F$ be its <Hessian curve of a plane cubic> equation. The <Hessian criterion for a flex> applies in the allowed <field characteristic>. Here is the local calculation, so no characteristic-zero argument is needed.
Move any point $P$ to $[0:1:0]$ and its <tangent line> to $Z=0$. Smoothness gives a nonzero coefficient $g$ of $Y^2Z$, while the coefficients of $Y^3$ and $XY^2$ vanish. Thus
$$
F(X,Y,0)=aX^3+bX^2Y,\qquad
\operatorname{Hess}F(P)=
\begin{pmatrix}2b&0&f\\0&0&2g\\f&2g&2i\end{pmatrix},
\qquad H_F(P)=-8bg^2.
$$
The <intersection multiplicity> with the <tangent line> is at least three precisely when $b=0$, which is precisely $H_F(P)=0$ because the <field characteristic> is not two. The tangent cannot be a component of a <nonsingular plane cubic>.
If $H_F$ vanishes identically, every point is a <flex> by this calculation. Otherwise it is a degree-three <homogeneous polynomial>. Its zero locus meets the cubic by <Bézout's theorem>; if they share a component, that also supplies a point of intersection. Every such point is smooth and satisfies the criterion, so a <flex> exists.
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