Bourbaki

Exercise 12, § 4

Commutative Algebra, VII, 574

Let AA be an integrally closed Noetherian domain and M,NM, N two torsion-free finitely generated AA-modules of the same rank rr such that N⊂MN \subset M; let j:N→Mj : N \to M be the canonical injection. (a) Show that, for all kk, ⋀kj:⋀kN→⋀kM\bigwedge^k j : \bigwedge^k N \to \bigwedge^k M is pseudo-injective, that Coker⁡(⋀kj)\operatorname{Coker}\left( \bigwedge^k j \right) is a torsion AA-module and that:

χ(Coker⁡(⋀kj))=(r−1k−1)χ(Coker⁡j). \chi\left( \operatorname{Coker}\left( \bigwedge^k j \right) \right) = \binom{r-1}{k-1} \chi(\operatorname{Coker} j).
(b) Using (a), show that, if MM is a torsion-free finitely generated AA-module, the torsion submodule of ⋀kM\bigwedge^k M is pseudo-zero for all kk and
c(⋀kM)=(r−1k−1)c(M), c\left( \bigwedge^k M \right) = \binom{r-1}{k-1} c(M),
where MM is of rank rr; in particular, c(⋀rM)=c(M)c\left( \bigwedge^r M \right) = c(M).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.