Algebraic curve

An algebraic curve is a one-dimensional algebraic variety. On a smooth algebraic curve, the local ring at every closed point is a discrete valuation ring, which allows orders of zeros and poles to be recorded as a divisor on an algebraic curve.

For $a>0$, this algebraic curve can be constructed using a circle of radius $a$ centred at $(a,0)$ and its tangent line $x=2a$. On a ray from the origin, subtract the distance to its nonzero circle intersection from the distance to the line. The resulting radial distance is $r=2a{\sin }^2\theta /\cos \theta$, for $-\pi /2<\theta <\pi /2$. Thus a parametric curve for it isEliminating $t$ gives the displayed cubic equation. It has a cusp at the origin, horizontal tangent direction there, and vertical asymptote $x=2a$. Its name honours Diocles.

An algebraic curve which is projective, reduced and irreducible. Over an algebraically closed field, its only global regular functions are constants and its arithmetic genus is $h^1({\mathcal{O}}_C)\ge 0$.

For an affine plane algebraic curve $F(x,y)=0$, a point over an algebraic closure is nonsingular when the displayed condition holds. Its Zariski tangent space has dimension one, and a local coordinate can be chosen in the direction of a nonzero partial derivative. A smooth algebraic curve has all points nonsingular. Over a complete local field, the Hensel lemma lifts a nonsingular residue point by solving for a coordinate with unit partial derivative.

For a projective plane curve $C=V(F)$ and a nonsingular projective point $p$, the tangent is the line $F_x(p)x+F_y(p)y+F_z(p)z=0$. The coefficients are not all zero, and the Euler theorem for homogeneous functions makes this line pass through $p$. On an affine chart this is precisely the first-order part of a local defining equation, so the definition agrees with the usual tangent line and is invariant under projective transformations.

An affine algebraic curve is an affine variety of Krull dimension one. An irreducible affine curve has a finitely generated one-dimensional domain as its coordinate ring. Quotienting this ring by a nonzero element gives a zero-dimensional Noetherian ring, hence an Artinian ring. Consequently a nonzero regular function has finitely many zeros. No nonsingularity assumption is part of the definition.

For nonzero real $a$, the folium of Descartes is the algebraic curve $x^3+y^3=3axy$, with rational parametrization of an algebraic curveIt is symmetric about $y=x$ and has a node at the origin, with the coordinate axes as tangents. For $a>0$, $0\le t<\infty$ traces a loop in the first quadrant, returning to the origin as $t\to \infty$. Its enclosed area, by Green's theorem, isThe other branches approach the asymptote $x+y+a=0$ as $t\to -1$. Negative $a$ rotates the positive-$a$ picture through $\pi$.

The projective algebraic curve $D=\{y^{2}z=x^3\}\subset {\mathbb{P}}^2$ is an irreducible cubic with a cusp at $[0:0:1]$. In the chart $z=1$, its equation is $y^2=x^3$, with parametrization $(x,y)=(t^2,t^3)$. Its affine version $y^2=z^3$, used below, differs only by renaming a coordinate. At the point $[0:1:0]$, the chart $y=1$ gives $z=x^3$, which is smooth and exhibits its unique smooth flex of a cuspidal cubic.

Parametrize the affine cuspidal cubic $x^2=y^3$ by $x=t^3,y=t^2$. Removing the zero set of $g=x-y$ removes exactly the parameter values zero and one. Its regular-function ring is the localization $k[t^2,t^3{]}_g$. To recover the normalization coordinate inside this ring, use $g^2=y(y^2-2x+y)$: inverting $g$ makes $y$ a unit, so $t=x/y$ lies in the localization. As $g=t^2(t-1)$ is a unit, so are $t$ and $t-1$, proving the displayed equality. This calculation is valid in every characteristic. At the smooth point $(1,1)$ of the original curve the local ring is instead $k[t{]}_{(t-1)}$: $y$ is a unit there, and every polynomial in $t$ nonzero at one is invertible. That point is absent from the displayed open subset, so its local ring is computed in the original curve.

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.

The algebraic set in ${\mathbb{A}}^3$ defined byis the disjoint union of the cuspidal cubic $V(x,y^2-z^3)$ and the two reduced points $(-1,0,0)$ and $(1,0,-1)$. Its radical vanishing ideal is the intersection of the three corresponding prime ideals.

The scheme $C^{(d)}$ parametrizes effective degree-$d$ Cartier divisors on a smooth algebraic curve. Near a Cartier divisor $\sum m_i{P}_i$, its completed local ring is a power-series ring in the elementary symmetric coordinates of $m_i$ local parameters at each $P_i$. It is therefore smooth of dimension $d$, even when the characteristic of a field divides $d!$. For projective $C$ it is projective.

Over a field $k$, an affine curve parametrized by positive integer powers of one parameter. Its coordinate ring is the subalgebra generated by those powers, and its defining prime ideal is the kernel of $k[x_1,\dots ,x_n]\to k[t]$ sending $x_i$ to $t^{a_i}$. Equal exponent sums yield binomial relations. Studying those relations connects an additive semigroup of exponents to the geometry of the curve.

The kernel of $x\mapsto t^3$, $y\mapsto t^4$, $z\mapsto t^5$ is generated by $y^2-xz$, $yz-x^3$, $z^2-x^{2}y$. Reducing these three relations leaves a unique normal form $A_0(x)+A_1(x)y+A_2(x)z$: its images have distinct exponents modulo three. The curve ring is therefore finite free of rank three over $k[x]$ and has dimension one. In a three-variable polynomial ring its ideal has height two, but its three independent quadratic initial forms force at least three generators. This is a concrete failure of generation by codimension many equations.

A rational parametrization expresses the coordinates of an algebraic curve as rational functions of a parameter, with a rational inverse on a dense open subset. It identifies the function field with $K(t)$ and the smooth projective normalization of an algebraic curve with the projective line. A nonconstant parametrizing map without an inverse is a weaker notion; even this cannot exist for an elliptic curve in characteristic zero.

The arithmetic genus of a proper curve $C$ is $p_a(C)=1-\chi (C,{\mathcal{O}}_C)$. For an integral projective curve over an algebraically closed field, $H^0(C,{\mathcal{O}}_C)=k$, so $p_a=h^1\ge 0$. Its finite normalization has $p_a(C)=g(C^{\nu })+\mathrm{length}({\nu }_{\ast }{\mathcal{O}}_{C^{\nu }}/{\mathcal{O}}_C)$. Hence arithmetic genus zero forces the curve to be smooth and rational.

For a projective complete intersection curve cut out by a homogeneous regular sequence of positive degrees $a,b$ in ${\mathbb{P}}_k^3$, the Koszul resolution gives Hilbert polynomial $abm+ab(4-a-b)/2$. Hence its arithmetic genus is $1+ab(a+b-4)/2$. The same resolution shows $H^0(C,{\mathcal{O}}_C)=k$, and the dimension of a scheme is one, so this genus equals ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$. Smoothness and irreducibility of the curve are not required. For degrees five and seven, the value is $141$.