Bourbaki

Exercise 2, § 4

Commutative Algebra, VII, 571

It is set in small type as supplementary.

Let AA be an integrally closed Noetherian domain and E,FE, F two finitely generated AA-modules. Suppose that EE is torsion-free, that HomA(E,F)\mathrm{Hom}_A(E, F) is reflexive and that ExtA1(E,F)=0\mathrm{Ext}_A^1(E, F) = 0. Show then that FF is reflexive. (Prove first that FF is torsion-free; if T=F∗∗/cF(F)T = F^{**}/c_F(F), calculate Ass(HomA(E,T))\mathrm{Ass}(\mathrm{Hom}_A(E, T)) in two ways using the exact sequence:

0→Hom(E,F)→Hom(E,F∗∗)→Hom(E,T)→0 0 \to \mathrm{Hom}(E, F) \to \mathrm{Hom}(E, F^{**}) \to \mathrm{Hom}(E, T) \to 0
and Chapter IV, § 1, no. 4, Proposition 10.)

Solution

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