Divisor on an algebraic curve

A divisor is a finite formal integer combination of closed points. For a divisor $D$,

The Riemann-Roch space of a divisor $D$ is the finite-dimensional vector spaceIts elements are the rational functions whose poles are bounded by the negative coefficients of $D$.

On a genus-one curve, for distinct $p,q,r$, the Riemann-Roch theorem gives $\ell (p+q+r)=3$. Choose $f$ with simple poles exactly at $p,q$ and $g$ with simple poles exactly at $q,r$. Pole comparison proves $1,f,g$ independent. Multiplying these coordinates by a vanishing local parameter extends $F$ across every pole, with images $[0:1:0]$, $[0:b:c]$, $[0:0:1]$, where $bc\ne 0$. The ten homogeneous cubic monomials evaluate in $L(3p+3q+3r)$, of dimension nine, so a nonzero homogeneous cubic vanishes on the image. The three chosen points map to the line with first coordinate zero. This connects linear systems to projective plane equations by a dimension count.

Let $P=p_1+\cdots +p_k$ and $Q=q_1+\cdots +q_k$ be disjoint sums of distinct points on a smooth projective curve. A nonzero function with principal divisor $P-Q$ exists exactly when $L(Q-P)\ne 0$: a nonzero member makes $(f)+Q-P$ an effective divisor of degree zero, forcing equality. Evaluation $f\mapsto (f(p_1),\dots ,f(p_k))$ on $L(Q)$ has kernel of a linear map $L(Q-P)$.

If $\ell (Q)=k$, a choice of basis $f_1,\dots ,f_k$ turns the criterion into $\det (f_j(p_i))=0$. This determinant is not identically zero. Starting with $L(Q)$, repeatedly choose a point where some nonzero function in the current space does not vanish; evaluation then lowers the dimension of that space by one. After $k$ choices the joint kernel is zero. Thus a generic tuple of points gives an invertible evaluation matrix, and no function with divisor $P-Q$. On a genus one curve, Riemann-Roch theorem gives $\ell (Q)=k$, so this obstruction applies for every $k>0$.

If a divisor on an algebraic curve $D$ on a smooth projective curve has negative degree of a divisor, then its Riemann-Roch space is zero. Otherwise a nonzero $f\in L(D)$ would make $(f)+D$ an effective divisor of degree $\mathrm{deg}D<0$, since a principal divisor has degree zero. This is impossible. In genus one the Riemann-Roch theorem consequently gives $\ell (D)=\mathrm{deg}D$ for every divisor of positive degree.

For finitely many distinct points $P_i$ on a smooth projective genus-$g$ curve, choose $N\ge 2g+1$. The Riemann-Roch theorem gives $\ell (N{P}_i)-\ell ((N-1)P_i)=1$, so choose $f_i$ with a sole pole of exact order $N$ at $P_i$. Then ${\sum }_i{f}_i$ has exactly those poles with that order: all other summands are regular at each $P_i$, so no principal part can cancel. This is an explicit construction rather than an unspecified generic choice of section.

For $D={\sum }_P{n}_{P}P$ on a curve over an algebraically closed field, its degree is $\mathrm{deg}D={\sum }_P{n}_P$.

Two divisors $D$ and $E$ are linearly equivalent, written $D\sim E$, when $D-E$ is a principal divisor. Multiplication by a rational function whose divisor is $D-E$ gives an isomorphism between their spaces of sections, so $\ell (D)=\ell (E)$.

A divisor class is an equivalence class of divisors under linear equivalence. Degree descends to divisor classes because every principal divisor has degree zero.

For a nonzero rational function $f$ on a smooth projective curve $C$, its principal divisor isIt records the zeros of $f$ with positive multiplicity and its poles with negative multiplicity. Every principal divisor has degree zero.

For $E=\mathbb{C}/\Lambda$, integrate the logarithmic derivative of an elliptic function and its product with $z$ around a fundamental cell. The first integral forces total degree zero; the second forces the weighted point sum to lie in the lattice. Conversely choose lifts of the points with equal zero and pole sums, and form the balanced ratio of translated Weierstrass sigma functions. Their quasi-period factors cancel, giving an elliptic function with exactly the prescribed divisor. Separate zero and pole divisors require disjoint support; common terms cancel when only their difference is prescribed.

Every degree-zero divisor on the projective line is principal. Indeed, if $D={\sum }_i{a}_i{p}_i$ and $L_i$ is a homogeneous linear form vanishing at $p_i$, then ${\sum }_i{a}_i=0$ makesa degree-zero rational function with $(f)=D$. Consequently, two divisors on ${\mathbb{P}}^1$ are linearly equivalent exactly when they have the same degree.

If distinct points $p,q$ on a smooth projective curve satisfy $(f)=p-q$, then $f:C\to {\mathbb{P}}^1$ is a finite morphism of degree one. Such a morphism between smooth projective curves is an isomorphism, so $C$ has genus zero. Hence no divisor $p-q$ on a curve of positive genus is principal.

For an elliptic curve with identity $O$, a divisor $D={\sum }_P{n}_P(P)$ is principal exactly whenThis is the identification of the degree-zero divisor class group with the elliptic curve.

For a smooth projective curve of genus $g$ and a canonical divisor $K$,

A nonidentity holomorphic automorphism of a compact connected Riemann surface of genus $g$ has at most $2g+2$ fixed points. If there were more, choose $g+1$ distinct fixed points and let $D$ be their sum. The Riemann-Roch theorem gives $\ell (D)\ge 2$, so there is a nonconstant meromorphic function with at most simple poles at these points and no other poles. The function $h=f-f\circ \phi$ has at most $g+1$ poles and at least $n-g-1$ zeros at the remaining fixed points. Since a nonzero principal divisor has degree zero, $n>2g+2$ forces $h=0$. At an actual pole fixed by $\phi$, the reciprocal $1/f$ is a local coordinate because its zero is simple. The identity $f=f\circ \phi$ therefore forces $\phi$ to be the identity near that pole, and the identity theorem on a Riemann surface makes it the identity everywhere.

For a closed point $p$ on a smooth projective curve, the quotient in $0\to L(-p)\to L\to L{|}_p\to 0$ has $k$-dimension $[k(p):k]$. Additivity of Euler characteristic therefore changes $\chi (L)$ by that residue-field degree. Every line bundle is associated with a divisor, so iteration proves the displayed formula. The dualizing sheaf on a smooth projective curve identifies $h^1(L)$ with $h^0({\Omega }_C^1\otimes L^{-1})$ and converts this Euler-characteristic formula into the usual divisor form of Riemann-Roch theorem.

On a smooth projective curve, the exact sequence $0\to \mathcal{O}(D-P)\to \mathcal{O}(D)\to k(P)\to 0$ increases the Euler characteristic by one. Iteration for positive and negative coefficients gives $\chi (\mathcal{O}(D))=\mathrm{deg}D+1-g$. Serre duality identifies $h^1(\mathcal{O}(D))$ with $h^0(\mathcal{O}(K-D))$, proving $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$.

If a line bundle on a smooth projective curve has both $h^0(\mathcal{L})>0$ and $h^0({\omega }_C\otimes {\mathcal{L}}^{-1})>0$, the displayed inequality holds. Write these dimensions as $r,s$. The product dimension bound for sections on a curve gives $r+s-1\le g$, and the Riemann-Roch theorem gives $r-s=\mathrm{deg}\mathcal{L}+1-g$. Adding gives $2r\le \mathrm{deg}\mathcal{L}+2$.

At a point $p$ of a smooth projective curve, a positive integer $n$ is a Weierstrass gap when no rational function has a pole of order exactly $n$ at $p$ and no other poles. Equivalently,

At every point of a smooth projective curve of genus $g$, exactly $g$ positive integers are Weierstrass gap. Indeed, each quotient $L(np)/L((n-1)p)$ has dimension zero or one, while Riemann--Roch gives $\ell (np)=n+1-g$ for all $n>2g-2$.

On a genus-two smooth projective curve, the gap set at any point is either $\{1,2\}$ or $\{1,3\}$. The integer $1$ is always a gap on a positive-genus curve. If $2$ is not a gap, then $3$ must be a gap, since pole orders are closed under addition and $2,3$ would otherwise generate every integer at least two.

A canonical divisor is the divisor of any nonzero rational differential. Its divisor class is independent of the differential and has degree $2g-2$.

A rational differential on a smooth projective curve is a rational section of its canonical bundle. In a local coordinate $t$ it has the form $f(t)dt$ for a rational function $f$.

At a smooth point with local parameter $u$, expand a rational differential as ${\sum }_j{a}_j{u}^{j}du$ in the completed local field; its residue is $a_{-1}$. Formal substitution shows this coefficient is independent of the parameter, in arbitrary characteristic. For a finite separable extension of these local fields, residues commute with the differential trace, with a sum over branches. The monogenic trace formula in the trace-dual module of a finite curve map proves this compatibility by coefficient extraction.

If $t$ is a uniformizer at a point $p$ of a smooth projective curve and a nonzero rational differential is $\omega =f(t)dt$, then ${\nu }_p(\omega )={\nu }_p(f)$. This integer is independent of the chosen uniformizer. Its positive and negative values are respectively the orders of the zero and pole of $\omega$ at $p$.

If $K_X=(\omega )$ is represented by a nonzero rational differential, multiplication by $\omega$ identifies the Riemann-Roch space $L(K_X)$ with the vector space of holomorphic differential forms. The Riemann-Roch theorem gives $\ell (K_X)=g$ for a smooth projective curve of genus $g$.

For the affine coordinate $t$ on ${\mathbb{P}}^1$, put $s=t^{-1}$ at infinity. Since $dt=-s^{-2}ds$,so $K_{{\mathbb{P}}^1}\sim -2\infty$. More generally,because $L(d\infty )$ consists of polynomials of degree at most $d$ when $d\ge 0$ and only the zero function when $d<0$.

A smooth plane curve of degree $d$ has genus

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 $d$.

A smooth plane quartic is a smooth degree-four curve in ${\mathbb{P}}^2$ and has genus three.

The smoothness condition excludes the singular degree-four curves considered in the general plane-quartic family.

The projective closure of the superelliptic curve $y^3=x^4+1$ is the smooth plane quarticIts unique point at infinity is $P_{\infty }=[0:1:0]$. The rational function $x=X/Z$ defines a degree-three morphism to ${\mathbb{P}}^1$ ramified with index three at $P_{\infty }$ and at the four affine points $(\alpha ,0)$ with ${\alpha }^4=-1$. Its genus is three, and

The Klein quartic has the smooth projective plane modelIts affine chart $Z=1$ is $x^{3}y+y^3+x=0$. If $\zeta$ is a primitive seventh root of unity, the projective transformationis an automorphism of order seven.

The projective completion ofis the smooth plane quarticThe coordinate relabelling $[A:B:C]=[Y:X:Z]$ turns this equation into the standard Klein quartic equation $A^{3}B+B^{3}C+C^{3}A=0$.

The coordinate function $x$ has degree three on the Klein quartic. It has ramification index three at $(0,0)$, index two at the seven affine points satisfying $y^7=-3/8$ and $x=2y^3$, and index two at $[0:1:0]$. The total ramification is ten, so the Riemann-Hurwitz formula gives genus three.

For a smooth plane quartic $C\subset {\mathbb{P}}^2$, the adjunction formula gives $K_C\cong {\mathcal{O}}_C(1)$. Hence every line section $C\cap \ell$ is a canonical divisor.

Projection from a point of a smooth plane quartic gives a degree-three map to the projective line. A degree-two map would make the curve hyperelliptic, but its canonical map is its plane embedding, whereas a hyperelliptic canonical map is not an embedding. Its gonality is therefore three.

For $g\ge 2$, a basis of the space of regular differentials defines the canonical map $X\to {\mathbb{P}}^{g-1}$. Equivalently, after choosing a differential with divisor $K$, one may use a basis of $L(K)$.

For a nonhyperelliptic smooth projective curve of geometric genus four, the canonical map is an embedding of degree six. Dimension counts $10>h^0(2K)=9$ and $20>h^0(3K)=15$ supply a quadric and a cubic not divisible by it. The quadric is irreducible because the curve is nondegenerate. Degree six and unmixedness of a complete intersection then identify the intersection with the curve as a scheme.