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# 3. Row-minimal matrices

(3.1) In virtue of Lemma 2 we shall look for presentation matrices which have a good chance of being equivalent to a matrix in block diagonal form.

Definition 3
For a given we call Q a J-row-minimal matrix if is minimal in the set of Rk-submodules , where is a permutation matrix such that the associated permutation maps J to J0.

Obviously, this notation is independent of the choice of . Moreover, Q is called row-minimal if it is J-row-minimal with respect to all strict subsets J of row indices .

(3.2) The test for whether a module M will split is based on the observation that a presentation matrix of M in block diagonal form Q0 is J0- and J'0-row-minimal (cf. Lemma 2). It is possible to check by a standard basis computation whether a row-minimal matrix Q is equivalent to a block diagonal matrix. To do that, we have to fix an arbitrary (local) monomial order of Rr, ordering first by components; that is, for any two monomials and the i-th unit column.

(3.3) Let be a matrix formed by an ordered standard basis of . From the above order we obtain an integer l such that all columns of Q' having index l'> l will have their first non-zero entry in the k'-th component, k'>k. Hence Q' will have the block structure

and the modules associated to A' and D' are characterized by Q:

Changing the roles of J0 and J'0 (resp. reversing the order of components) we may write a corresponding standard basis Q'' in analogous block form

Comparing these two different standard bases of Q, we obtain as a trivial consequence

Lemma 4
With the notation from above

(3.4) Usually we are not in this situation and we first have to apply row operations. But, assuming J0-row-minimality, it is enough to apply only those row-operations belonging to :

Proposition 5
If for some and
if Q' is J0-row-minimal, then
(i1)
(i2) ; that is, .

(i0) First note that given an isomorphism and a -invariant submodule , then N is -invariant, too (otherwise would produce an infinite ascending chain!).
(i1) With the notation of (3.3) let

Letting , , we obtain

. Because of the minimality assumption,

Again, by Lemma 1(i3), we have a factorization

Consequently, we obtain

and, by the initial remark, .
(i2) An easy computation shows that

Thus, , which shows that . On the other hand implies the existence of a such that . Because , there is an such that ; that is, and .

Next: 4. Splitting criteria Up: Splitting algorithm for vector Previous: 2. Block type Bruhat
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