Bézout's theorem

Two projective plane curves of degrees $m$ and $n$ with no common irreducible component have total intersection multiplicity $mn$. In particular, two nonempty projective plane curves always intersect.

Suppose two degree-$n$ projective plane curves meet at $n^2$ distinct points, and $nm$ of them lie on an irreducible degree-$m$ curve $R=0$, with $m<n$. Pick an extra point of $R=0$ outside the finite base locus. A nonzero linear combination $H=\lambda P+\mu Q$ vanishes at that point. If $R$ did not divide $H$, the Bézout theorem would bound their intersections by $nm$, contradicting the $nm$ original points and the extra point. Therefore $H=RS$ with $\mathrm{deg}S=n-m$. All remaining base points lie off $R=0$ and hence on $S=0$. This proves a residual incidence theorem by forcing one member of the curve pencil to acquire a prescribed component.

For projective plane curves without a common irreducible component, project from a point outside both curves. In coordinates where that point is $[1:0:0]$, their defining homogeneous polynomials have constant nonzero leading coefficients in $X$. The resultant in $X$ is nonzero by Gauss's lemma, and is homogeneous of degree $\mathrm{deg}C\mathrm{deg}D$ in the remaining two coordinates. Every intersection projects to one of its zeros. Each projection line has only finitely many intersections with the curves, so their intersection is finite. Choose the center again outside all joining lines of pairs of intersection points; the projection is now injective on that finite set, proving the bound.

A projective plane curve is a projective curve embedded in the projective plane, usually given by one homogeneous equation.

This pencil consists of projective cubic curves generated by $x^3+y^3+z^3$ and $xyz$. The displayed finite parameter gives a nonsingular cubic exactly when ${\lambda }^3\ne -1$; the parameter at infinity gives the triangle $xyz=0$. Multiplying the three gradient equations proves the finite singularity criterion. The Hessian curve of a member has equation ${\lambda }^2(x^3+y^3+z^3)-({\lambda }^3+4)xyz=0$, so the Hessian stays in the same pencil.

For ${\omega }^3=1$ primitive and $k=0,1,2$, these nine points are the inflection points of every nonsingular finite member of the Hesse pencil. On the cubic, substituting the equation into its Hessian curve gives $-4({\lambda }^3+1)xyz=0$. Nonsingularity forces one coordinate to vanish, and the remaining two cubes sum to zero, producing exactly the nine displayed points. They are the fixed base configuration of the pencil, independently of its smooth parameter.

For six distinct points on an irreducible conic in the complex projective plane, the three intersections of opposite sides of the resulting hexagon are collinear. Form the two cubics given by the products of alternate side equations. They meet at the six conic points and the three opposite-side intersections. Applying residual intersection through a curve pencil with degrees three and two places the last three points on a line. This is the conic counterpart of Pappus's hexagon theorem for two lines.

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.

A projective plane curve is reduced when its defining homogeneous polynomial is a squarefree polynomial, so no irreducible component is repeated. Its degree is the sum of the degrees of its distinct irreducible components. This convention is essential when bounding the number of singular points: the nonreduced equation $L^2=0$ has vanishing first partial derivatives at every point of its supporting line.

A degree-two projective plane curve, allowing singular and nonreduced cases. In characteristic different from two, write its equation as $z^{\mathsf{T}}Mz$ with $M$ a symmetric matrix. It is smooth exactly when $M$ has matrix rank three. Over an algebraically closed field, rank two gives two distinct projective lines, while rank one gives a double line.

A degree-four projective plane curve. Its arithmetic genus is three by the genus-degree formula; singularities can lower the genus of its normalization of an algebraic curve.

Translate the point to the origin of an affine chart and expand a local defining polynomial as $\sum a_{ij}{x}^i{y}^j$. Its lowest nonzero total degree is the multiplicity. A smooth point of a variety has multiplicity one; a double point has multiplicity two. On a point blowup of a smooth algebraic surface, the total transform is the strict transform of an algebraic subvariety plus this multiplicity times the algebraic exceptional divisor.

For a nonzero homogeneous polynomial of degree $d>0$, multiplication gives $0\to {\mathcal{O}}_{{\mathbb{P}}^2}(-d)\to {\mathcal{O}}_{{\mathbb{P}}^2}\to i_{\ast }{\mathcal{O}}_X\to 0$. Intermediate cohomology of the twisting sheaves vanishes. The top cohomology of $\mathcal{O}(-d)$ has Laurent basis $t_0^{-b_0}{t}_1^{-b_1}{t}_2^{-b_2}$ with $b_i\ge 1$ and $\sum b_i=d$. There are $(d-1)(d-2)/2$ such monomials, with zero basis for $d=1,2$. The long exact sequence gives the displayed dimensions. The calculation allows singular, reducible and nonreduced hypersurfaces; it is an arithmetic-genus calculation, not an assertion about a normalization's genus.

Two distinct singular points would impose at least two zeros each on the restriction of the cubic to their joining line. A nonzero homogeneous polynomial of degree three cannot have total zero multiplicity four, so that restriction vanishes identically. The line would be a component, contradicting the irreducible polynomial hypothesis. This argument works in every characteristic.

In its classical nine-point cubic form, the Cayley-Bacharach theorem says that a cubic through eight of the nine distinct intersections of two plane cubics must contain the ninth. The eight-point cubic completion for two triples of lines is the elementary case needed for triangular-strip propagation in incidence geometry. The phenomenon explains why interpolation conditions at such a complete intersection are dependent.

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.

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 smooth complex degree-two projective plane curve is a projective linear image of the degree-two Veronese map and is diffeomorphic to a sphere. Its class in ${\mathbb{CP}}^2$ is twice the class of a projective line. Its self-intersection number is four, so its oriented normal bundle has Euler class evaluating to four.

The tangent lines of a smooth plane conic form a smooth plane conic in the dual projective space. If the original equation is $v^{\mathsf{T}}Qv=0$, its tangent has coefficient vector $Qv$, giving the displayed equation. The matrix $Q$ is symmetric and invertible, so this correspondence is a projective linear isomorphism of the conics.

If two smooth plane conics over $\mathbb{C}$ meet at four distinct points, their tangent incidence curve is a smooth projective genus one curve. Projection to $C_2$ is a double cover branched at those four intersections. The tangent equation becomes a quadratic whose discriminant cuts out $C_1\cap C_2$; its four simple zeros make the cover smooth and connected. The Riemann-Hurwitz formula then gives genus one.

For two smooth plane conics meeting transversely, the operation of crossing a chord of the second conic tangent to the first and choosing the other tangent through the new endpoint acts as a translation on an elliptic curve on their incidence curve. Each switch is an involution of a degree-two map from a genus one curve. One periodic orbit means that the translating point is torsion, so every orbit is periodic with the same least period. The construction extends through coincident choices at ramification points using the regular involutions.

The complement of a closed tubular neighborhood of a smooth plane conic has the displayed integral homology. The Excision theorem and Thom isomorphism theorem reduce the calculation to the relative long exact sequence: the ambient degree-four fundamental class restricts with coefficient one, while the degree-two map is intersection with the conic and has coefficient two. A collar neighborhood identifies the open and compact exterior homotopy equivalence types. The boundary instead has first homology $\mathbb{Z}/4$, from the Gysin sequence and the normal Euler class four.

A plane cubic is a projective plane curve defined by a nonzero homogeneous polynomial of degree three. Its arithmetic genus is one by the genus-degree formula, and its Hilbert polynomial is $3m$. A singular cubic can have normalization of smaller genus; the Hilbert polynomial measures the arithmetic genus.

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.

This smooth projective plane curve has no real rational points: a sum of three real squares is zero only when all three coordinates vanish, which is forbidden in projective space. It is therefore not isomorphic over $\mathbb{R}$ to the projective line, which has rational points. Over $\mathbb{C}$ it becomes a nonsingular conic isomorphic to a projective line. Thus the ground field changes whether a geometric genus-zero curve admits a projective-line parametrization.

The arithmetic genus of a degree-$d$ projective plane curve isEquivalently, its structure sheaf has Euler characteristic $\chi ({\mathcal{O}}_X)=1-p_a$.

Blow up each distinct singular point of an integral projective curve in the projective plane once. The arithmetic genus drop under a point blowup and the genus-degree formula give arithmetic genus $(d-1)(d-2)/2-{\sum }_P{m}_P(m_P-1)/2$ for the strict transform. Its arithmetic genus is nonnegative, proving the bound in every characteristic. Infinitely near singularities need not have been resolved.

Three distinct points of multiplicity two use the entire arithmetic genus three of an integral projective curve of degree four. One point blowup at each gives arithmetic genus zero. The normalization of an algebraic curve then has genus zero and is the projective line over an algebraically closed field, by rationality of a smooth projective genus-zero curve. This allows double points with one branch as well as ordinary nodes.

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

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 $\sum d_i=d$.

Two projective plane curves with no common irreducible component meet transversely at $p$ if both are nonsingular there and their projective tangent lines differ. This holds exactly when their intersection multiplicity is one. Indeed, in the local ring of the ambient smooth surface, independent linear terms of the two equations generate the maximal ideal and leave a quotient of length one. Conversely, length one forces those linear terms to span the two-dimensional cotangent space, proving both nonsingularity and distinctness of the tangents.

A smooth point of an algebraic plane curve is a flex if its tangent line has intersection multiplicity at least three there. This definition also applies to smooth points of a singular curve.

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.

For a plane curve defined by a homogeneous polynomial $F$ of degree $d\ge 2$, assume the field characteristic is different from two and $d-1$ is nonzero in the field. A smooth point of a variety is a flex exactly when the determinant of the homogeneous Hessian matrix vanishes there, provided its tangent line is not a component. Put the point at $[0:0:1]$ with tangent $y=0$ and $F_y(p)\ne 0$. The homogeneous identities give $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, so the determinant is $-(d-1{)}^2{F}_y(p{)}^2{F}_{xx}(p)$. In the completed local ring, write the curve as a formal power series $y=h(x)$. Its linear coefficient is zero and its quadratic coefficient is $-F_{xx}(p)/(2F_y(p))$. Vanishing of that coefficient is exactly intersection multiplicity at least three with the tangent, proving the criterion in these positive characteristics as well as in characteristic zero.

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.

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.