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 $J \subset \{ 1,...,r \}$ we call Q a J-row-minimal matrix if ${\cal M}(P_{\sigma} Q)$ is minimal in the set of R^k-submodules $\{ {\cal M}(UP_{\sigma} Q) \mid U \in {I\!\!B}_+ \}$, where $P_{\sigma}$ is a permutation matrix such that the associated permutation $\sigma$ maps J to J₀.

Obviously, this notation is independent of the choice of $\sigma$. Moreover, Q is called row-minimal if it is J-row-minimal with respect to all strict subsets J of row indices $\{ 1,...,r \}$.

 

(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 R^r, ordering first by components; that is, $m \underline{e}_i > m' \underline{e}_j$ for any two monomials $m,\ m'$ and $i<j, \ \underline{e}_i$ the i-th unit column.

 

(3.3) Let $Q' \sim Q$ be a matrix formed by an ordered standard basis of $\langle Q\rangle$. 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

$$Q' = \left( \begin{array}{rr} A' & 0 \\ C' & D' \end{array} \right) , \quad A' \in \mbox{Mat}\, (k,l;R),$$

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

$$\langle A' \rangle = {\cal M}(Q), \quad \langle D' \rangle = ... ...ne{y} \end{array} \right) \in \langle Q \rangle \right\}.$$

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

$$Q'' = \left( \begin{array}{cr} A'' & B'' \\ 0 & D'' \end{array} \right).$$

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

Lemma 4  
With the notation from above

$$Q \sim Q_0 \quad \mbox{if and only if} \quad \langle A'\rangl... ...\quad \mbox{and} \quad \langle D'\rangle = \langle D''\rangle.$$

 

(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 $I\!\!B_+$:

Proposition 5  
  If $Q' \sim UQ_0$ for some $U \in {\,I\!\!\!\!G}$ and
if Q' is J₀-row-minimal, then

(i1) $\langle A'\rangle = {\cal M}(Q) \cong {\cal M}(Q_0) = \langle A\rangle$
(i2) ${\underline{0} \choose \underline{y}} \in \langle Q \rangle \Longleftrightarrow \underline{y} \in {\cal M}' (Q_0)$; that is, $\langle D'\rangle = \langle D \rangle$.

(i0) First note that given an isomorphism $\varphi :R^l \rightarrow R^l$ and a $\varphi^{-1}$-invariant submodule $N \subset R^l$, then N is $\varphi$-invariant, too (otherwise $N \subset \varphi (N) \subset \varphi ^2 (N) \ldots$ would produce an infinite ascending chain!).
(i1) With the notation of (3.3) let

$$Q \sim Q' = \left( \begin{array}{rc} A' & 0 \\ C' & D' \e... ...eft( \begin{array}{rl} A & 0 \\ 0 & D \end{array} \right).$$

Letting $\tilde A := SA, \tilde D := WD$, $\tilde I := I+VT$, we obtain

$$\left( \begin{array}{cc} A' & 0 \\ C' & D' \end{array} \... ... \\ V \tilde A & \tilde I \tilde D \end {array} \right),$$

${\cal M}(Q) = \langle A'\rangle = \langle \tilde A \rangle + \langle T \tilde D \rangle \supseteq \langle \tilde A \rangle \cong \langle A \rangle$. Because of the minimality assumption,

$$\langle \tilde A \rangle \subseteq {\cal M}(Q) \subseteq {\ca... ...- \tilde T \\ 0 & I \end{array} \right) \cdot Q \right).$$

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

$$\left( \begin{array}{rr} I & 0 \\ V & I \end{array} \rig... ...rray}{rl} \tilde S & 0 \\ 0 & \tilde W \end{array} \right).$$

Consequently, we obtain

$${\cal M}\left( \left( \begin{array}{rc} I & - \tilde T \\ ... ... \right) \right) = \langle \tilde S \tilde A \rangle,$$

and, by the initial remark, $\langle \tilde A \rangle = {\cal M}(Q) = \langle \tilde S \tilde A \rangle$.
(i2) An easy computation shows that

$$\begin{eqnarray*}\underline{y} \in \langle D' \rangle & \Longleftrightarrow ... ...{y} = V \tilde A \underline f + \tilde I \tilde D \underline g . \end{eqnarray*}$$

Thus, $\underline{y} = (\tilde I - VT) \tilde D g = (I + VT - VT) \tilde D \underline g = \tilde D g$, which shows that $\langle D'\rangle \subseteq \langle \tilde D \rangle$. On the other hand $\underline{y} \in \langle \tilde D \rangle$ implies the existence of a $\underline{g}$ such that $\underline{y} = \tilde D \underline{g}$. Because $\langle T \tilde D \rangle \subseteq \langle A' \rangle = \langle A \rangle$, there is an $\underline{f}$ such that $\ -T \tilde Dg = \tilde A \underline{f}$; that is, $\left( \begin{array}{r} \underline{0} \\ \underline{y} \end{array} \right) \in \langle Q \rangle$ and $\underline{y} \in \langle D' \rangle$.