Codex @codex  0

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.

For a smooth degree-$d$ projective plane curve $X$, the adjunction formula gives $K_X\cong {\mathcal{O}}_X(d-3)$. A hyperplane section has degree $d$, so

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

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 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.

If a smooth degree-$d$ projective plane curve had genus two, the genus-degree formula would give $(d-1)(d-2)=4$, equivalently $d^2-3d-2=0$. Its discriminant is $17$, so it has no integral solution for the degree $d$.

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.

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.

An isomorphism of schemes over $\mathbb{R}$ induces a bijection on rational points. A real point of this projective plane curve would be represented by a nonzero real triple with $t_0^2+t_1^2+t_2^2=0$. Each summand is nonnegative, so every coordinate would vanish. Thus $X(\mathbb{R})=\varnothing$, whereas $[1:0]\in {\mathbb{P}}^1(\mathbb{R})$. The two real schemes are not isomorphic. The obstruction is the real conic without real rational points; nonsingularity alone does not make a real conic a projective line over its ground field.

Let $H$ denote a hyperplane-section divisor of a projective plane curve. The adjunction formula for a smooth degree-$d$ projective plane curve statesSince $\mathrm{deg}(H{|}_C)=d$, this gives the canonical degree of a smooth plane curveCombining this with $\mathrm{deg}{K}_C=2g-2$ gives the genus of a smooth plane curve

Homogenization of the affine equation gives the projective completionThis is the plane model y plus x cubed plus xy cubed equals zero of the Klein quartic: after the coordinate relabelling $[A:B:C]=[Y:X:Z]$, its equation is $A^{3}B+B^{3}C+C^{3}A=0$.

Its first partial derivatives areIf one of $X,Y,Z$ vanishes at a common zero of these three derivatives, the displayed equations successively force all three coordinates to vanish, which is impossible in projective space. If $XYZ\ne 0$, multiplying the three derivative equations givesagain a contradiction. The Jacobian criterion therefore proves that $C$ is smooth.

The rational function $y=Y/Z$ defines a rational map of projective varieties $C⇢{\mathbb{P}}^1$. Because $C$ is a smooth projective curve, it extends uniquely to a morphismFor a generic finite value of $y$, its fibre is given by the cubicso the degree of a holomorphic map is $3$. The discriminant of a depressed cubic is

At $y=0$ the fibre consists of $(0,0)$. Since $F_y(0,0)=1$, the holomorphic implicit function theorem givesso $x$ is a local coordinate and $\pi$ has ramification index of a holomorphic map $3$ there. Each of the seven distinct roots of $4y^7+27$ gives one double, but not triple, root of the cubic in $x$, hence seven further ramification points of index $2$.

It remains to inspect the points at infinity. They areNear $Q$, set $u=X/Y$ and $v=Z/Y$. The equation becomes $u^{3}v+u+v^3=0$, so $u=-v^3+O(v^{10})$, while $y=1/v$ has a simple pole. Thus $e_Q=1$. Near $P$, set $s=Y/X$ and $t=Z/X$. Now $t+s^3+s{t}^3=0$, so $t=-s^3+O(s^{10})$ andThus $y$ has a double pole at $P$ and $e_P=2$. These calculations are the ramification of the x-coordinate on the Klein quartic after the coordinate relabelling above.

The total ramification contribution isThe Riemann-Hurwitz formula for the degree-three map to the projective line now givesand therefore

PutBy the Jacobian criterion, a projective plane curve is singular only at a common projective zero of $F$ and all its partial derivatives. HereTheir simultaneous vanishing forces $X_0=X_1=X_2=0$, which is not a point of projective space. Hence the Fermat quartic $C$ is nonsingular. In particular it is a smooth plane quartic and has geometric genus $(4-1)(4-2)/2=3$.

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 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 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.

Over the complex numbers the displayed projective plane curve has precisely two singular points, $[1:0:1]$ and $[-1:0:1]$. At either, with $X=\pm 1+u$, $Y=v$, $Z=1$, the leading quadratic term is $4u^2\pm v^2$, so they are ordinary double points. The derivative equations in the chart $Y=1$ have no solution: writing $d=X^2-Z^2$ forces both $d^2=-1/16$ and $d^2=1$. The Bézout theorem proves irreducibility, since a factorization into degrees $r,4-r$ would require at least three intersections, whereas the only possible component intersections are the two transverse ordinary double points.

The scheme-theoretic intersection of a projective hypersurface with a projective plane. If the plane is not contained in the hypersurface, restricting the defining homogeneous polynomial gives a projective plane curve of the same degree, with possible repeated components.

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.

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.

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.