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# 6. Finding a J-row-reduction

(6.1) In looking for a strict J-row-reduction of a presentation matrix Q, we have to find a matrix T from such that for the associated constant matrix

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 z of the ideal of maximal minors

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 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.

Proposition 12
Let be the ideal of i-minors from the first i rows of N*. Then
(i1) ,
(i2) .
(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).

In other words, if is dependent of the choice of a zero from minor(s), then Q does not J-split!
(i1) and (i2) hold by our choice of s. If is a zero of minor(s), then for

(6.5) If Q is J-split, the set of zeros of the ideal minor(s) form a linear subspace with respect to Z:

Proposition 13
If Q is J-split, the radical of minor(s) is generated by linear polynomials.

Note that the radical of an ideal may be computed by just taking annihilators of certain Ext-groups (cf. [EV]). This method is implemented as a procedure in SINGULAR, using only syzygy- and standard-basis-computations. Hence finding a J-row-reduction reduces to solving linear equations if there exists a unique minimal.

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 minor(s), the same holds for . Therefore, is a zero of minor(s) if and only if if and only if the s-th row of is identically zero. By construction, all entries of N *, and so the entries of , are linear polynomials in Z.

(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 , T=(1). Then , and . Thus , and minor(N*) = (1+2Z+Z2) = (1+Z)2 is not reduced.

Next: 7. Finiteness Up: Splitting algorithm for vector Previous: 5. Criteria for row-minimality
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