Exercise 10, § 9

Let $A$ be a noetherian ring. Let $I$ be a bounded injective $A$-complex, whose homology is of finite type; for every $A$-complex $C$, denote by $\mathbf{D}(C)$ the complex $\mathrm{Homgr}_A(C, I)$.

a) Prove that the following conditions are equivalent:

(i) for every finitely generated $A$-module $M$, the canonical morphism of complexes $\alpha_M : M \to \mathbf{D}(D(M))$ (No. 5) is a homology isomorphism ; (ii) for every bounded $A$-complex $C$ whose homology is of finite type, the morphism $\alpha_C : C \to \mathbf{D}(D(C))$ (loc. cit.) is a homology isomorphism ; (iii) the canonical morphism $\alpha_A : A \to \mathrm{Homgr}_A(I, I)$ is a homology isomorphism.

(Adapt the proof of Theorem 1.)

The complex $I$ is said to be dualizing if it is bounded, injective, if $H(I)$ is of finite type and if it satisfies the equivalent conditions above.

b) If $A$ admits a dualizing module $\Omega$, every finite length injective resolution $(I, \delta)$ of $\Omega$ defines a dualizing complex (loc. cit.).

c) Let $B$ be an $A$-algebra which is a finitely generated $A$-module, $I$ a dualizing complex of $A$-modules; prove that the complex of $B$-modules $\mathrm{Hom}_{\mathbf{A}}(B, I)$ is dualizing. Thus every quotient ring of a Gorenstein ring of finite dimension admits a dualizing complex.

d) Let $I$ be a bounded injective $A$-complex. For $I$ to be dualizing, it is necessary and sufficient that the complex of $A_m$-modules $I_m$ be dualizing for every maximal ideal $m$ of $A$.

e) Let $I$ be a dualizing $A$-complex, $P$ a projective $A$-module of rank 1, $n$ an integer. The complex $I \otimes_A P(n)$ is dualizing.

f) Let $I, J$ be bounded injective $A$-complexes, whose homology is of finite type, and let $u : I \to J$ be a homology isomorphism; for $J$ to be dualizing it is necessary and sufficient that $I$ be so.