8
$\begingroup$

Let $m$ and $d$ be two positive integers.

Consider the polynomial ring $R = \mathbb{C}[x_1 , \dots , x_m]$. Let $I$ be an ideal of $R$ generated by a finite family of polynomials of degree $d$, and finally let $H(t)$ be the Hilbert series of $R/I$.

I say that the Hilbert series $H(t)$ is standard if there exists an integer $n$ such that $$H(t) = \frac{(1-t^d)^n}{(1-t)^m} \, . $$

My question

I am looking for a (ideally, necessary and sufficient) condition on $I$ for its Hilbert series to be standard.

An example

Consider the case $m=3$ and $d=2$. Then

  • For $I=\langle x_1 x_2 \rangle$, we have $H(t) = \frac{1-t^2}{(1-t)^3}$, so $I$ satisfies the condition.
  • For $I=\langle x_1 x_2 , x_1 x_3 \rangle$ we have $H(t) = \frac{t^3-2 t^2+1}{(1-t)^3}$ and this can not be put in standard form, so $I$ does not satisfy the condition.
$\endgroup$

1 Answer 1

2
$\begingroup$

I assume you mean that I is generated by homogeneous polynomials of degree d. If so, a necessary and sufficient condition is that I be generated by a regular sequence of n polynomials of degree d. See Corollary 3.3 in Stanley (Advances in Math. 28 (1978) 57-83).

$\endgroup$
2
  • 5
    $\begingroup$ My theorem deals only with sufficiency, not necessity. A necessary and sufficient condition seems very unlikely in view of examples like $\langle x_1^2,x_1x_2,x_2^3\rangle$, which is not generated by a regular sequence. $\endgroup$ Mar 22, 2018 at 22:14
  • $\begingroup$ I stand corrected. It's still sufficient though and probably the only answer available to date. $\endgroup$
    – pjr
    Mar 24, 2018 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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