sidebar

A Generators of the Module of Syzygies

We require the 4-tuple solutions (q3,q′1,q2′,q′3) to the equation
(1 − xyz )q + (xz − y)q ′+ x(1 − z2)q ′+ (1 − x2)q′ = 0, (114 ) 3 1 2 3
where for convenience we have substituted x = 𝒟1, y = 𝒟2, z = 𝒟3. q3, q1′, q′2, q3′ are polynomials in x, y, z with integral coefficients, i.e., in Z [x, y,z].

We now follow the procedure in the book by Becker et al. [3].

Consider the ideal in Z[x,y,z ] (or 𝒬 [x, y,z] where 𝒬 denotes the field of rational numbers), formed by taking linear combinations of the coefficients in Eq. (114*), f1 = 1 − xyz, f2 = xz − y, f3 = x (1 − z2), f4 = 1 − x2. A Gröbner basis for this ideal is

2 2 𝒢 = {g1 = z − 1,g2 = y − 1,g3 = x − yz }. (115 )
The above Gröbner basis is obtained using the function GroebnerBasis in Mathematica. One can check that both the fi, i = 1,2,3, 4, and gj, j = 1,2,3, generate the same ideal because we can express one generating set in terms of the other and vice-versa:
fi = dijgj, gj = cjifi, (116 )
where d and c are 4 × 3 and 3 × 4 polynomial matrices, respectively, and are given by
( ) − 1 − z2 − yz ( 2 ) | y 0 z | 0 0 − x z − 1 d = |( |) , c = ( − 1 − y 0 0 ) . (117 ) − x 0 2 0 0 z 1 0 − 1 − z − (x + yz)
The generators of the 4-tuple module are given by the set A ⋃ B ∗, where A and B ∗ are the sets described below:

A is the set of row vectors of the matrix I − d ⋅ c where the dot denotes the matrix product and I is the identity matrix, 4 × 4 in our case. Thus,

a1 = (z2 − 1,0,x − yz,1 − z2), 2 2 a2 = (0,z (1 − z ),xy − z,y (1 − z )), (118 ) a3 = (0,0, 1 − x2,x(z2 − 1)), 2 2 a4 = (− z ,xz, yz,z ).
We thus first get four generators. The additional generators are obtained by computing the S-polynomials of the Gröbner basis 𝒢. The S-polynomial of two polynomials g1,g2 is obtained by multiplying g1 and g2 by suitable terms and then adding, so that the highest terms cancel. For example in our case g1 = z2 − 1 and g2 = y2 − 1, and the highest terms are z2 for g1 and y2 for g2. Multiply g1 by y2 and g2 by z2 and subtract. Thus, the S-polynomial p12 of g1 and g2 is
p = y2g − z2g = z2 − y2. (119 ) 12 1 2
Note that order is defined (x ≫ y ≫ z) and the y2z2 term cancels. For the Gröbner basis of 3 elements we get 3 S-polynomials p12, p13, p23. The pij must now be re-expressed in terms of the Gröbner basis 𝒢. This gives a 3 × 3 matrix b. The final step is to transform to four-tuples by multiplying b by the matrix c to obtain ∗ b = b ⋅ c. The row vectors ∗ bi, i = 1,2,3, of ∗ b form the set ∗ B:
b∗1 = (z2 − 1,y (z2 − 1),x (1 − y2),(y2 − 1)(z2 − 1)), ∗ 2 2 2 b2 = (0,z(1 − z ),1 − z − x (x − yz),(x − yz) (z − 1)) , (120 ) b∗3 = (− x + yz, z − xy,1 − y2,0).
Thus, we obtain three more generators, which gives us a total of seven generators of the required module of syzygies.
  Go to previous page Scroll to top Go to next page