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. $\endgroup$ – Sam Hopkins Apr 5 at 19:38
  • $\begingroup$ Are you saying, as we vary $<$, can we get infinitely many different initial ideals $\mathrm{in}_{<}(I)$? $\endgroup$ – Sam Hopkins Apr 5 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 at 19:44
  • $\begingroup$ it's clear now, thanks. $\endgroup$ – Sam Hopkins Apr 5 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 at 0:25

Your Answer

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

Browse other questions tagged or ask your own question.