Inflection point of a plane cubic

A flex of a smooth plane cubic is a point whose tangent line has intersection multiplicity three with the cubic. Over an algebraically closed field, choose a flex $O$ as the elliptic curve identity. A line cuts out a divisor linearly equivalent to $3O$, so $P$ is a flex exactly when $3P=O$: the tangent divisor is then $3P$. Thus the flexes are precisely the geometric 3-torsion points.

If the characteristic is not three, prime-to-characteristic geometric torsion gives nine distinct flexes. In characteristic three, the scheme-theoretic fibre $E[3]$ still has length nine, but need not have nine distinct points: an ordinary elliptic curve has three geometric 3-torsion points, and a supersingular elliptic curve has only the identity. A flex chosen over the algebraic closure need not be rational over the original field.

For a smooth homogeneous cubic in characteristic zero, its plane cubic flexes are precisely its intersections with the determinant of its Hessian matrix. Choose coordinates with a given smooth point $(0:0:1)$ and its tangent $y=0$. Up to a nonzero coefficient, the cubic starts as $y{z}^2+a{x}^{2}z+bxyz+c{y}^{2}z$ plus cubic terms in $x,y$. Its tangent contact has order three exactly when $a=0$, and the Hessian determinant at the point is a nonzero constant times $a$. This proves the criterion.

For $c\ne 0$ and $d^3\ne 27c$ in characteristic zero, the displayed plane cubic is smooth. Its Hessian determinant is $-6d^2(x^3+y^3+c{z}^3)+(216c-2d^3)xyz$. On the cubic this becomes $8(27c-d^3)xyz$. Hence all nine plane cubic flexes lie on the coordinate axes: $(1:-\zeta :0)$, $(-\sqrt[3]{c}\zeta :0:1)$ and $(0:-\sqrt[3]{c}\zeta :1)$ for ${\zeta }^3=1$.