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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.

Intersection multiplicity measures the local order of contact of two algebraic subvarieties. Transverse intersections have multiplicity one.

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.

Write $C=V(P)$, where $P\in \mathbb{C}[x,y,z]$ is a squarefree polynomial and a homogeneous polynomial of degree $d$. A projective point $p\in C$ is a singular point precisely whenIf $p=[p_0:p_1:p_2]$ is a nonsingular point of an algebraic curve, its projective tangent line isThe coefficients are not all zero, and the Euler theorem for homogeneous functions gives $p_0{P}_x(p)+p_1{P}_y(p)+p_2{P}_z(p)=dP(p)=0$, so the line passes through $p$. Changing the representative of the projective point only rescales the equation.

The strong form of Bézout's theorem says that two projective plane curves $C,D$ of degrees $d,e$, with no common irreducible component, have finitely many intersections andEach intersection multiplicity is a positive integer. At $p\in C\cap D$,This is the criterion for a transverse intersection of plane curves. In particular, an intersection at a singular point of either curve contributes at least two.

If $d=1$, $C$ is a line and has no singular points. Suppose $d>1$ and take $P$ to be an irreducible polynomial. First, $P_x$ is not the zero polynomial: otherwise $P\in \mathbb{C}[y,z]$, and a homogeneous polynomial in two variables of degree greater than one factors into linear factors over $\mathbb{C}$, contrary to irreducibility.

Let $D=V(P_x)$ be the polar curve in this coordinate direction. Since $\mathrm{deg}{P}_x=d-1<d$, the irreducible polynomial $P$ cannot divide $P_x$. Thus $C,D$ have no common irreducible component, and Bézout's theorem applies. One may count repeated factors of $P_x$ with their multiplicities; alternatively, passing to its reduced equation only decreases $\mathrm{deg}D$.

Every singular point $p$ of $C$ satisfies $P_x(p)=0$ and has $I_p(C,D)\ge 2$ by part (a). Consequently, if $s$ is the number of distinct singular points,This argument also proves that the singular locus is finite.

There is a nonsingular point of an algebraic curve on $C$, since part (b) makes its singular locus finite whereas a projective curve has infinitely many complex points. Apply a projective transformation taking one such point to $p=[1:0:0]$, and use $P$ for the transformed homogeneous polynomial. As in part (b), $P_x$ is nonzero and $D=V(P_x)$ has no common irreducible component with $C$.

The Euler theorem for homogeneous functions and its differentiated identities giveandSince $p$ is nonsingular on $C$, at least one of $P_y(p),P_z(p)$ is nonzero. Because $d>1$, the polar curve $D$ is also nonsingular at $p$, and these identities show that $T_{p}D=T_{p}C$. Hence $I_p(C,D)\ge 2$.

This additional intersection is separate from all the singular points of $C$. Thus Bézout's theorem givesThe improvement comes from making the polar curve tangent at a chosen nonsingular point, thereby reserving at least two of the available intersections.

Let $d_i=\mathrm{deg}{C}_i$. Under the reduced plane curve convention, the distinct irreducible components have ${\sum }_i{d}_i=d$. Away from intersections with other components, a point of $C_i$ is nonsingular on the union exactly when it is nonsingular on $C_i$. For example, a local equation of the union is the equation of $C_i$ multiplied by a nonvanishing factor. At an intersection of two components, differentiating their product makes every first partial derivative vanish. Therefore

Part (b) bounds the number of singular points of $C_i$ by $d_i(d_i-1)/2$. For each distinct pair of irreducible components, Bézout's theorem bounds the number of distinct intersections by $d_i{d}_j$. The cardinality of a union is at most the sum of the cardinalities, even if some points have been counted several times. Hence

The reduced convention matters. If one instead allows the equation $L^2=0$ as a nonreduced curve and uses its equation to test singularity, every point on its supporting line has vanishing first partial derivatives. A finite bound would then be false.

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 $d=1$.

Bézout's theorem says that two projective plane curves over $\mathbb{C}$ of degrees $d,e$ with no common irreducible component have total intersection multiplicity $de$, so at most $de$ distinct intersection points. Let the irreducible curve have reduced homogeneous equation $F$ of positive degree $d$. In characteristic zero some partial derivative is nonzero. For $d>1$ this derivative has degree $d-1$ and cannot share a component with the irreducible degree-$d$ polynomial $F$. Every singular point belongs to their intersection. Bézout bounds their number by $d(d-1)$, proving there are finitely many singular points. For $d=1$ a partial is a nonzero constant and there are none.

Write $L_i=a_{i}x+b_{i}y+c_{i}z$ and $S={\prod }_i{L}_i$. The numerator and denominator have the same homogeneous degree, so their ratio is well-defined on projective points away from its poles and extends as a meromorphic function on $C$. Locally the pole order cannot exceed the zero order of $S{|}_C$, whose divisor is $D$. Thus $f_Q\in L(D)$.

Equality of two ratios is equivalent to $Q-Q^{\prime }$ vanishing on $C$. A nonsingular projective plane curve is irreducible: distinct positive-degree components would intersect by Bézout's theorem, producing a singular point. Its homogeneous vanishing ideal is therefore generated by its irreducible defining polynomial $P$. One can also see this using Bézout's theorem: a homogeneous polynomial not divisible by $P$ has only finitely many intersections with $C$, so cannot vanish on the whole curve. ConsequentlyMultiplication by the nonzero polynomial $P$ is injective. The linear map $V_m\to L(D)$ therefore has kernel $P{V}_{m-d}$ and image dimensionThe assumption $m\ge d$ makes both polynomial-space dimensions applicable directly.

Choose homogeneous polynomials $F,G$ defining the two projective plane curves. Their absence of a common irreducible component makes them coprime. For a point $O=[1:0:0]$ outside both curves, they have degrees $n,m$ in $X$ and nonzero constant leading coefficients. The resultantis therefore nonzero. Indeed, a vanishing resultant over $k(Y,Z)$ would give a common factor there, and Gauss's lemma would give a common component of the original curves.

The homogeneity of $F,G$ makes $R$ a homogeneous polynomial of degree $nm$: scaling $(Y,Z)$ by $a$ scales their $X$-roots by $a$, and the product of the $nm$ differences of roots scales by $a^{nm}$. Every point in the intersection projects from $O$ to a zero of $R$ in the projective line. Each such projection line contains only finitely many points of either curve, since it passes through $O$ outside them. Thus the intersection is finite already.

We may now choose $O$ also outside the finitely many joining lines of distinct intersection points. Projection is then injective on the intersection. A nonzero degree-$nm$ homogeneous polynomial on the projective line has at most $nm$ distinct zeros, so the resultant bound for intersections of plane curves givesThis derives the required cardinality bound without assuming the full intersection-multiplicity statement of Bézout's theorem.

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.

For a complex reduced plane curve of degree $d$, the number $s$ of distinct singular points is at most $d(d-1)/2$. For an irreducible polynomial $F$ of degree $d>1$, $F_x$ is nonzero, since a binary homogeneous polynomial over $\mathbb{C}$ factors into linear factors. It has no common factor with $F$. Each singular point contributes at least two to the intersection with the polar curve $F_x=0$, so Bézout's theorem gives $2s\le d(d-1)$.

For an irreducible curve with $d>1$, one can improve this to $s\le d(d-1)/2-1$. Place a nonsingular point at $p=[1:0:0]$. The Euler theorem for homogeneous functions gives $F_x(p)=0$ and $\nabla F_x(p)=(d-1)\nabla F(p)$, so the polar curve is nonsingular and tangent to the curve at $p$. This consumes at least two further units of intersection multiplicity. For a reducible curve with component degrees $d_i$, add the irreducible bounds and at most $d_i{d}_j$ intersections per component pair. The sum is $d(d-1)/2$ because $\sum d_i=d$.

For a homogeneous polynomial $F$ of degree $d$, its first polar with respect to $[a:b:c]$ is the equation $a{F}_x+b{F}_y+c{F}_z=0$, of degree $d-1$ when this polynomial is nonzero. Every singular point of $V(F)$ lies on every such polar. Polars turn the Bézout's theorem intersection count into bounds on singularities and tangencies. The equation is covariant under a linear change of homogeneous coordinates, by the chain rule.