Two transverse cubics meet in nine points

Given: AC The Axiom of Choice, an algebraically closed field $k$ of characteristic not three, the Fermat cubic $C=V(x_0^3+x_1^3+x_2^3)$ and the triangle $D=V(x_0{x}_1{x}_2)$.

A repeated irreducible factor of the Fermat form would divide all three derivatives $3x_0^2,3x_1^2,3x_2^2$, which have no common nonconstant factor because $3\mathrm{\ne }0$. Thus it is square-free; the triangle is a product of three distinct prime linear factors. None of those factors divides the Fermat form (setting each coordinate to zero leaves a nonzero binary cubic), so the two degree-three forms have no common factor, so $C,D$ are plane projective curves of degree three with no common component Plane projective curves and their components. In particular their intersection is nonempty and finite Two plane projective curves meet, Curves without a common component meet finitely often, and Bezout gives ${\sum }_p{I}_p=9$ Bezout's theorem for plane projective curves.

A point lies on $D$ exactly when one of its coordinates vanishes. If, say, $x_0=0$, then the cubic equation reads $x_1^3+x_2^3=0$ with $x_1{x}_2\mathrm{\ne }0$, so $[0:a:b]$ with $a^3+b^3=0$; over the algebraically closed field and in characteristic not three the ratio $b\mathrm{/}a$ solves $1+t^3=0$, which splits over $k$ A field is algebraically closed exactly when every nonconstant polynomial splits, equivalently when it has no nontrivial finite extension. No root is repeated: a repeated root would annul the derivative $3t^2$, whereas every root is nonzero and $3\mathrm{\ne }0$. Thus the ratio takes three distinct values, so this coordinate line contributes three points, and the same holds for the other two coordinate lines; their intersection subsets with $C$ are disjoint, because the pairwise intersections of the coordinate lines are the coordinate vertices and no such vertex satisfies the cubic equation, giving exactly nine points Evaluation and roots of a polynomial in a commutative target ring.

At each intersection point $C$ is smooth: not all of $3x_0^2,3x_1^2,3x_2^2$ vanish for a nonzero point, and the characteristic is not three. At a point with exactly one vanishing coordinate $D$ is also smooth, with tangent line the corresponding coordinate line, and the tangent line of $C$ there is not that coordinate line. Hence the two curves meet transversally at each of the nine points and every local multiplicity equals one Transversal smooth curves meet with multiplicity one, Local intersection multiplicity of two plane curves.