### 4.3 The module of syzygies

Equation (20) has non-trivial solutions. Several solutions have been exhibited in [17]. We merely mention these solutions here; in the forthcoming text we will discuss them in detail. The solution is given by . The solution is described by and . The solutions and are obtained from by cyclically permuting the indices of , , and . 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 (20).

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 , can be multiplied by polynomials in , , (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 , , . First we present the general methodology for obtaining the solutions to Equation (20) and then apply it to Equation (20).

There are three linear constraints on the polynomials given by Equation (20). 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 with 1 in the -th place and 0 everywhere else; 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 (20) explicitly component-wise:

The first step is to use Gaussian elimination to obtain and in terms of ,

and then substitute these values in the third equation to obtain a linear implicit relation between , , , . We then have:
Obtaining solutions to Equation (27) amounts to solving the problem since the remaining polynomials , have been expressed in terms of , , , in Equation (26). Note that we cannot carry on the Gaussian elimination process any further, because none of the polynomial coefficients appearing in Equation(27) have an inverse in the ring.

We will assume that the polynomials have rational coefficients, i.e. the coefficients belong to , the field of the rational numbers. The set of polynomials form a ring - the polynomial ring in three variables, which we denote by . The polynomial vector . The set of solutions to Equation (27) is just the kernel of the homomorphism , where the polynomial vector is mapped to the polynomial . Thus the solution space is a submodule of . 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. [2], and Kreuzer and Robbiano [16] below. The details have been included in Appendix A.