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For a cubic homogeneous polynomial $F$, its Hessian matrix has linear entries, so its determinant is a cubic homogeneous polynomial, possibly zero. On a nonsingular plane cubic in characteristic different from two, its zeros are exactly the flexes, by the Hessian criterion for a flex. If the determinant is nonzero, Bézout's theorem supplies intersections; if it is zero, the local criterion itself supplies flexes.

For a homogeneous polynomial $F$ of degree $d\ge 2$, the determinant of its three-by-three Hessian matrix is homogeneous of degree $3(d-2)$ if nonzero. Its zero locus is the Hessian curve; it can be reducible or degenerate, and a nonzero constant for $d=2$ has empty zero locus. Over $\mathbb{C}$, its intersections with the nonsingular locus of $V(F)$ are precisely the flexes, by the Hessian criterion for a flex, as long as the tangent is not a component.

A nonsingular complex projective plane curve of degree $d\ge 2$, with no common irreducible component with its Hessian curve, has at most $3d(d-2)$ distinct flexes. The Hessian criterion for a flex identifies the flexes with the intersection, and Bézout's theorem counts its total intersection multiplicity. Higher contact can make this bound on distinct points strict.

For a general nonsingular curve of degree $d\ge 3$, all flexes have tangent contact exactly three, making the bound an equality. For a fixed point and line through it, contact at least four imposes four independent linear conditions on the coefficients; the space of such point-line pairs has dimension three. Thus degree-$d$ equations with contact at least four form a proper closed exceptional locus. In local graph coordinates $y=h(x)$, the Hessian determinant along the curve is a nonvanishing factor times $h^{\prime \prime }(x)$, so contact exactly three makes the intersection simple. A nonsingular plane conic has no flexes; a line is excluded because its tangent is a component and its Hessian determinant is identically zero.

A nonsingular point of an algebraic curve $p$ is an inflection point of an algebraic plane curve, or flex, if its projective tangent line has contact of order at least three:Here the projective tangent line must not be a component of $C$. Ordinary tangency has contact order two.

Let $C=V(F)$, where $F$ is a homogeneous polynomial of degree $d\ge 2$. Its Hessian curve is defined byEach entry has degree $d-2$, so a nonzero $H_F$ has degree $3(d-2)$. By the Hessian criterion for a flex, the flexes are exactly the points of $C\cap V(H_F)$. The stated no-common-component assumption and Bézout's theorem therefore giveThis counts distinct points; higher contact can use more than one unit of intersection multiplicity at a single point.

For completeness, the Hessian criterion for a flex follows by putting $p=[0:0:1]$ and $T_{p}C=\{y=0\}$, with $F_y(p)\ne 0$. The Euler theorem for homogeneous functions gives $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, whenceThe restriction $F(x,0,1)$ already has vanishing constant and linear terms. Its quadratic term vanishes exactly when the intersection multiplicity with the tangent is at least three, proving the criterion.

For $d\ge 3$ the bound is the actual maximum number of flexes of a smooth plane curve: a general nonsingular curve has only contact of order three, and the corresponding intersections with its Hessian curve are simple. One way to see genericity is to vary a point and a line through it, a three-dimensional family. Contact of order at least four imposes four independent linear conditions on the coefficients of a degree-$d$ equation for $d\ge 3$. Thus the locus of equations with such contact is a proper exceptional locus. Locally, if the curve is $y=h(x)$, its Hessian curve restricted to $C$ is a nonvanishing factor times $h^{\prime \prime }(x)$; contact exactly three gives a simple zero.

A nonsingular plane conic has no flexes, agreeing with the formula at $d=2$. A line is excluded from this formulation: its tangent is itself and its Hessian matrix is zero, so the no-common-component hypothesis fails. The formula is therefore intended for $d\ge 2$, rather than giving the meaningless negative value at $d=1$.

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 $F$ define the nonsingular plane cubic and let $H_F=\det \mathrm{Hess}F$ 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 $Y^{2}Z$, while the coefficients of $Y^3$ and $X{Y}^2$ vanish. ThusThe 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.

Apply a projective transformation to put the cuspidal cubic in the form $D=\{F=y^{2}z-x^3=0\}$. Since $L$ meets $D$ at only one smooth point $p$, the Bézout theorem forces intersection multiplicity three there. Hence $L$ is the tangent line at a flex.

The homogeneous Hessian matrix has determinantBy the Hessian criterion for a flex, every smooth flex lies in $D\cap \{x{y}^2=0\}$. These intersections are $[0:0:1]$, the cusp, and $[0:1:0]$, which is smooth. Thus the unique smooth flex of a cuspidal cubic is $p=[0:1:0]$, with tangent line $L=\{z=0\}$. Conversely the pair in part (b) has exactly the required properties. Therefore there is one projective-equivalence class, represented by

Over a field of characteristic different from two and three, let $D=\{y^{2}z=x^3\}\subset {\mathbb{P}}^2$. Its homogeneous Hessian matrix has determinant $24x{y}^2$. Its common zeros with $D$ are the cusp $[0:0:1]$ and the smooth point $p=[0:1:0]$. The Hessian criterion for a flex therefore leaves exactly one smooth flex.

At $p$, the projective tangent line is $L=\{z=0\}$. Restricting the cubic equation to $L$ gives $x^3=0$, so its intersection multiplicity at $p$ is three. Consequently a triple projective line and a cuspidal cubic meeting only at a smooth point are projectively equivalent to $(3L,D)$ under these field hypotheses.

The characteristic restriction matters. In characteristic three, every smooth point can be written $[a:1:a^3]$ and has projective tangent line $z=a^{3}y$. Substitution gives $(x-ay{)}^3=0$, so every smooth point has triple contact. The unique-flex conclusion then fails.