On the degree of Hilbert polynomials associated to the torsion functor

Abstract

Abstract. Let R be a local, Noetherian ring and I ⊆ R an ideal. A question of Kodiyalam asks whether for fixed i > 0, the polynomial giving the ith Betti number of I n has degree equal to the analytic spread of I minus one. Under mild conditions on R, we show that the answer is positive in a number of cases, including when I is divisible by m or I is an integrally closed m-primary ideal.

Hilbert-Samuel polynomials for the contravariant extension functor

“…The proof of this for each of the stated conditions follows closely the proofs given for [5,Theorems 3.3 and 3.4]. We will try to give a convincing account without repeating all of the details from [5].…”

“…In [5], the second and fourth authors considered the Hilbert polynomial giving the lengths of Tor i (M, R/I n ). In that paper, various assumptions were made on I and M which forced τ…”

“…Roughly speaking, the assumptions on M were made so that the ith syzygy has maximal dimension. The assumptions on the filtrations given in [5] were made in order to replicate some of the properties satisfied by the m-adic filtration. The reason for this is now clear in light of Theorem 3.12.…”

“…We will try to give a convincing account without repeating all of the details from [5]. A crucial point in each case is that the annihilator of T i is nilpotent, by Lemma 4.6.…”

“…Finally, in Section 4, we address our third purpose, to show that the results of Section 3 can, in many cases, be extended to more general filtrations to give results for the extension functor parallel to those given in [5] for the torsion functor. be a complex of finitely generated R-modules with Y = 0.…”