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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.

Distinct negative curves cannot have the same integral cohomology class: their mutual intersection would equal a negative self-intersection, contradicting nonnegative intersection multiplicities for distinct curves. A smooth complex projective surface has finitely generated integral cohomology groups, so its negative curves form an at most countable set. Zero-self-intersection curves need not obey this restriction: one ruling of ${\mathbb{P}}^1\times {\mathbb{P}}^1$ gives an uncountable family.

The degree of a projective algebraic variety of dimension $r$ is the length, with intersection multiplicities, of its intersection with a sufficiently general linear subspace of complementary codimension. Equivalently, its Hilbert polynomial has leading coefficient $\mathrm{deg}X/r!$. For a projective curve, a general hyperplane section measures this degree.

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.

If a projective line $H$ does not contain the projective plane curve $X$, its hyperplane-section divisor iswhere $I_p(X,H)$ is the intersection multiplicity. The Bézout theorem gives $\mathrm{deg}[X\cap H]=(\mathrm{deg}X)(\mathrm{deg}H)=\mathrm{deg}X$.

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.

The intersection number $D\cdot C$ is the degree of the restriction of ${\mathcal{O}}_X(D)$ to the complete integral curve $C$. On a smooth surface, proper effective intersections are sums of local intersection multiplicities.

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.

The movable part is $|D-F|$, where $F$ is the fixed part. It has no fixed components. On a smooth algebraic surface its associated divisor is nef: for each curve choose a member avoiding that curve and use nonnegative local intersection multiplicities.

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.

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 $d=1$.

Let $S=k[x_0,\dots ,x_n]$ and let $B=S/I(X)$ be the homogeneous coordinate ring of a nonempty projective variety. Its Hilbert function is $h_X(t)={\dim }_k{B}_t$. For all sufficiently large integers $t$, this function agrees with a polynomial $P_X(t)$, the Hilbert polynomial.

The dimension $r$ is the largest length of a strict chain of nonempty irreducible closed subvarieties of $X$. Equivalently $r=\dim B-1=\mathrm{deg}{P}_X$. The degree of the embedded variety is the integer specified by the leading termGeometrically it is the number of intersection points with a general linear subspace of codimension $r$, counted with intersection multiplicities. The polynomial's degree is the dimension $r$; the variety's degree is the leading coefficient multiplied by $r!$. Keeping those two meanings separate is important in the next computation.

Work with the smooth projective cubic, including its points at infinity. For two points $P,Q$, draw their chord, using the tangent when they coincide, and let $R$ be its third intersection counted with intersection multiplicity. Draw the line through $R$ and $o$, again using the tangent if necessary, and let $S$ be its third intersection. Define $P+Q=S$. When $o$ is a flex, this is the familiar reflection of the first third-intersection point across the Weierstrass involution. The two-line construction also works when the chosen identity is not a flex.

To prove associativity without assuming it in the geometry, use divisor class. A nonsingular plane cubic has genus one. By the Riemann-Roch theorem, every divisor class of degree one contains exactly one effective divisor on an algebraic curve of degree one: for such a divisor on an algebraic curve $D$,and $\mathrm{deg}(K_C-D)=-1$, so $\ell (K_C-D)=0$. Thus $\ell (D)=1$, and the nonzero section is unique up to scalar. Over $K$, a degree-one effective divisor on an algebraic curve defined over $K$ is a $K$-rational point. This also proves uniqueness of that point. In particular the mapis injective and identifies these points with the corresponding degree-zero Picard group of a curve: to recover the point from a class, add $[o]$ and take its unique effective representative. This is the elliptic curve group law from Riemann-Roch.

Let $H$ denote the divisor class cut out by a line. The two lines in the chord-and-tangent group law giveTheir difference gives $P+Q\sim o+S$, henceAddition of divisor class is associative. Therefore both $(P+Q)+R$ and $P+(Q+R)$ map to $[P-o]+[Q-o]+[R-o]$. Injectivity of $\iota$ proves the chord-and-tangent group law is associative. The same argument gives identity $o$, commutativity and inverses, and covers tangencies through the use of intersection multiplicities. All constructions are defined over $K$.

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.

An inflection point is a smooth point where the tangent line has intersection multiplicity at least three with the curve. For a nonsingular plane curve of degree $n\ge 3$, these are the intersections with its Hessian curve, whose degree is $3(n-2)$. The original curve is not a component of its Hessian: otherwise in each graph chart the tangent direction would have zero derivative identically, making the curve locally a line and hence a line globally. Bézout therefore gives the upper bound $3n(n-2)$, counting multiplicity. A nonsingular conic has no inflection points; lines are excluded from this finite-contact definition.

Take $F=y^{2}z-x^3+x{z}^2$. Its Hessian determinant isAt infinity the curve has only $O=[0:1:0]$, which is an inflection point: its tangent is $z=0$ and the restricted equation is $-x^3=0$. In the chart $z=1$, substitute $y^2=x^3-x$ into the Hessian equation to obtainThus the full answer isThe four complex values of $x$ are distinct and none is $0,1,-1$, so each gives two distinct nonzero values of $y$. Together with $O$ these are all nine inflection points, exhausting the Bézout bound.

An inflection point of an algebraic plane curve is a smooth point where the tangent has intersection multiplicity at least three. For a homogeneous defining polynomial $P$ of degree $d$, its Hessian curve is defined by the determinant of the matrix of second partial derivatives, of degree $3(d-2)$. In characteristic zero, the smooth inflection points are exactly the smooth points of $C\cap \mathrm{Hess}(C)$.

For a local verification, move a smooth point to $[0:0:1]$ and its tangent to $y=0$, and scale so $P(x,y,1)=y+a{x}^2+bxy+c{y}^2+\cdots$. The tangent contact is at least three exactly when $a=0$. The homogeneous Euler identities give the Hessian matrix at that point as $\left(\begin{array}{ccc}2a& b& 0\\ b& 2c& d-1\\ 0& d-1& 0\end{array}\right)$, whose determinant is $-2a(d-1{)}^2$. This proves the criterion for $d>1$.

For the given cubic, the singularity equations are $x^2+\lambda yz=y^2+\lambda xz=z^2+\lambda xy=0$. No coordinate can vanish at a projective solution, since that would force all three to vanish. Multiplying the equations therefore gives $1=(-\lambda {)}^3$, so singularity requires ${\lambda }^3=-1$. Conversely, under that condition the point $[1:1:-1/\lambda ]$ satisfies all three equations, and the Euler identity places it on the cubic. Thus

The strong Bézout theorem says that if two projective plane algebraic curves $C,D$ over $\mathbb{C}$ have degrees $m,n$ and no common irreducible component, their intersection is finite andEach intersection multiplicity is a positive integer. Scheme-theoretic multiplicities, rather than just distinct intersection points, are essential: tangencies and singularities can make individual terms larger than one.

Write the irreducible curve as $F=0$, with $F$ homogeneous of degree $n$. For $n=1$ there are no singular points. Otherwise choose a nonzero partial derivative, say $F_x$, which has degree $n-1$. Such a derivative exists in characteristic zero. Irreducibility and its smaller degree show that $F$ and $F_x$ have no common component.

At every singular point $p$ of $C$, $F_x(p)=0$ and the curve multiplicity of $F$ is at least two. The standard local lower bound for intersection multiplicity gives $I_p(F,F_x)\ge 2$. Applying the Bézout theorem,ThereforeThis is the polar bound for singular points of a plane curve.

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.

Let $O=(0:1:0)$ be the point at infinity of the Weierstrass equation of an elliptic curve. It is the identity element for the elliptic curve group law. For an affine rational point $P=(x,y)$, the inverse isIndeed, the two affine intersections with the vertical line have these ordinates, since their sum as roots of the quadratic in $y$ is $-a_{1}x-a_3$. This reflection is valid for a generalized Weierstrass equation of an elliptic curve in every characteristic of a field.

For distinct $P,Q$, draw their line; for $P=Q$, draw the tangent line. Let $R$ be its third intersection with the elliptic curve, counted with intersection multiplicity. Then the sum is the reflection of the third intersection:A vertical line has third intersection $O$, so $P+(-P)=O$. If a tangent has third intersection $O$, then $2P=O$. Define addition with $O$ by $P+O=P$; these conventions include all exceptional cases of the affine construction.

The construction is defined over the coefficient field: after two known intersections, the remaining root is rational over that field. It gives a commutative group operation. To see why it is associative, use the elliptic curve group law from Riemann-Roch: the map $P\mapsto [P-O]$ identifies the elliptic curve with its degree-zero divisor class group. A line has intersection divisor $P+Q+R$ linearly equivalent to $3O$, the divisor of the line at infinity. Thus $[P-O]+[Q-O]+[R-O]=0$. The geometric sum agrees with addition of divisor classes, which is associative.