Consider the polynomial ring $R=\mathbb Z[x\_1,\ldots,x\_n]$ and an ideal $I\subset R$. Let $<$ be a monomial order, i.e. a total order on the set of monomials in $R$ such that for any monomials $a$, $b$ and $c$ we have $ab<ac$ whenever $b<c$. For such an order the initial ideal $\mathrm{in}\_<(I)$ is the linear space spanned by the monomials $\mathrm{in}\_<(p)$ with $p\in I$ where $\mathrm{in}\_<(p)$ is the $<$-greatest monomial occurring in $p$. This linear space is easily seen to be a monomial ideal.


Now, the definition of a monomial order often requires that we also have $1<x\_i$ for all $i$, which is equivalent to $<$ being Artinian and also to $<$ being a well-order. It is well known that with this assumption in place the number of initial ideals is finite: any ideal $I$ has only finitely many distinct initial ideals $\mathrm{in}\_<(I)$ with respect to *Artinian* monomial orders $<$. See page 1 in [this book by Sturmfels](https://play.google.com/books/reader?id=K-bxBwAAQBAJ&printsec=frontcover&pg=GBS.PA1) for a neat proof.


I'm, however, interested in what happens when we *do not* require the order to be Artinian, since this condition is not needed to define the initial ideal itself. Question: may the set of the distinct $\mathrm{in}\_<(I)$ provided by various arbitrary (not necessarily Artinian) monomial orders $<$ be infinite for some ideal $I$? (As always, references are very welcome. Quite surprisingly, I have not been able to find anything that would be truly relevant here.)


**Update.** Chris Manon has recently told me how this is handled and the answer is no, the set may not be infinite. The trick is to homogenize the ideal by adding an extra variable in the right way. Anyhow, thanks, Chris!



Infinitely many initial ideals for non-Artinian monomial orders?
The OP has already figured out the answer since the question was posted. I am writing a slightly more expanded version here because I haven't been able to find this written down somewhere.


Let $\mathcal T\_n$ denote the set of all possible term orders on monomials in the variables $\{x\_1,x\_2,\dots,x\_n\}$. These are total orders that satisfy $\mathbb x^{\alpha}< \mathbb x^{\beta}\implies x^{\alpha+\gamma}< \mathbb x^{\beta+\gamma}$, but we don't require that they be well-orders (or, equivalently, Artinian). Let's also define $\mathcal T^{\deg}\_n\subset \mathcal T\_n$ as the set of those total orders that satisfy $\deg \mathbb x^{\alpha}< \deg \mathbb x^{\beta} \implies \mathbb x^{\alpha}<\mathbb x^{\beta}$. By definition, all the orders in $\mathcal T^{\deg}\_n$ are well-orders.


Given an order $<$ from $\mathcal T\_n$ we can construct an order $<\_h$ in $\mathcal T\_{n+1}^{\deg}$ (on variables $\{h,x\_1,x\_2,\dots,x\_n\}$) by requiring
$$h^{\alpha\_0}x\_1^{\alpha\_1}\cdots x\_n^{\alpha\_n}<\_h h^{\beta\_0}x\_1^{\beta\_1}\cdots x\_n^{\beta\_n} \iff x\_1^{\alpha\_1}\cdots x\_n^{\alpha\_n}< x\_1^{\beta\_1}\cdots x\_n^{\beta\_n}$$
for all tuples satisfying $\alpha\_0+\alpha\_1+\cdots+\alpha\_n=\beta\_0+\beta\_1+\cdots+\beta\_n$. The map sending $<$ to $<\_h$ is, in fact, a bijection between $\mathcal T\_n$ and $\mathcal T\_{n+1}^{\deg}$.


To every ideal $I\subset R$ generated by $\{f\_1, f\_2,\dots,f\_k\}$ we can associate the ideal $I\_h=\langle h^{\deg f\_1}f\_1\left(\frac{x\_1}{h},\dots,\frac{x\_n}{h}\right),\dots, h^{\deg f\_k}f\_k\left(\frac{x\_1}{h},\dots,\frac{x\_n}{h}\right)\rangle$ (this depends on the choice of generating set). Given a term order $<$ from $\mathcal T\_n$ we can look at the monomial ideal $\mathrm{in}\_{<\_h}I\_h$. One can check that for each monomial $h^{\alpha\_0}\mathbb x^{\alpha}\in \mathrm{in}\_{<\_h}I\_h$ we have $\mathbb x^{\alpha}\in \mathrm{in}\_< I$ and that all monomials of $\mathrm{in}\_{<} I$ are obtained this way.


Since there are only finitely many possibilities for $\mathrm{in}\_{<\_h} I\_h$ as $<\_h$ ranges over all Artinian term orders, then there are only finitely many possibilities for $\mathrm{in}\_< I$ as $<$ ranges over all term orders.


