Go to previous page Go up Go to next page

4.3 The module of syzygies

Equation (20View Equation) has non-trivial solutions. Several solutions have been exhibited in [1Jump To The Next Citation Point7Jump To The Next Citation Point]. We merely mention these solutions here; in the forthcoming text we will discuss them in detail. The solution z is given by T 'T - q = q = (D1,D2, D3). The solution a is described by T q = -(1,D3, D1D3) and 'T q = (1,D1D2, D2). The solutions b and g are obtained from a by cyclically permuting the indices of Dk, q, and q'. These solutions are important, because they consist of polynomials with lowest possible degrees and thus are simple. Other solutions containing higher degree polynomials can be generated conveniently from these solutions. Since the system of equations is linear, linear combinations of these solutions are also solutions to Equation (20View Equation).

However, it is important to realize that we do not have a vector space here. Three independent constraints on a six-tuple do not produce a space which is necessarily generated by three basis elements. This conclusion would follow if the solutions formed a vector space but they do not. The polynomial six-tuple q, ' q can be multiplied by polynomials in D1, D2, D3 (scalars) which do not form a field. Thus the inverse in general does not exist within the ring of polynomials. We therefore have a module over the ring of polynomials in the three variables D1, D2, D3. First we present the general methodology for obtaining the solutions to Equation (20View Equation) and then apply it to Equation (20View Equation).

There are three linear constraints on the polynomials given by Equation (20View Equation). Since the equations are linear, the solutions space is a submodule of the module of six-tuples of polynomials. The module of six-tuples is a free module, i.e. it has six basis elements that not only generate the module but are linearly independent. A natural choice of the basis is fm = (0, ...,1,...,0) with 1 in the m-th place and 0 everywhere else; m runs from 1 to 6. The definitions of generation (spanning) and linear independence are the same as that for vector spaces. A free module is essentially like a vector space. But our interest lies in its submodule which need not be free and need not have just three generators as it would seem if we were dealing with vector spaces.

The problem at hand is of finding the generators of this submodule, i.e. any element of the submodule should be expressible as a linear combination of the generating set. In this way the generators are capable of spanning the full submodule or generating the submodule. In order to achieve our goal, we rewrite Equation (20View Equation) explicitly component-wise:

' ' q1 + q1- D3q 2- D2q3 = 0, q2 + q'2- D1q'3- D3q1 = 0, (25) ' ' q3 + q3- D2q 1- D1q2 = 0.

The first step is to use Gaussian elimination to obtain q 1 and q 2 in terms of q ,q',q',q' 3 1 2 3,

q = - q'+ D q'+ D q , 1 1 3 2 2 3 q2 = - q'2 + D1q'3 + D3q1 (26) = - D q' - (1- D2)q' + D q'+ D D q , 3 1 3 2 1 3 2 3 3
and then substitute these values in the third equation to obtain a linear implicit relation between q3, q'1, q' 2, q' 3. We then have:
(1- D D D )q + (D D - D )q'+ D (1 - D2 )q' + (1 - D2 )q' = 0. (27) 1 2 3 3 1 3 2 1 1 3 2 1 3
Obtaining solutions to Equation (27View Equation) amounts to solving the problem since the remaining polynomials q1, q2 have been expressed in terms of q3, q'1, q'2, q'3 in Equation (26View Equation). Note that we cannot carry on the Gaussian elimination process any further, because none of the polynomial coefficients appearing in Equation(27View Equation) have an inverse in the ring.

We will assume that the polynomials have rational coefficients, i.e. the coefficients belong to Q, the field of the rational numbers. The set of polynomials form a ring - the polynomial ring in three variables, which we denote by R = Q[D1, D2, D3]. The polynomial vector ' ' ' 4 (q3,q1,q2,q3) (- R. The set of solutions to Equation (27View Equation) is just the kernel of the homomorphism f : R4 --> R, where the polynomial vector (q3,q'1,q'2,q'3) is mapped to the polynomial (1 - D1D2D3)q3 + (D1D3 - D2)q'1 + D1(1 - D23)q'2 + (1 - D21)q'3. Thus the solution space kerf is a submodule of R4. It is called the module of syzygies. The generators of this module can be obtained from standard methods available in the literature. We briefly outline the method given in the books by Becker et al. [2Jump To The Next Citation Point], and Kreuzer and Robbiano [16Jump To The Next Citation Point] below. The details have been included in Appendix A.

  Go to previous page Go up Go to next page