 MatrixCombine - Maple Help

RegularChains

 MatrixCombine
 equiprojectable decomposition of a list of regular chains Calling Sequence MatrixCombine(lrc, R) MatrixCombine(lrc, R, lm) Parameters

 R - polynomial ring lrc - list of regular chains of R lm - list of matrices with coefficients in R Description

 • The function call MatrixCombine(lrc, R, lm) returns the equiprojectable decomposition of the variety given by lrc, and the corresponding combined matrices.
 • The variety encoded by lrc is the union of the regular zero sets of the regular chains of lrc.
 • It is assumed that every regular chain in lrc is zero-dimensional and strongly normalized, and that all matrices in lm have the same format.
 • It is also assumed that lrc and lm have the same number of elements.
 • This command is part of the RegularChains package, so it can be used in the form MatrixCombine(..) only after executing the command with(RegularChains).  However, it can always be accessed through the long form of the command by using RegularChains[MatrixCombine](..). Examples

 > $\mathrm{with}\left(\mathrm{RegularChains}\right):$$\mathrm{with}\left(\mathrm{ChainTools}\right):$

Consider a polynomial ring with three variables

 > $R≔\mathrm{PolynomialRing}\left(\left[x,y,z\right]\right):$
 > $\mathrm{rc}≔\mathrm{Empty}\left(R\right):$

Consider the following four regular chains of R

 > $\mathrm{rc1}≔\mathrm{Chain}\left(\left[z,y,x-1\right],\mathrm{rc},R\right):$
 > $\mathrm{rc2}≔\mathrm{Chain}\left(\left[z,y-1,x\right],\mathrm{rc},R\right):$
 > $\mathrm{rc3}≔\mathrm{Chain}\left(\left[z-1,y,x\right],\mathrm{rc},R\right):$
 > $\mathrm{rc4}≔\mathrm{Chain}\left(\left[{z}^{2}+2z-1,y-z,x-z\right],\mathrm{rc},R\right):$

Consider the following four matrices over R

 > $\mathrm{m1}≔\mathrm{Matrix}\left(\left[\left[x+y-z,x-y+1\right],\left[x-2y+1,x-2+z\right]\right]\right)$
 ${\mathrm{m1}}{≔}\left[\begin{array}{cc}{x}{+}{y}{-}{z}& {x}{-}{y}{+}{1}\\ {x}{-}{2}{}{y}{+}{1}& {x}{-}{2}{+}{z}\end{array}\right]$ (1)
 > $\mathrm{m2}≔\mathrm{Matrix}\left(\left[\left[x-z-1,x+z-2\right],\left[x-3+y,x+y+z-1\right]\right]\right)$
 ${\mathrm{m2}}{≔}\left[\begin{array}{cc}{x}{-}{z}{-}{1}& {x}{-}{2}{+}{z}\\ {x}{-}{3}{+}{y}& {x}{+}{y}{+}{z}{-}{1}\end{array}\right]$ (2)
 > $\mathrm{m3}≔\mathrm{Matrix}\left(\left[\left[x-y-1,x+y-2z\right],\left[x+y-z-3,x+y+z-2\right]\right]\right)$
 ${\mathrm{m3}}{≔}\left[\begin{array}{cc}{x}{-}{y}{-}{1}& {x}{+}{y}{-}{2}{}{z}\\ {x}{+}{y}{-}{z}{-}{3}& {x}{+}{y}{+}{z}{-}{2}\end{array}\right]$ (3)
 > $\mathrm{m4}≔\mathrm{Matrix}\left(\left[\left[x-z-y+1,x+2z-2y\right],\left[x+y-z+1,x+y-z-1\right]\right]\right)$
 ${\mathrm{m4}}{≔}\left[\begin{array}{cc}{x}{-}{z}{-}{y}{+}{1}& {x}{+}{2}{}{z}{-}{2}{}{y}\\ {x}{+}{y}{-}{z}{+}{1}& {x}{+}{y}{-}{z}{-}{1}\end{array}\right]$ (4)

We view each matrix as a result obtained modulo the corresponding regular chain in the given order. We combine these four results as follows

 > $\mathrm{clr}≔\mathrm{MatrixCombine}\left(\left[\mathrm{rc1},\mathrm{rc2},\mathrm{rc3},\mathrm{rc4}\right],R,\left[\mathrm{m1},\mathrm{m2},\mathrm{m3},\mathrm{m4}\right]\right)$
 ${\mathrm{clr}}{≔}\left[\left[\left[\begin{array}{cc}{-}{2}{}{y}{+}{1}& {-}{4}{}{y}{+}{2}\\ {-}{4}{}{y}{+}{2}& {y}{-}{1}\end{array}\right]{,}{\mathrm{regular_chain}}\right]{,}\left[\left[\begin{array}{cc}{-}\frac{{1}}{{2}}{}{{z}}^{{2}}{-}{2}{}{z}{+}\frac{{3}}{{2}}& {-}\frac{{3}}{{2}}{}{{z}}^{{2}}{-}{2}{}{z}{+}\frac{{3}}{{2}}\\ {-}{3}{}{{z}}^{{2}}{-}{5}{}{z}{+}{4}& {-}\frac{{{z}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}\end{array}\right]{,}{\mathrm{regular_chain}}\right]\right]$ (5)

The four cases cannot be combined into a single one. In fact, we obtained the following two cases

 > $\mathrm{rc1}≔\mathrm{clr}\left[1\right]\left[2\right];$$\mathrm{Equations}\left(\mathrm{rc1},R\right)$
 ${\mathrm{rc1}}{≔}{\mathrm{regular_chain}}$
 $\left[{x}{+}{y}{-}{1}{,}{{y}}^{{2}}{-}{y}{,}{z}\right]$ (6)
 > $\mathrm{rc2}≔\mathrm{clr}\left[2\right]\left[2\right];$$\mathrm{Equations}\left(\mathrm{rc2},R\right)$
 ${\mathrm{rc2}}{≔}{\mathrm{regular_chain}}$
 $\left[{2}{}{x}{+}{{z}}^{{2}}{-}{1}{,}{2}{}{y}{+}{{z}}^{{2}}{-}{1}{,}{{z}}^{{3}}{+}{{z}}^{{2}}{-}{3}{}{z}{+}{1}\right]$ (7)

The two ideals generated by rc1 and rc2 are obviously relatively prime (no common roots in z) so the Chinese Remaindering Theorem applies. However, if we try to recombine them, we create a polynomial in y with a zero-divisor as initial. This is forbidden by the properties of a regular chain.