In class, I was given the following useful corollary in judging whether a given ordering ">" is a monomial ordering or not.
Let > be a relation on $\mathbb{Z}_{\geq 0}^n$ that satisfies
i) > is a total ordering on $\mathbb{Z}_{\geq 0}^n$.
ii) if $\alpha > \beta$ and $\gamma \in \mathbb{Z}_{\geq 0}^n$, then $\alpha + \gamma > \beta + \gamma$.
Then > is a well-ordering if and only if
(Here comes the statement I don't understand)
$\alpha \geq 0$ for all $\alpha \in \mathbb{Z}_{\geq 0}^n$.
What I don't understand is, $\alpha \in \mathbb{Z}_{\geq 0}^n$ simply, does it not already mean that $\alpha \geq 0$? Because $\mathbb{Z}_{\geq 0}^n$, the $_{\geq 0}$ I thought it means that "every entry of the $n$-tuple in $\mathbb{Z}_{\geq 0}^n$ takes integer values greater than or equal to $0$."
So I cannot think of any $\alpha \in \mathbb{Z}_{\geq 0}^n$ that would violate $\alpha \geq 0$. The possibilities are
I am misunderstanding what is meant by $\mathbb{Z}_{\geq 0}^n$.
I am misunderstanding what is meant by $\alpha \geq 0$ (I am taking it to mean that every entry is $\geq 0$).
Please kindly elaborate on what this condition is saying and maybe give me an example of when it is violated. Is there a type of order > that would make this false?
The $0$ here is the zero multi-index $(0, \ldots, 0) \in \mathbb{Z}_{\geq 0}^n$, so the third axiom means that $\alpha \geq (0, \ldots, 0)$ for any $\alpha \in \mathbb{Z}_{\geq 0}^n$. This is just a condition we place on the partial order $\geq$ on $\mathbb{Z}_{\geq 0}^n$: we insist, that the zero tuple must be the smallest multi-index with respect to $\geq$. In particular, it does not mean $\alpha_i \geq 0$ for all $i$.
Alternatively, since the multi-indices are exponents of monomials, i.e., $\alpha \leftrightarrow x^\alpha = x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}$, the third axiom is equivalent to $m \geq 1$ for all monomials $m \in k[x_1, \ldots, x_n]$.