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 ${\cal N}_Q^{(J)}$ such that for the associated constant matrix

$$N=N_T := \left( \tilde{T}_0', \tilde{T}_0'' \right) \in \mbox... ...ne{m} \right): \ \ \mbox{rk} \left( \tilde{I} + N \right) < l;$$

here $\tilde{I} := ( I_l, 0)$. The set $\left\{ N_T \vert T \in {\cal N}_Q^{(J)} \right\}$ forms a linear subspace of $\mbox{Mat} \, (l, n;R/\underline{m})$. Choose a basis $N_1, \ldots , N_q$ and consider the generic matrix

$$N_* = N_* (Z) := \tilde{I} + \sum_{i=1}^q N_i \cdot Z_i .$$

Finding a reduction is equivalent to obtaining a zero z of the ideal of maximal minors

$$minor(N_*) \subset K[Z_1,...,Z_q] ,$$

which, in general, seems to be not possible. (For simplicity we assume $K=R/\underline{m}$.)

 

(6.2) If our presentation matrix Q splits with respect to J, then by Lemma 2 there exists a unique minimal module $\langle A_0 \rangle$ in $\{ {\cal M}\left( U_T Q \right) \vert T \in {\cal N}_Q^{(J)} \}.$ 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 $m_1=ld (\underline {a}_1),..., m_l=ld(\underline{a}_l)$ be the set of leading monomials of the minimal generating system of ${\cal M}(Q)=\langle A \rangle$ , formed by the columns of A, cf. (5.2). Take the first index $s \leq l$ such that m_s is not a leading term of $\langle A_0 \rangle$. 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, $\langle A_0 \rangle$ and ${\cal M}\, (Q)$ have the same image in R^k(m_s) and $\langle A_0 \rangle$ is still the unique minimal a an R^k₊ (m_s)-submodule. It is therefore of K-codimension 1 with respect to ${\cal M}(Q)$. Subsequentely, any strict reduction ${\cal M}(U_TQ)$ of ${\cal M}(Q)$ in R^k₊(m_s) will induce $\langle A_0 \rangle$ in R^k₊(m_s).

 

(6.4) By construction we have $a_j \equiv 0$ for j>s in R₊(m_s); hence, ${\cal M}(U_TQ)$ is determined by the submatrix N_T^(s) of the first s columns of N_T.

Proposition 12  
Let $minor(i)=minor(N_*^{(i)}) \subset K[Z]$ be the ideal of i-minors from the first i rows of N_*. Then

(i1) $\ minor(s-1)=(1)$,
(i2) $\ minor(s) \not= (1)$.
(i3) $\ \mbox{If} \ Q$ splits with respect to J, then the linear space spanned by the columns of $N_*^{(s)} (\underline{z})$ : $\langle N_*^{(s)} (\underline{z}) \rangle$ is either K^s or a fixed hyperplane $H \subset K^s$ in case $\underline{z}$ is a zero of the ideal minor(s).

In other words, if $\langle N_*^{(s)} (\underline{z}) \rangle$ is dependent of the choice of a zero $\underline{z}$ from minor(s), then Q does not J-split!
(i1) and (i2) hold by our choice of s. If $\underline{z}$ is a zero of minor(s), then for $N_* (\underline z) = N _T$

$${\cal M}(U_T Q)\equiv \langle A_0 \rangle \ \ \mbox{over} \ \ R_+ (m_s).$$

 

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

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 $N_0 = N^{(s)} (\underline {z})$ and $\underline{z}$ be a zero of minor(s). Then $\mbox{rk}\, N_0 = s-1$, and consequently Then regular constant matrices $\Lambda \in Gl_s(K)$, $\tilde \Lambda \in Gl_n(K)$ exist such that

$$\Lambda N_0 \tilde \Lambda = \left( \begin{array}{cc} I_{s-1} & 0 \\ 0 & 0 \end{array} \right) =: \tilde I_{s-1}.$$

Moreover, $minor(s) = minor( \Lambda N_*^{(s)} \tilde \Lambda )$. Because $\langle N_*^{(s)} (\underline{z}) \rangle$ is independent of the choice of a zero $\underline{z}$ of minor(s), the same holds for $\langle \Lambda N _*^{(s)} \tilde \Lambda \rangle$. Therefore, $\underline{z}$ is a zero of minor(s) if and only if $\langle \Lambda {\cal N}_*^{(s)} ( \underline {z}) \tilde \Lambda \rangle = \langle \Lambda N_0 \tilde \Lambda \rangle = \langle \tilde I_{(s-1)} \rangle$ if and only if the s-th row of $\Lambda N _*^{(s)} (\underline {z}) \tilde \Lambda$ is identically zero. By construction, all entries of N _*, and so the entries of $\Lambda {\cal N}_*^{(s)} \tilde \Lambda$, are linear polynomials in Z.

 

(6.6) Note that the minor ideal need not to be reduced, as demonstrated by the following example:
Given $Q = \left( \begin{array}{cc} x & -y \\ y & x-2y \end{array} \right)$, then a representative in the sense of (5.2) is given by

$$\tilde{Q} = \left( \begin{array}{ccc} x & -y & 0\\ y & x-2y & x^2-2xy+y2 \end{array} \right).$$

None of the matrices of $I\!\!B_+$ increase ${\cal M}(Q)$; that is, ${\cal N}^{(1)}_Q = \mbox{Mat}\,(1,1;K[[x,y]])$.
A generator is given by $U_T = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$, T=(1). Then $\tilde{T} = \left( \begin{array}{ccc} 0 & 1 & x+y\\ -1& 2 & x-y\\ 0 & 0 & 0 \end{array} \right)$, $\tilde{T}'_0 = \left( \begin{array}{cc} 0 & 1\\ -1 & 2 \end{array} \right)$ and $\tilde{T}''_0 = \left( \begin{array}{c} 0\\ 0 \end{array} \right)$. Thus $N_* = \left( \begin{array}{cc} 1 & Z\\ -Z & 1+2Z \end{array} \right)$, and minor(N_*) = (1+2Z+Z²) = (1+Z)² is not reduced.