Let $R=k[t_1,\ldots,t_m]$ be a polynomial ring over a field $k$ and $I=(f_1,\ldots,f_r)$ an ideal of R. The $f_i$ shall be homogeneous for the natural grading of R and of degree greater than 1. Let $S=R/I$ and consider the series

$$ \sum_{i=0}^\infty \dim_k Tor^S_i(k,k) z^i = g(z) $$

In computer experiments with Macaulay2 the function $g(z)$ always came out rational in $z$.

Especially, if the polynomials $f_\nu$ are chosen "absolutely randomly" and $r \leq m$ (what counts is probably that they are forming a regular sequence in $R$) then one seems to have


with the $r$ from above and $r+s = m$.

My questions are

  1. Is it true that $g(z)$ is always rational?
  2. If 1. is true, is there an algorithm to compute $g(z)$ from $R$ and the $f_i$?
  3. If 1. or 2. is unknown, is it at least true, that $S$ is exactly then a polynomial ring, when $g(z)$ is a polynomial? (With $g(z)=(1+z)^n$ being the polynomial then.)

You will find a counter-example for (1) in

Roos, Jan-Erik; Sturmfels, Bernd. A toric ring with irrational Poincaré-Betti series. C. R. Acad. Sci. Paris Sér. I Math. 326 (1998), no. 2, 141--146

  • $\begingroup$ The corresponding problem for local rings is called the Serre-Kaplanski problem. This was solved on the negative by Anick (I do not recall if he was first, though) $\endgroup$ Sep 14 '13 at 23:43
  • $\begingroup$ For a little of the history of Anick's result, see math.mit.edu/~rstan/hcmr.pdf. $\endgroup$ Sep 16 '13 at 1:11

(1) As answered by Mariano, it is not true in general. The first example was probably due to Anick, I think you can even find his MIT thesis available for free online.

The formula you stated is true for complete intersections, see Proposition 3.3.5 of Avramov's note "Infinite free resolutions" available here (his paper on the work of Roos also contains a discussion Serre's question and some concrete examples).

As for (3), if $g(z)$ is a polynomial then $k$ has finite projective dimension, so in the graded case $S$ would have to be a polynomial ring.


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