# Has anyone used the isomorphism with $\Bbb{N}_{\gt 0}$ as a monomial ordering?

Let $R[x_1, x_2, \dots]$ be a ring of formal polynomials in a countably $\infty$ number of indeterminates $x_i$, over a commutative ring $R$.

The commutative monoid $X = \{ x^e = x_1^{e_1} x_2^{e_2} \cdots \vert e_i \in \Bbb{N}, \ e_i \neq 0 \text{ only finitely many times }\}$ is isomorphic to $(\Bbb{N}_{\gt 0}, \cdot)$. An ismorphism is given by any bijection between $\{x_1, x_2, \dots\}$ and the primes $\{p_1, p_2, \dots \} \subset \Bbb{N}_{\gt 0}$ extended homomorphically. Stated another way, any commutative free monoid on a countably $\infty$ set is isomorphic to $\Bbb{N}_{\gt 0}$.

So let $f: X \to \Bbb{N}_{\gt 0}$ be any such isomorphism. Then the order on $X$ defined as $a \lt b \iff f(a) \lt f(b)$ is an admissible ordering (one that can be used for Gröbner bases).

Additionally, since $X$ is countable we have that $R[x_1, \dots]$ is an $R$-module with countable basis $X$.

• Probably not. There are lots of such suitable orderings, and it seems unnecessary to try this one. Why do you think it might be interesting, compared to any other possible ordering? – Thomas Andrews May 24 '14 at 18:32
• @ThomasAndrews because it has a simple definition, uses the simple isomorphism to $\Bbb{N}_{\gt 0}$. It seems equally arbitrary for the researchers to use the lex, degrevlex, etc, just because the 1-variable monomials are sometimes ordered that way. – Shine On You Crazy Diamond May 24 '14 at 18:36
• The ones that are used are quite non-abitrary, really! In fact, they usual orders have very good properties in several ways. – Mariano Suárez-Álvarez May 24 '14 at 18:37
• it's not clear to me how you would "use" a monomial ordering for infinitely many variable to figure out the basis for an ideal, when the ideal could also be infinitely generated. As far as I know, the only reason to use the specific ordering is to compute an actual basis, so I'm not sure where you'd use this. You could certainly use it for finite polynomial rings, but the primes are so irregular, the order of $x_1^n$ and $x_2^m$ (assuming $x_1\mapsto 2$ and $x_2\mapsto 3$) is fairly complicated compared to lexical order. – Thomas Andrews May 24 '14 at 18:45
• But way too many questions you ask here are of the form, "this is a case of $X$. Is that interesting?" If you have a motivating example, tell, don't ask. There is huge amounts of literature out there; before you ask people to dig into them, give motivation for digging. Otherwise you get a lot of unanswered questions. – Thomas Andrews May 24 '14 at 18:50