Bourbaki

Exercise 2, § 3

Commutative Algebra, X, p. 157

Let AA be a noetherian ring, MM an AA-module admitting a finite length resolution (L,p)(L, p) by finite type free modules.

a) Now prove that one has, for every prime ideal pp of AA,

∑i≥0(−1)irg⁡A(Li)=∑i≥0(−1)i[Tor⁡iA(M,κ(p))]=∑i≥0(−1)i[Ext⁡Ai(M,κ(p))]; \sum_{i \geq 0} (-1)^i \operatorname{rg}_A(L_i) = \sum_{i \geq 0} (-1)^i [\operatorname{Tor}_i^A(M, \kappa(p))] = \sum_{i \geq 0} (-1)^i [\operatorname{Ext}_A^i(M, \kappa(p))] ;

we denote this integer by χ(M)\chi(M).

b) Prove that one has χ(M)≥0\chi(M) \geq 0 (consider a prime ideal pp associated with AA).

c) Prove that the following conditions are equivalent:

(i) χ(M)=0\chi(M) = 0 ; (ii) the annihilator of MM is not reduced to zero ; (iii) the annihilator of MM contains a cancellable element.

(If χ(M)=0\chi(M) = 0, prove that MpM_p is zero for every p∈Ass⁡(A)p \in \operatorname{Ass}(A) ; if χ(M)>0\chi(M) > 0, prove analogously that the annihilator of Ann⁡(M)\operatorname{Ann}(M) contains a cancellable element, which implies Ann⁡(M)=0\operatorname{Ann}(M) = 0.)

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-6. Not checked by a person.