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A two-cell-thick strip of triangles with degree-six vertices in a dual arrangement of a planar point set produces three indexed primal point families with collinearities $a_i,b_j,c_k$ whenever $i+j+k=0$. A plane cubic can be fitted to nine initial points because its homogeneous coefficient space has dimension ten. Overlapping nine-point configurations then force each next point onto the same cubic by eight-point cubic completion for two triples of lines. Safe neighbourhoods guarantee the distinctness needed in these local completion steps. A possible seed is $a_{-1},a_0,a_1,a_2,b_{-3},b_{-2},b_{-1},c_1,c_2$. Adjacent completion blocks force $c_3,b_{-4},c_4,a_{-2}$, followed by the alternating continuation along the two long transverse families.

A projective line $L=0$ is contained in the zero locus of the nonzero cubic $L^3$. The union of three lines is given by their product, and a conic together with a line is likewise a plane cubic. Reducibility and repeated factors are permitted in polynomial coverings used in incidence geometry; a covering by cubics need not consist of smooth irreducible curves.

Let two triples of projective lines meet in nine distinct points. Any plane cubic containing eight contains all nine. To prove this, write the triples as $F=L_1{L}_2{L}_3$ and $G=M_1{M}_2{M}_3$, with the missing point on $L_3$. The candidate cubic $H$ agrees with a scalar multiple of $G$ on $L_1$, so $H-\lambda G=L_{1}Q$. Its three known zeros on $L_2$ force $Q=L_{2}L$. Two remaining known zeros on $L_3$ force the linear factor $L$ to be a multiple of $L_3$. Hence $H=\lambda G+\mu F$ and vanishes at the missing point.

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 ${\zeta }^3=1$.

For $F=x^3+y^3+3x{z}^2$, the first partial derivatives areTheir simultaneous vanishing would force $y=0$, $xz=0$ and $x^2+z^2=0$, hence $x=y=z=0$, which is not a projective point. This also verifies that the given plane cubic is nonsingular.

The Hessian matrix and its determinant areThus the Hessian curve is the union of $y=0$, $x=z$ and $x=-z$. On $y=0$, the equation of $C$ reduces to $x(x^2+3z^2)=0$, givingOn $x=z$ one must have $z\ne 0$, so put $z=x=1$ and obtain $y^3=-4$. On $x=-z$ similarly put $z=1$, $x=-1$, obtaining $y^3=4$. Choose $r\in \mathbb{C}$ with $r^3=4$ and let $\omega =\exp (2\pi \mathrm{i}/3)$. The remaining points areAll nine listed projective points are distinct. They exhaust the intersection with the Hessian curve, so the Hessian criterion for a flex proves that these, and only these, are the flexes in this example. There are none at infinity: $z=0$ and $H_F=0$ would force $x=y=0$ on $C$.

Let $(r_0,r_1,r_2)$ be any ordering of the three distinct numbers $0,1,\lambda$. Set $h=r_1-r_0$ and choose $s\ne 0$ with $s^2=h^3$. The projective transformationhas the required matrix form. As $x=hX+r_{0}Z$ and $y=sY$, substitution givesThe choice $s^2=h^3$ is exactly what cancels the coefficient introduced by the change in $x$.

The six orderings give the Legendre parameter orbit:Every entry is different from $0,1$ when $\lambda \ne 0,1$.

There are no further possibilities for a matrix of the stated form. Its determinant is $acd$, so $a,c,d$ must be nonzero. On $z=1$ it induces the affine change $u\mapsto (a/d)u+b/d$ and scales $v=y/z$ by $c/d$. It preserves the line $y=0$, whose three intersections with the plane cubic have $u$-coordinates $0,1,\lambda$. Their images must have coordinates $0,1,\tilde{\lambda }$. Choosing the two roots sent to $0,1$ determines the affine change and leaves exactly the six orderings above.

For general $\lambda$ the six values are distinct. They coincide in some special cases: $\lambda \in \{-1,2,1/2\}$ gives three values, and ${\lambda }^2-\lambda +1=0$ gives two. In particular, for the value obtained in part (c), the two possible values are $(1\pm \mathrm{i}\sqrt{3})/2$. This does not contradict the paper's qualification that there are six for general $\lambda$.

Every projective line on the smooth algebraic surface through $P$ lies in the projective tangent plane $T_{P}S$, since its tangent direction is contained in the surface's Zariski tangent space. The plane section $S\cap T_{P}S$ is a nonzero plane cubic. It cannot be the entire plane, since a smooth cubic surface is irreducible and has no plane component.

Each distinct projective line through $P$ is therefore a distinct degree-one factor of the cubic restricted to $T_{P}S$. A degree-three polynomial has at most three such factors. Hence there are at most three lines through any point. This is the lines through a point of a smooth cubic surface bound; equality is allowed when the tangent section consists of three concurrent lines.

Eliminate $t_3$ using the linear equation. The homogeneous coordinate ring becomesThe cubic is a nonzero element of a polynomial integral domain, hence a non-zero-divisor. It gives the exact sequence of graded modulesThus $\dim B_m=\left(\genfrac{}{}{0ex}{}{m+2}{2}\right)-\left(\genfrac{}{}{0ex}{}{m-1}{2}\right)$ for $m\ge 3$. Equivalently, use the Hilbert series $(1-z^3)/(1-z{)}^3$, or the sheafified exact sequence on ${\mathbb{P}}^2$. The Hilbert polynomial of the plane cubic isIts degree is three and its arithmetic genus is one, although the curve is singular: its affine equation near $[1:0:0]$ is the cusp $t_2^2+t_1^3=0$. The Hilbert polynomial records the arithmetic genus, rather than the genus of the normalization.