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isLexModule -- whether a monomial module over an exterior algebra is lex

Synopsis

Description

A monomial module M is lex if for all monomials $u,v$ of F of the same degree with $v\in M$ and $u>v$ (> lex order) then $u\in M$. An equivalent definition of a lex submodule is the following one: a monomial submodule $M=\oplus_{i=1}^{r}{I_ig_i}$ of F is lex if $I_i$ is a lex ideal of E for each $i,$ and $(e_1,\dots, e_n)^{\rho_i + f_i - f_{i-1}} \subseteq I_{i-1}$ for $i = 2, \dots, r$ with $\rho_i = \mathrm{indeg}\ I_i.$

Example:

i1 : E = QQ[e_1..e_4, SkewCommutative => true]

o1 = E

o1 : PolynomialRing, 4 skew commutative variable(s)
i2 : F=E^{0,0}

      2
o2 = E

o2 : E-module, free
i3 : I_1=ideal(e_1*e_2,e_1*e_3,e_2*e_3)

o3 = ideal (e e , e e , e e )
             1 2   1 3   2 3

o3 : Ideal of E
i4 : I_2=ideal(e_1*e_2,e_1*e_3)

o4 = ideal (e e , e e )
             1 2   1 3

o4 : Ideal of E
i5 : M=createModule({I_1,I_2},F)

o5 = image | e_2e_3 e_1e_3 e_1e_2 0      0      |
           | 0      0      0      e_1e_3 e_1e_2 |

                             2
o5 : E-module, submodule of E
i6 : Malex=almostLexModule M

o6 = image | e_1e_4 e_1e_3 e_1e_2 e_2e_3e_4 0      0      |
           | 0      0      0      0         e_1e_3 e_1e_2 |

                             2
o6 : E-module, submodule of E
i7 : isLexModule Malex

o7 = false
i8 : L=createModule({ideal(e_1*e_2,e_1*e_3*e_4),ideal(e_1*e_2*e_3*e_4)},F)

o8 = image | e_1e_2 e_1e_3e_4 0            |
           | 0      0         e_1e_2e_3e_4 |

                             2
o8 : E-module, submodule of E
i9 : isLexModule L

o9 = true

See also

Ways to use isLexModule :

For the programmer

The object isLexModule is a method function.