In this tutorial we describe one way to represent divisors on a smooth projective subvariety $X$ of $\PP^r$, and show methods for computing the group operations, computing the vector space of sections, and determining whether two divisors are linearly equivalent. We also construct the canonical divisor on $X$.
We consider smooth varieties only, although most of this can be extended to normal varieties. Cartier and Weil divisors on normal varieties might be the subject of a further tutorial.
Other possible future topics would be: intersection numbers, determining whether a divisor is very ample, and finding the base point locus of the divisor class.
The simplest case is when the homogeneous coordinate ring $S_X$ of $X$ satisfies the $S_2$ condition of Serre: We say that a domain $R$ is $S_2$ if every proper nonzero principal ideal has pure codimension 1 (all associated primes of the ideal are of codimension 1).
In this tutorial, we consider the case when this holds (e.g., this holds for complete intersections). In a further tutorial, we will make the necessary extensions to handle the non $S_2$case.
An example that we will use throughout is the plane cubic curve $E$, whose homogeneous coordinate ring is SE:


The sections in this tutorial are A. Representation of divisors
B. Group operations on divisors
C. Global Sections
D. Linear Equivalence
E. The canonical divisor
Let $X$ be a smooth irreducible variety. A (Weil) divisor on $X$ is an integral linear combination of irreducible subvarieties of $X$ of codimension $1$. The divisor is called effective if all the coefficients are nonnegative. To any ideal $I$ in the homogeneous coordinate ring $S_X$ of $X$ we associate the effective divisor that is the sum of the pure codimension $1$ components of $I$, each taken with the multiplicity it has in the primary decomposition of $I$.
Let $D = E  F$ be a divisor, where $E$ and $F$ are effective. Because $X$ is normal, there is a unique homogeneous ideal $I$ in $S_X$ such that $V(I) = E$, and $I$ is either $(1)$, or has pure codimension one. Similarly, there is a unique such ideal $J$ with $V(J) = F$. Our plan is to represent the divisor $D$ by the pair of ideals $(I,J)$.
This representation is not unique. If $(I,J)$ and $(I',J')$ are two pairs of ideals (such that each ideal is either $(1)$ or has pure codimension one), then $(I,J)$ and $(I',J')$ represent the same divisor iff $$sat(I J') = sat(I' J),$$ where $sat(K)$ is the saturation of $K$ (the largest ideal $L$ such that a power of the irrelevant ideal times $L$ is in $K$) Write $(I,J) \equiv (I',J')$ if $sat(I J') = sat(I' J)$.
This correspondence defines a bijection between $Div(X)$ and $\{(I,J) \mid I,J$ are homogeneous ideals in $S_X$ either trivial, or pure codim one$\}/\equiv$.
As we will often have to saturate ideals of codimension 1, we give here the most efficient method we know, which has the additional advantage of throwing away all components not of codimension 1. That is, we define purify1S2(I), a function that takes an arbitrary ideal $I$ in a ring satisfying $S_2$, and returns the ideal which is the intersection of the codimension 1 primary components of $I$. In the next divisor tutorial (not yet written), we will write a routine purify1(I) which does not require the ring to be $S_2$.

For example, in the ring

we have

B Unfortunately at this moment the quotient code B makes this slow. Another possibility is B purify1S2 = saturation B which is (9/23/96) slightly faster (88 vs 106 sec).
Throughout this tutorial, we will treat divisors as equivalence classes of pairs, and our operations will operate on pairs. So let's define a divisor type in Macaulay2. The following declaration provides a new data type, the Divisor.

Let's write a routine to create a divisor, from either a single ideal, or a pair of ideals. (This routine should check that its arguments are pure codimension one, or trivial, and in the same ring, but we will ignore that).
Defining divisor to be a method allows us to define different versions of this routine which take different arguments.

The following allows us to define an object of class Divisor from a pair of ideals.

The following routine defines an (effective) divisor from a single ideal.

The divisors of some rational points on the elliptic curve $E$ include





Testing equality of divisors is often made simpler by having a ``normal form'' for divisors. The normal form of a divisor $D$ is $E  F$ where $E$ and $F$ are both effective and have disjoint support. It is easy to see that the normal form of $(I,J)$ is $(I:J, J:I)$.
In the following code, the expressions D\#0 and D\#1 refer to the first and second ideals in the list representing $D$. (D\#0 is the first because Macaulay2 counts everything starting from 0.)


Two pairs $(I,J), (I',J')$ define the same divisor exactly when their normal forms are equal. The following code establishes a method for testing the equality of divisors. The function "toList" converts a divisor to a list of ideals, and then we let the built in method for comparing lists take over: it compares corresponding elements.

An important point here is that the built in method for comparing two ideals used by the operator "==" succeeds even if the given generators differ. We shall later show that with R1 and {\tt R2} as above, the divisor (R1 + R2)  R1 is represented by

so that the normal form of $D$ is {\tt R2}:

and we can directly test equality by

To add divisors we multiply the corresponding ideals and then saturate. This may be coded as follows (the products are saturated in the divisor routine):

Negation is even simpler, since all we need do is exchange the two ideals. We don't use the divisor routine, since our ideals are already saturated.

Let's also include functions to compute differences and to multiply by integers.


Some arithmetic of divisors on our elliptic curve


Notice that $3P$ is the hyperplane section $z=0$, which is the equation of the flex line to the cubic at the point $P$.


Since we have assumed $X$ smooth, Weil divisors can all be represented by Cartier divisors, that is, by sections of an invertible sheaf. If $D = (I,J)$ is a divisor, and $sheaf(I)$ denotes the sheaf of $O_X$modules corresponding to $I$, then we put $$O_X(D) = sheaf(I)^{1} \otimes sheaf(J).$$
We define $L(D)$ to be the space of global sections of the sheaf $O_X(D)$. Note that a global section is the same as a sheaf homomorphism $O_X \rightarrow{} O_X(D)$. If we write $D = EF$, where $E$ and $F$ are effective, then global sections of $O_X(EF)$ can be identified with homomorphisms $O_X(E) \rightarrow{} O_X(F)$.
If we write $D = (I,J)$, then $L(D)$ and $Hom(I,J)$ can be identified with subsets of the field of fractions of $S_X$. Since $S_X$ satisfies $S_2$, these sets are equal. The following proposition allows us to compute $Hom(I,J)$: Proposition. Suppose $X$ is a smooth projective variety whose homogeneous coordinate ring $S_X$ is $S_2$. If $D$ is the divisor $(I,J)$ and $f$ is any nonzero element of $I$, then $L(D)$ is the degree zero part of $${sat((f*J) : I) \over f}.$$
Proposition. If $s = g/f$ is section of the divisor $D = (I,J)$ as above, then the zero scheme of $s$ is defined by the ideal $$ sat(f I : g) : J.$$
Consider the divisor $2P$ on our curve $E$:

In this case, $I = (x^2, z)$, and $J = (1)$. Compute the vector space of sections $L(2P)$:



The degree 0 part in the proposition is the degree $d$ part of $sat((fJ) : I)$, divided by $f$, where $d = deg f$.
We can use the command basis to obtain a vector space basis of a module or ideal in a given degree and thus compute the global sections (For an explanation of this use of the basis routine, see the tutorial on canonical embeddings of plane curves and gonality)


so the vector space $L(2P)$ is generated by $1=z/z$, and $x/z$. Since $J = (1)$, the zero locus of the section $(z+x)/z$ is defined by the ideal

and its degree is:

Let's now package this into a routine globalSections which takes an argument D of class Divisor, and computes a basis of $L(D)$, represented as fractions with a common denominator. The output is a row vector of the numerators, followed by the denominator.


Another important task is to find the ideal of zeros of a section $s = f/g$ of a divisor $D$.

Let's find the image of the elliptic curve $E$ under the linear system $4P$. To do this we define a ring homomorphism from the global sections with the command map. Its kernel defines the image of $E$.




The image in $\PP^3$ is a complete intersection of two quadrics: the elliptic normal curve in $\PP^3$.
For a less obvious example, consider the divisor $4P  R$, which is not effective. Since it has degree 3 as a divisor on an elliptic curve, the Riemann Roch theorem tells us that it is equivalent to an effective divisor; in fact that it has three sections. We can check this as follows:




Testing whether two divisors $E$ and $F$ are linearly equivalent boils down to testing whether $D = EF$ is principal and thus linearly equivalent to 0.
One method to determine whether $D$ is principal is to compute the global sections of $D$. A divisor $D$ is principal iff $L(D)$ has dimension one, and the zero locus of its generator is the empty set.
For example, on the elliptic curve $E$, consider $P  R$:

$PR$ has no global sections, so it is not equivalent to 0. Now consider $2 P  2 R$


Since the divisor $D = 2P2R$ has degree 0 and has a section, $D$ is linearly equivalent to 0. The result shows that the rational function $x/z$ has divisor $2P2R$.
To check that a divisor of unknown degree is equivalent to 0, we attempt to find a section and show it does not vanish anywhere. We include this in the routine below.
Remember that in this tutorial we are assuming that $S_X$ is $S_2$ and that $X$ is smooth. These computations are easily modified in the non$S_2$ case. See the corresponding tutorial, once it is written!

We get the same answers as before:


We now look at the group law on the cubic: We take the point $P$ to be 0; we can then identify the natural group of divisor classes of degree 0 with the set of points on the curve. With this identification, the group law $++$ on points of the curve is defined by: $R ++ S =$ the unique point $T$ for which the divisor $(RP)+(SP)$ is linearly equivalent to $(TP)$. i.e. $R ++ S := $ unique effective divisor in $R+SP$.
What we need to do is: given a divisor $R+SP$, find an effective divisor equivalent to it.




Some points are in the torsion subgroup:




So the point $Q_1 = (3,6,1)$ is a point of order 4 in the group.
Exercise: Write a routine that computes $n$ times a point in this group law.
The most important divisor class on a variety is the canonical class. For example, consider the twisted cubic curve whose ideal is the ideal of $2\times 2$ minors of the ``catalecticant'' matrix




As a graded module, the canonical class is given as $K_X = Ext^c(S_X, S(r1))$, where $c = codim X$, $X \subset \PP^r$, and $S = k[x_0,\ldots,x_r]$ is the polynomial ring.



Thus we need a routine that takes a rank 1 torsion free module over a domain and finds an ideal isomorphic to it. We wish to compute homomorphisms from the canonical module into $S_X$, and take the divisor whose first ideal is the image of a homomorphism of nonnegative degree, and whose second ideal is an arbitrary nonzero element of $S_X$ whose degree is equal to the degree of the homomorphism. First we find a homomorphism of lowest degree:

The degree is

We need to balance the degree dg with a power of the first nonzero generator of the ring. This is done in the following packaged version.

We start from a module over the ring SX:


Some tests:



Here is the canonical divisor routine in packaged form:


There are other ways of computing the canonical class. Perhaps we have already written a tutorial on this subject.