research33 min studyLesson 567 in Mathematics

Divisors, line bundles, and linear systems

Relate local equations and invertible sheaves while keeping Cartier divisors distinct from codimension-one Weil data.

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What you’ll understand

  • Check the nonzerodivisor condition for an effective Cartier divisor.
  • Distinguish Cartier local equations, Weil multiplicities, and their equivalence on smooth curves.
  • Use sections of a line bundle to describe a linear system and its base points.

A divisor can describe where a rational function vanishes, but several different definitions coexist because singularities and zero divisors matter. Cartier divisors record locally compatible meromorphic equations, while Weil divisors record codimension-one components with integer multiplicities. A line bundle connects these descriptions to global sections and maps into projective space. We will state the setting of each construction rather than assume that every codimension-one subset has one good equation. The elementary projective-line calculation will also show how a nontrivial-looking list of zeros and poles can determine a trivial line bundle.

A principal equation must not be a zero divisor

An effective Cartier divisor is a closed subscheme whose ideal sheaf is invertible. Locally it is defined by one nonzerodivisor f, so multiplication by f embeds the local ring into itself. Being locally principal is insufficient: a zero divisor can vanish along an entire irreducible component and does not give the required invertible ideal. The condition concerns the ambient local ring, not whether a derivative of f vanishes or whether a drawing looks like a hypersurface.

A general Cartier divisor is specified by invertible meromorphic local equations whose ratios on overlaps are regular units. Its associated invertible sheaf O(D) glues using these ratios. For an effective divisor there is a regular section of O(D) whose zero scheme recovers it, and the ideal sheaf is O(−D). The zero scheme can be nonreduced: a square of a parameter in a domain is a nonzerodivisor even though its vanishing has multiplicity two.

0⟶OX(−D)⟶OX⟶OD⟶0,OX(−D)∣U=(f)≅OU(f a nonzerodivisor)0\longrightarrow\mathcal O_X(-D)\longrightarrow\mathcal O_X\longrightarrow\mathcal O_D\longrightarrow0,\qquad \mathcal O_X(-D)|_U=(f)\cong\mathcal O_U\quad(f\text{ a nonzerodivisor})
This exact sequence is for an effective Cartier divisor. The ideal (f) is free of rank one because multiplication by f is injective, not merely because it has one generator.↔ Scroll wide equations horizontally.

Weil multiplicities need an appropriate comparison setting

On a Noetherian integral scheme a Weil divisor is a finite integer combination of codimension-one integral closed subschemes. If the scheme is only locally Noetherian, locally finite combinations are needed instead. On a normal such scheme the local rings at codimension-one points are discrete valuation rings, giving orders of zeros and poles of rational functions. Normality does not imply that every Weil divisor is Cartier: a codimension-one combination can fail to admit compatible local principal equations near singular points.

On a smooth integral curve over an algebraically closed field, the local rings at closed points are discrete valuation rings and the two divisor notions agree. A divisor is a finite sum of points, and its degree is the sum of coefficients in this field setting. Over a non-algebraically-closed field, closed-point residue degrees must weight that sum. Principal divisors have degree zero on a smooth projective curve. Linear equivalence quotients by such principal divisors and identifies divisors that give isomorphic associated line bundles.

H0(C,OC(D))={g∈k(C)×:div⁡(g)+D≥0}∪{0},OC(D+div⁡(h))≅OC(D)H^0(C,\mathcal O_C(D))=\{g\in k(C)^\times:\operatorname{div}(g)+D\ge0\}\cup\{0\},\qquad \mathcal O_C(D+\operatorname{div}(h))\cong\mathcal O_C(D)
Here C is a smooth projective integral curve. A section may be represented by a rational function with poles bounded by D; the zero element is included separately from nonzero rational functions.↔ Scroll wide equations horizontally.

Global sections produce maps only after checking base points

Choose a finite-dimensional vector space V of global sections of a line bundle L. A point is a base point when every section in V vanishes in its fiber. If V generates L everywhere, the sections define a morphism into a projective space, with the usual choice of dual convention. Without that generation property they define only a rational map on the open where not all values vanish. Counting sections alone does not prove that the map separates points, separates tangent directions, or is an embedding.

For the projective line, the sections of O(2) are homogeneous quadratics. A basis gives the familiar conic map [s:t]↦[s²:st:t²], illustrating the relation between a line bundle and a projective presentation. Choosing a smaller subspace can introduce base points or identify points, even though the underlying line bundle has not changed. Consequently the divisor class, the line bundle, the chosen space of sections, and the resulting morphism are related objects with distinct information.

Recognize a principal degree-zero divisor

On P¹ over an algebraically closed field of characteristic zero, with affine coordinate t, let D=2[0]+[2]−3[∞]. Find a rational function with divisor D and describe H⁰(P¹,O(D)).

  1. The function h=t²(t−2) has a zero of order two at 0 and a zero of order one at 2. Using u=1/t near infinity, h=u^(−3)(1−2u), so its pole there has order three and there are no further zeros or poles.

  2. Therefore div(h)=D and the degree is 2+1−3=0. A principal Cartier divisor gives a trivial associated line bundle, even though its coefficients are not individually zero.

  3. A section g of O(D) satisfies div(g)+div(h)≥0, so gh is a global regular function on P¹ and is constant. Hence every section is c/h, including c=0, and H⁰(P¹,O(D)) has dimension one.

The result

The line bundle is trivial, with the rational-function representative 1/[t²(t−2)] as a nowhere-vanishing generator in its local trivializations.

Explore the model

Interactive lab

Interactive lab · function

Inspect multiplicity without declaring a singular ambient space

Pause and Reset; set Coefficient a=1,Position x=0 and predict value 0 and derivative 0 for x². For a fresh comparison Reset and set a=1,x=1, predicting value 1 and derivative 2. Algebraically x² is a nonzerodivisor in k[x], so its double zero defines an effective Cartier divisor despite the vanishing derivative at zero. The real graph displays values and slopes, not ideal-sheaf invertibility, Weil classes, or a general divisor algorithm.

Inspect multiplicity without declaring a singular ambient spaceThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forMultiplicity of a zero and the nonzerodivisor condition on its ambient ring are different tests.

Check your understanding

Why does the equation x fail to define an effective Cartier divisor on Spec k[x,y]/(xy)?

Try it yourself

In A=k[x,y]/(xy), decide whether x+y is a nonzerodivisor and compute A/(x+y). What does this show about reducedness of an effective Cartier divisor?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Restrict to the two component domains, or substitute y=−x after imposing the additional equation.

Compare with the solution

If (x+y)g=0, reducing modulo (x) gives yg(0,y)=0 in k[y], and reducing modulo (y) gives xg(x,0)=0 in k[x]. Thus g lies in both (x) and (y), whose intersection is zero in A; x+y is a nonzerodivisor. The quotient is k[x]/(x²), by substituting y=−x. It defines an effective Cartier divisor with nonreduced length-two structure. Cartier does not mean reduced.

Take it with you

  • An effective Cartier equation must be a nonzerodivisor, not merely a single generator.
  • Normal Weil divisors need not be Cartier; on smooth curves the local valuation descriptions agree.
  • A linear system needs a base-point check before its sections define a global projective morphism.

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