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## The vector space of reduced polynomials

For simplicity let us assume that the set of monomials not divisible by the leading term of any element of G is finite, as in our first example (where the monomials were ). In this case we say the quotient ring is finite-dimensional, and this corresponds to a `zero-dimensional' variety (it just means that there are only a finite set of solutions to the polynomial system).

Then we can express any given f and g, modulo the ideal, as say and respectively. It follows immediately that the representation of f+g is and somewhat less immediately that the representation of fg is the remainder on division by I of .

This produces a linear vector space as is easily verified. Let us consider the effects of multiplying by in this vector space.

Suppose , where the are the monomials in A and the are coefficients. Suppose also that reduces (on division by I) to , for i = 1 to s. Then it is easily seen that

That is, multiplication by acts on the space spanned by these monomials in a linear way, which we can represent by a matrix.

Obviously there's nothing special about , and each variable is represented by such a matrix. Note that since multiplication by and in the polynomial ring commutes, these matrices and must also commute.

It should also be clear that knowledge of the action of these multiplication matrices should tell us everything possible about the ideal, in some sense. So the following miracle is actually explainable.

Robert Corless
Tue Mar 12 21:09:19 EST 1996