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 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!

  • $\begingroup$ What does "may the set of $\mathrm{in}_{<}(I)$ [...] be infinite for an ideal $I$" mean? The way you have defined it, $\mathrm{in}_{<}(I)$ is an ideal. $\endgroup$ Apr 5 '19 at 19:38
  • $\begingroup$ Are you saying, as we vary $<$, can we get infinitely many different initial ideals $\mathrm{in}_{<}(I)$? $\endgroup$ Apr 5 '19 at 19:40
  • $\begingroup$ @SamHopkins Yes, that is exactly what I mean. I thought this would be clear from the overall setting but I guess I better reword it if it isn't. $\endgroup$
    – imakhlin
    Apr 5 '19 at 19:44
  • $\begingroup$ it's clear now, thanks. $\endgroup$ Apr 5 '19 at 19:47
  • $\begingroup$ Somewhat unexpectedly for myself I found a simple argument showing that the set of initial ideals is indeed finite when $n=2$. I still have no idea what happens when $n\ge 3$, however. $\endgroup$
    – imakhlin
    Apr 30 '19 at 0:25

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.

  • $\begingroup$ Instead of $I_h$ you can consider something invariant, namely the $\mathbb Z[h]$-module generated by the homogenizations of all polynomials in $I$. This contains all of the $I_h$ and is, in fact, equal to the $I_h$ obtained for a Gröbner basis of $I$ with respect to ordering by degree. $\endgroup$
    – imakhlin
    Jan 23 at 8:48
  • $\begingroup$ Also, using this construction you can generalize another fundamental (and related) property from Artinian orders to arbitrary orders. That is the fact that $\mathrm{in}_<I$ always has the form $\mathrm{in}_{<_w} I$ for some $w\in\mathbb R^n$ where $\mathbf{x}^\alpha<_w\mathbf{x}^\beta$ iff $\alpha\cdot w<\beta\cdot w$. For Artinian orders this is Proposition 1.11 in Sturmfels's GBCP. $\endgroup$
    – imakhlin
    Jan 23 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.