here . The set forms a linear subspace of . Choose a basis and consider the generic matrix

Finding a reduction is equivalent to obtaining a zero

which, in general, seems to be not possible. (For simplicity we assume .)

**(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 *m*_{s} is not a leading term of
.
Let *R*^{k}(*m*_{s}), resp.
*R*^{k}_{+}(*m*_{s}), denote the quotient of *R*^{k}
by the submodule of
all monomials greater than or equal to *m*_{s}, resp. strictly greater than *m*_{s}, with
respect to the monomial order of *R*^{k}.

By our choice,
and
have the same image in
*R*^{k}(*m*_{s}) and
is still the unique minimal a an
*R*^{k}_{+} (*m*_{s})-submodule. It is therefore
of K-codimension 1
with respect to
.
Subsequentely, any strict reduction
of
in
*R*^{k}_{+}(*m*_{s})
will induce
in
*R*^{k}_{+}(*m*_{s}).

**(6.4)**
By construction we have
for *j*>*s* in *R*_{+}(*m*_{s}); hence,
is determined by the submatrix *N*_{T}^{(s)} of the first *s*
columns of *N*_{T}.

Let be the ideal of

(i1) ,

(i2) .

(i3) splits with respect toJ, then the linear space spanned by the columns of : is eitherK^{s}or a fixed hyperplane in case is a zero of the idealminor(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*:

Note that the radical of an ideal may be computed by just taking annihilators of certain

Let
and
be a zero
of *minor*(*s*). Then
,
and consequently
Then regular constant matrices
,
exist such that

Moreover, . Because is independent of the choice of a zero of

**(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

None of the matrices of increase ; that is, .

A generator is given by ,