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Hi,

Let $I$ be an ideal in a polynomial ring $S = k[x_1, \ldots, x_n]$, where $k$ is an algebraically closed field of characteristic zero. Fix a term order on $S$ (e.g. a lexicographic order) and let $in(I)$ denote the initial ideal of $I$, that is, the monomial ideal generated by all the initial terms of elements of $I$.

It is clear that for any $k > 0$ one has $in(I)^k \subset in(I^k)$, with equality if $I$ itself is a monomial ideal.

  • Is there an example of $I$ where for large $k$, the quotient $in(I^k) / in(I)^k$ is NOT finite dimensional (over $k$)?

  • Is there a "good" sufficient condition on $I$ and the term order to guarantee that for large $k$, $in(I^k) / in(I)^k$ is always finite dimensional? Clearly if $I$ is primary or is a monomial ideal this is the case.

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  • $\begingroup$ Hello. Can you show the proof or a reference for the case when $I$ is a primary (or prime) ideal? Thank you. $\endgroup$
    – Youngsu
    Commented Mar 25, 2013 at 1:04
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    $\begingroup$ Youngsu: It follows from the following observations: $I$ is primary iff $S/I$ is finite dimensional as a vector space over $k$. Since $\dim(S/I) = \dim(S/ in(I))$ it follows that $in(I)$ is also primary. Now for any $k > 0$, $in(I)^k$ should be primary. Thus $S/in(I)^k$ is finite dimensional which implies that $in(I^k) / in(I)^k$ is also finite dimensional. $\endgroup$
    – Kiu
    Commented Mar 25, 2013 at 5:13
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    $\begingroup$ Here by "primary" I mean m-primary where $\mathfrak{m}$ is the maximal ideal generated by $x_1, \ldots, x_n$. Nevertheless if $\mathfrak{n}$ is another maximal ideal and $I$ is $\mathfrak{n}$-primary then $S/I$ is finite dimensional over ${\bf k}$. $\endgroup$
    – Kiu
    Commented Mar 25, 2013 at 5:30
  • $\begingroup$ Kiumars: Hi. I thought by primary you meant an ideal which is primary to an arbitrary prime ideal. Now, it is clear to me too. Thank you. $\endgroup$
    – Youngsu
    Commented Mar 25, 2013 at 19:21
  • $\begingroup$ Sorry Youngsu for confusion. $\endgroup$
    – Kiu
    Commented Mar 25, 2013 at 20:26

1 Answer 1

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Hal Schenck kindly provided a counter-example to the claim in question. Consider the twisted cubic: (the code is in Macaulay)

o3 I = ideal (- y^2 + x*z, - y*z + x*w, - z^2 + y*w)

i4 : J = ideal leadTerm I

o4 = ideal (z^2 , y*z, y^2 )

i5 : hilbertPolynomial coker gens ideal leadTerm I^2

o5 = - 16*P + 9*P 0 1

i6 : hilbertPolynomial coker gens ideal leadTerm J^2

o6 = - 20*P_0 + 10*P_1

So, the quotient is actually positive dimensional.

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