A flex of a cubic has contact order three

Given: AC The Axiom of Choice, an algebraically closed field $k$ of characteristic not three, the Fermat cubic $C=V(F)$ with $F=x_0^3+x_1^3+x_2^3$, the point $p=[1:-1:0]$, and the line $T=V(x_0+x_1)$.

If $F$ had a repeated irreducible factor, it would divide all of $3x_0^2,3x_1^2,3x_2^2$, impossible since $3\mathrm{\ne }0$ and these polynomials have no common nonconstant divisor. Thus $F$ is square-free of degree three, and $p\in C$ because $1^3+(-1{)}^3+0^3=0$; the gradient of $F$ at $p$ is $(3x_0^2,3x_1^2,3x_2^2)=(3,3,0)$, nonzero since the characteristic is not three, so $p$ is a smooth point Plane projective curves and their components, Multiplicity one characterises smooth points with a unique tangent.

The tangent line at the smooth point $p$ is computed from the gradient: $3x_0+3x_1+0\cdot x_2=0$, i.e. $x_0+x_1=0$, so $T=T_{p}C$ Tangent cone and tangent lines at a point, Multiplicity one characterises smooth points with a unique tangent.

For a smooth point $p$ whose tangent line $T$ is not a component of $C$, $I_p(C,T)={\mathrm{ord}}_p(F{\mathrm{\mid }}_T)$, the order of vanishing of the nonzero restricted form, and $p$ is a flex exactly when that order is at least three; an ordinary flex is the case of order exactly three Flexes are contacts of order at least three with the tangent line, Intersection with a smooth curve is a vanishing order.