Betti numbers of modules over Noetherian rings with ... - IPM

Chapter 3: Growth of the Betti sequence of the canonical module 50
Lemma 3.4.1. Let J be an ideal of R generated by a regular sequence f1, · · · , fn and
let I be an ideal containing J. Then for j ≥ 0
1. Ext j
R (R/J, R/I) ∼ = (R/I) (nj)
.
2. Tor R j (R/J, R/I) ∼ = (R/I) (nj)
.
3. More generally, Ext j
R (R/J, N) ∼ = N (nj)
R
and Torj (R/J, N) ∼ = N (nj)
for each R-
module N which is killed by J.
Proof. (1) & (2) Since J is generated by a regular sequence we may use the Koszul
complex K•(f1, · · · , fn; R) as a free resolution of R/J : if U1, · · · , Un is a basis for the
free module in degree one such that Uj maps to fj, then the elements Ui1∧· · ·∧Uij such
that i1 < · · · < ij are a free basis for the free R-module Kj = Kj(f1, · · · , fn; R). Since
I contains J, when we apply HomR(−, R/I) (resp. − ⊗R R/I) to K•, the maps in the
complex all become 0, and the jth module may be identified with HomR(Kj, R/I) ∼ =
HomR(Kj, R/J) ⊗R R/I (resp. Kj ⊗R R/I ∼ = (Kj ⊗R R/J) ⊗R R/I). We may take as
an (R/I)-free basis the image of the basis for HomR(Kj, R/I) (resp. (Kj ⊗R R/I)). In
particular, Ext 1 R(R/J, R/I) ∼ = (R/I) n (resp. Tor R 1 (R/J, R/I) ∼ = (R/I) n ). The same
idea works for higher Ext and Tor modules.
(3) The idea is essentially the same. Replace R/I by any R-module N which
is killed by J. Then we get Ext j
R (R/J, N) ∼ = N (nj)
∼ j
= ExtR (R/J, R/J) ⊗R N and
Tor R j (R/J, N) ∼ = N (nj)
∼ R
= Torj (R/J, R/J) ⊗R N as well.
The formulation of case (‡†) is given in the following:
Theorem 3.4.2. Let R be a local ring and let M be a non zero R-module for which
0 → (R/I) a → M → (R/J) b → 0 is an exact sequence where J is an ideal of R
generated by a regular sequence f1, · · · , fn and let I ⊇ J. Let t be an integer.