4. Splitting criteria

 

(4.1) Next we want to characterize those row-minimal matrices belonging to a decomposable module. The idea comes from the fact that the canonical homomorphism $R^r \longrightarrow R^k / {\cal M}(Q) \oplus R^{r-k} / {\cal M}'(Q)$ factors via M and that the induced surjection $M \longrightarrow R^k / {\cal M}(Q) \oplus R^{r-k} / {\cal M}'(Q)$ is an isomorphism. Hence, it reflects a direct sum decomposition of M if and only if Q is equivalent to a block diagonal matrix Q₀. Then Q is necessarily row-minimal with respect to J₀ and J'₀.

Proposition 6  
  If Q is row-minimal, then $M= \mbox{coker}\; Q$ is decomposable if and only if for some permutation $\sigma , \quad P_{\sigma}Q \sim Q_0$.

Assume M splits. Then we know $\tilde U Q \sim Q_0$ and, for some permutation $\sigma$, $\tilde UP_{\sigma^{-1}} \in {\,I\!\!\!\!G}$; i.e., $P_{\sigma} Q \sim UQ_0$ and $U = P_{\sigma} \tilde U^{-1} \in {\,I\!\!\!\!G}.$ By Proposition 5(i2), we obtain

$${\cal M}'(P_{\sigma}Q) \supseteq \left\{ \underline y \in R^... ... \langle P_{\sigma} Q \rangle \right\} \cong {\cal M}' (Q_0).$$

The same holds for J'₀ , because Q is J'₀-row-minimal, too; that is,

$${\cal M}(P_{\sigma}Q) \supseteq \left\{ \underline{x} \in R^... ...n \langle P_{\sigma} Q \rangle \right\} \cong {\cal M}(Q_0).$$

But then the composition of the canonical surjection

$$M \longrightarrow R^k / {\cal M}(P_{\sigma}Q) \oplus R^{r-k} / {\cal M}'(P_{\sigma}Q)$$

with

$$R^k / {\cal M}(P_{\sigma}Q) \oplus R^{r-k} / {\cal M}'(P_{\si... ... R^k / {\cal M}(Q_0) \oplus R^{r-k} / {\cal M}'(Q_0) \cong M$$

must be an isomorphism, showing $P_{\sigma}Q \sim Q_0$.
Note: ${\cal M}(P_{\sigma} Q)$ and ${\cal M}' (P_{\sigma} Q)$ are computed by a standard basis computation with respect to a certain module ordering in the components. So we simply have to check whether

$$P_{\sigma} Q = \left( \begin{array}{rr} A & 0 \\ 0 & D ... ...,\ A:= {\cal M}(P_{\sigma} Q), \ D:= {\cal M}' (P_{\sigma} Q)$$

has a solution X, which corresponds to a simple syzygy computation.

 

(4.2) Before computing a solution X, it is necessary for Q to be row-minimal, which may be tested using methods discussed in the next section. Usually Q is not row-minimal, in which case we shall use an algorithm to transform Q into an equivalent and J₀-row-minimal matrix. But we do not know a procedure to obtain row-minimality for all subsets immediately. Hence we shall formulate a second version of the splitting criteria.

Proposition 7  
If Q is J₀-row-minimal, then

$$Q \sim U Q_0 , \ U \in {\,I\!\!\!\!G} \qquad \mbox{if and only if} \qquad Q \sim UQ_0 , \ U \in {I\!\!B_-}.$$

By Proposition 5, it follows that

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

then $\langle A' \rangle = \langle \tilde A \rangle, \ \tilde A :=SA$ and $\langle D' \rangle = \langle \tilde D \rangle , \ \tilde D:= WD$. Therefore

$$\left( \begin{array}{cc} A' & 0 \\ C'-VA' & D' \end{array}... ... \tilde A & T \tilde D \\ 0 & \tilde D \end{array} \right)$$

and $\langle T\tilde D \rangle \subseteq \langle A' \rangle = \langle \tilde A\rangle$, giving

$$\left( \begin{array}{cc} \tilde A & T \tilde D \\ 0 & \til... ...( \begin{array}{cc} A' & 0 \\ 0 & D' \end{array} \right),$$

from which it may be concluded that

$$Q \sim \left( \begin{array}{rr} I & 0 \\ V & I \end{arra... ...ray}{rr} I & 0 \\ V & I \end{array} \right) \in I\!\!B_-.$$

The other direction is obvious. Note:
$\left( \begin{array}{cc} A' & 0 \\ C'-VA' & D' \end{array} \right) \sim \left( \begin{array}{cc} A' & 0 \\ 0 & D' \end{array} \right)$ if and only if C' - VA' = D'V' for some V'; i.e. we obtain

Corollary 8  
 If Q is J₀-row-minimal, then for an ordered standard basis $Q' = \left( \begin{array}{cc} A' & 0 \\ C' & D' \end{array} \right)$ of Q we have:
$Q \sim UQ_0 , \ U \in {\,I\!\!\!\!G} \quad$ if and only if $\quad$ the equation VA'+D' V'=C' has a solution (V, V').

Deciding the solvability of the above equation is reduced to a lifting computation: Let $A^{\ast}$ and $D^{\ast}$ be matrices corresponding to $\mbox{Hom}\,(-,A')$ and $\mbox{Hom}\,(D',-)$, and let [V], [V'], [C'] denote flattenings to a column vector of V, V', C', respectively. Then we obtain

$$(A^{\ast}\ D^{\ast}) \cdot \left( \begin{array}{r} \mbox{}[V] \\ \mbox{}[V'] \end{array} \right) = [C'];$$

that is, [C'] can be lifted to $(A^{\ast}\ D^{\ast})$ if and only if the equation has a solution.