3
$\begingroup$

Macaulay's lemma states:

Let R be a polynomial ring and I a homogeneous ideal. Then the Hilbert function of I is the same as the Hilbert function of in(I).

(Schenck, Computational Algebraic Geometry, p55)

(Where in(I) / lt(I) is the ideal consisting of leading terms of elements of I, and by the Hilbert function of I we mean the Hilbert function of R/I.)

Is there a counterexample showing that this isn't necessarily true for inhomogeneous ideals? (In all the cases that I have tried, it seems also to be true for inhomogeneous ideals.)

  • 0
    @DownvotemeifIanswer9-5 thanks!2013-08-02

1 Answers 1

1

An answer to repeat what I wrote in the comments: There is no standard way of defining the Hilbert function of an inhomogeneous ideal, so the original question is undefined. Moreover, there is no definition that would make the OP's question true. Consider the ideal $\langle y-x^2 \rangle$ in $k[x,y]$. If we choose a term order such that $y$ is the leading term, then the initial ideal is $\langle y \rangle$ with Hilbert function $(1,1,1,1,\ldots)$; if we choose a term order such that $x^2$ is the leading term, then the initial ideal is $\langle x^2 \rangle$ with Hilbert function $(1,2,2,2,\ldots)$.

One could imagine ways to salvage the statement. For example, one could ask that our term order always obey $\prod x_i^{d_i} > \prod x_i^{e_i}$ if $\sum d_i > \sum e_i$. However, the OP seemed to be satisfied with the observations in the first paragraph, so I never thought through the details of such a salvage.