Theorem 47.18.3. Let (A, \mathfrak m, \kappa ) be a Noetherian local ring. Let \omega _ A^\bullet be a normalized dualizing complex. Let E be an injective hull of the residue field. Let Z = V(\mathfrak m) \subset \mathop{\mathrm{Spec}}(A). Denote {}^\wedge derived completion with respect to \mathfrak m. Then
R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet )^\wedge \cong R\mathop{\mathrm{Hom}}\nolimits _ A(R\Gamma _ Z(K), E[0])
for K in D(A).
Proof.
Observe that E[0] \cong R\Gamma _ Z(\omega _ A^\bullet ) by Lemma 47.18.1. By More on Algebra, Lemma 15.91.13 completion on the left hand side goes inside. Thus we have to prove
R\mathop{\mathrm{Hom}}\nolimits _ A(K^\wedge , (\omega _ A^\bullet )^\wedge ) = R\mathop{\mathrm{Hom}}\nolimits _ A(R\Gamma _ Z(K), R\Gamma _ Z(\omega _ A^\bullet ))
This follows from the equivalence between D_{comp}(A, \mathfrak m) and D_{\mathfrak m^\infty \text{-torsion}}(A) given in Proposition 47.12.2. More precisely, it is a special case of Lemma 47.12.3.
\square
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