What is the way to after finding Cohen-Macaulay semigroup as ring of monomial?

http://people.missouristate.edu/lesreid/reu/2007/PPT/robin.ppt

it said missing part inside and not missing part outside is non Cohen-Macaulay semigroup

1. which determine whether is missing in the most left hand side?

it seems once design the most left hand side's things, other layers can be calculated from last layer

1. What is the way to after finding Cohen-Macaulay semigroup as ring of monomial? or what is the application to finding this or using Cohen-Macaulay semigroup as ring of monomial?

2. where is the end ? i mean the most right hand side

i can only think the next step is to use normal form when using these monomials as ring

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you misunderstand the question, i mean the most left hand side has some missing black circle in the grid which is like randomly or depend on designer's wish –  Mark Mar 26 '13 at 10:29
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1 Answer

The question is nearly incomprehensible, but I'll answer what I think might have been intended by the first and last numbered items. The middle one is probably asking for an expository article on Cohen-Macaulayness, which I'm not qualified to write (and which wouldn't be appropriate for MO anyway).

For the first question: The lowest tier (not counting the 0 vector) consists of the generators, which are chosen arbitrarily (except that the vectors on the coordinate axes are always present.)

For the third question (labeled 2): There is no end; the semigroups continue arbitrarily far to the right.

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is it the black circle on the most left hand side in the grid missing just depend on designer's wish? –  Mark Mar 26 '13 at 10:30
any lemma stated which coordinate should be missing? –  Mark Mar 26 '13 at 10:31
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