Bourbaki

Exercise 8, § 4

Commutative Algebra, VII, 572

Let AA be an integrally closed Noetherian domain and M1,M2M_1, M_2 two finitely generated AA-modules. (a) Show that the AA-modules Tor⁡iA(M1,M2)\operatorname{Tor}_i^A(M_1, M_2), Ext⁡Ai(M1,M2)\operatorname{Ext}_A^i(M_1, M_2) are pseudo-zero for i>2i > 2 (reduce it to the case where AA is a principal ideal domain). (b) If MiM_i is torsion-free, show that Tor⁡1A(M1,Mi)\operatorname{Tor}_1^A(M_1, M_i) and Ext⁡A1(M1,M2)\operatorname{Ext}_A^1(M_1, M_2) are pseudo-zero (same method). (c) If MiM_i is a torsion AA-module, show that (in the notation of no. 5):

χ(M1⊗AM2)−χ(Tor⁡1A(M1,M2))=r(M2)χ(M1) \chi(M_1 \otimes_A M_2) - \chi(\operatorname{Tor}_1^A(M_1, M_2)) = r(M_2)\chi(M_1)
χ(Hom⁡A(M1,M2))−χ(Ext⁡A1(M1,M2))=−r(M2)χ(M1) \chi(\operatorname{Hom}_A(M_1, M_2)) - \chi(\operatorname{Ext}_A^1(M_1, M_2)) = -r(M_2)\chi(M_1)
χ(Hom⁡A(M2,M1))−χ(Ext⁡A1(M2,M1))=r(M2)χ(M1). \chi(\operatorname{Hom}_A(M_2, M_1)) - \chi(\operatorname{Ext}_A^1(M_2, M_1)) = r(M_2)\chi(M_1).

(d) Show that for every ordered pair of finitely generated A-modules M1,M2M_1, M_2:

c(M1⊗AM2)−c(Tor1A(M1,M2))=r(M1)c(M2)+r(M2)c(M1) c(M_1 \otimes_A M_2) - c(\mathrm{Tor}_1^A(M_1, M_2)) = r(M_1)c(M_2) + r(M_2)c(M_1)
c(HomA(M1,M2))−c(ExtA1(M1,M2))=r(M1)c(M2)−r(M2)c(M1).(*) c(\mathrm{Hom}_A(M_1, M_2)) - c(\mathrm{Ext}_A^1(M_1, M_2)) = r(M_1)c(M_2) - r(M_2)c(M_1). \tag{*}

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.