Page 131 - Geometric Modeling and Algebraic Geometry

132 T. Beck and J. Schicho
7.5.4 Inverting the canonical divisor

(−K ∅ ), the linear system associated to the

We want to compute a K-basis for L C
anticanonical divisor. Therefore choose l 1 ,k 1 ,l 2 ,k 2 ∈ Z/nZ s.t. l 2 = k 1 +1,
[l 1 ,k 2 ] =[1,n] and for each pair d [l j ,k j ] ≥ 2. By what was just said we can compute
(K [l j ,k j ] ). For every valuation ν we define the twisted orders

elements 0 = g j ∈L C
β j,ν := ν [l j ,k j ] (g j ).Now let S ⊂ C be the image of the support of the (effective) di-

visor (g 1 ) [l 1 ,k 2 ] +(g 2 ) [l 1 ,k 2 ] . This means the twisted orders are zero for all valuations
except those of V S . More precisely we have

(g j )= −D [l j ,k j ] +(g j ) [l j ,k j ] = −D [l j ,k j ] + β j,ν P ν .

ν∈V S
(K [l j ,k j ] ) we further have the inequality β j,ν ≥ α ν . We will show that

Since g j ∈L C
(−K ∅ ) in the sense that the according

g 1 g 2 is a good denominator for computing L C
(K [l 1 ,k 2 ] ), a space that can again be computed by

numerators are all elements of L C
means of the above theorem (for an explicit computation see 7.7.4 in the example
section).
Theorem 12. With the choices from above define the subspace
(K [l 1 ,k 2 ] ) | ν [l 1 ,k 2 ] (h) ≥ β 1,ν + β 2,ν − α ν ∀ν ∈V S }.

V := {h ∈L C
(−K ∅ ) → V : k → kg 1 g 2 is an isomorphism of K-vector spaces.

Then L C
h
Proof. Any rational function k can be written as k = g 1 g 2 for some other rational
(−K ∅ ) if and only if

function h.Now k ∈L C

h

0 ≤ g 1 g 2 − K ∅

=(h) − (g 1 ) − (g 2 ) − K ∅

=(h)+(D [l 1 ,k 1 ] + D [l 2 ,k 2 ] − D ∅ )+ (−β 1,ν − β 2,ν + α ν )P ν

ν∈V S

=(h)+ D [l 1 ,k 2 ] + (−β 1,ν − β 2,ν + α ν )P ν

ν∈V S

H:=
(H) ⊂

if and only if h ∈L C (H). But −β 1,ν − β 2,ν + α ν ≤−α ν and therefore L C

(H).
L C (K [l 1 ,k 2 ] ). The exact calculation above shows that V = L C
(−K ∅ )) = 3 and if {b 1 ,b 2 ,b 3 }⊂ V is a K-basis

K K
In particular dim (V )=dim (L C
!
then g 1 g 2 , g 1 g 2 , g 1 g 2 is a K-basis of L C (−K ∅ ). In an actual computation, we will

b 3
b 1
b 2
of course again start from L S (D [l 1 ,k 2 ] ) and impose directly the vanishing conditions
of this theorem. When executing the algorithm in section 7.7 we will see that the
output basis will already be in K(C).