ideal(RWeilDivisor) -- calculate the corresponding module of a divisor and represent it as an ideal

Synopsis

Function: ideal
Usage:

ideal( D1 )
Inputs:
- D1, an instance of the type RWeilDivisor,
Outputs:
- an ideal,

Description

Get an ideal associated to $O(-D)$ of a Weil Divisor $D$. Recall that on an affine scheme, $O(-D)$ is by definition the subset of the fraction field made up of elements such that $Div(f) - D \geq 0$. If $D$ is effective, this will produce the ideal corresponding to $O(-D)$. Otherwise, it will produce some ideal isomorphic to a module corresponding to $O(-D)$.

i1 : R = QQ[x, y, u, v] / ideal(x * y - u * v);

i2 : D1 = divisor({1, -2, 3, -4}, {ideal(x, u), ideal(x, v), ideal(y, u), ideal(y, v)})

o2 = -4*Div(y, v) + Div(x, u) + -2*Div(x, v) + 3*Div(y, u)

o2 : WeilDivisor on R

i3 : ideal( D1 )

             4 2     3 2   2 2 2     4    2 4
o3 = ideal (u v , y*u v , y u v , x*u v, x u )

o3 : Ideal of R

i4 : D2 = divisor(ideal(x,u))

o4 = Div(x, u)

o4 : WeilDivisor on R

i5 : ideal(D2)

o5 = ideal (u, x)

o5 : Ideal of R

Note, if the divisor has non-integer coefficients, their ceilings will be taken, since $O(-D) = O(floor(-D)) = O(-ceiling(D))$.

i6 : R = QQ[x,y,z]/(ideal(x^3 + y^3 - z^3));

i7 : D1 = 1.3*divisor(ideal(x, y-z))

o7 = 1.3*Div(x, y-z)

o7 : RWeilDivisor on R

i8 : ideal D1

                    2
o8 = ideal (y - z, x )

o8 : Ideal of R

i9 : I1 = ideal (ceiling(D1))

                    2
o9 = ideal (y - z, x )

o9 : Ideal of R

i10 : I2 = ideal (-ceiling(D1))

              2          2   2
o10 = ideal (y  + y*z + z , x )

o10 : Ideal of R

i11 : reflexify(I1*I2)

             2
o11 = ideal x

o11 : Ideal of R

The output value of this function is stored in the divisor's cache.

Ways to use this method: