(6.2) If our presentation matrix Q splits with respect to J, then by Lemma 2 there exists a unique minimal module in This fact will reduce (6.1) to a linear problem! Otherwise we could stop and try another subset J of rows of Q.
(6.3) Let be the set of leading monomials of the minimal generating system of , formed by the columns of A, cf. (5.2). Take the first index such that ms is not a leading term of . Let Rk(ms), resp. Rk+(ms), denote the quotient of Rk by the submodule of all monomials greater than or equal to ms, resp. strictly greater than ms, with respect to the monomial order of Rk.
By our choice, and have the same image in Rk(ms) and is still the unique minimal a an Rk+ (ms)-submodule. It is therefore of K-codimension 1 with respect to . Subsequentely, any strict reduction of in Rk+(ms) will induce in Rk+(ms).
(6.4) By construction we have for j>s in R+(ms); hence, is determined by the submatrix NT(s) of the first s columns of NT.
(i3) splits with respect to J, then the linear space spanned by the columns of : is either Ks or a fixed hyperplane in case is a zero of the ideal minor(s).
(6.5) If Q is J-split, the set of zeros of the ideal minor(s) form a linear subspace with respect to Z:
be a zero
of minor(s). Then
Then regular constant matrices
exist such that
(6.6) Note that the minor ideal need not to be reduced, as demonstrated by the following example:
Given , then a representative in the sense of (5.2) is given by