Problem:
Solution:
All we needed to do for this problem is to verify the three rules.
(i) > is a total (or linear order) on $ \mathbf{Z}^n_{\ge 0} $
This is simple to verify according to the definition. As long as two monomials are not the same, then either one is strictly larger than the other.
(ii) If $ \alpha > \beta $ and $ \gamma \in \mathbf{Z}^n_{\ge 0} $, then $ \alpha + \gamma > \beta + \gamma $.
This is also simple to verify. if $ \alpha $ has a total degree larger than $ \beta $, then so is $ \alpha + \gamma $ and $ \beta + \gamma $. Otherwise if the rightmost element of $ \alpha - \beta $ is negative, then so is $ \alpha + \gamma $ and $ \beta + \gamma $ as well.
(iii) > is a well ordering
For any set of $ n $ tuple, their total order is a set of non-negative integer and therefore there exist a smallest total degree, consider that subset of $ n $ tuples with smallest total degree.
For any set of $ n $ tuple with the same total order, their last element is an integer, and therefore exists a smallest first element, consider that subset of $ n $ tuples with smallest total degree and smallest last element. (*)
Repeat the above until we find the one with every element smallest. Now we have found a smallest element.
Suppose we have two elements $ \alpha $ and $ \beta $ with same total degree to compare with and $ \alpha $ has a smallest last element, then $ \alpha - \beta $ will have a negative rightmost non-zero element.
Solution:
All we needed to do for this problem is to verify the three rules.
(i) > is a total (or linear order) on $ \mathbf{Z}^n_{\ge 0} $
This is simple to verify according to the definition. As long as two monomials are not the same, then either one is strictly larger than the other.
(ii) If $ \alpha > \beta $ and $ \gamma \in \mathbf{Z}^n_{\ge 0} $, then $ \alpha + \gamma > \beta + \gamma $.
This is also simple to verify. if $ \alpha $ has a total degree larger than $ \beta $, then so is $ \alpha + \gamma $ and $ \beta + \gamma $. Otherwise if the rightmost element of $ \alpha - \beta $ is negative, then so is $ \alpha + \gamma $ and $ \beta + \gamma $ as well.
(iii) > is a well ordering
For any set of $ n $ tuple, their total order is a set of non-negative integer and therefore there exist a smallest total degree, consider that subset of $ n $ tuples with smallest total degree.
For any set of $ n $ tuple with the same total order, their last element is an integer, and therefore exists a smallest first element, consider that subset of $ n $ tuples with smallest total degree and smallest last element. (*)
Repeat the above until we find the one with every element smallest. Now we have found a smallest element.
Suppose we have two elements $ \alpha $ and $ \beta $ with same total degree to compare with and $ \alpha $ has a smallest last element, then $ \alpha - \beta $ will have a negative rightmost non-zero element.
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