Lemma 47.15.7. Let A be a Noetherian ring. Let f_1, \ldots , f_ n \in A generate the unit ideal. If \omega _ A^\bullet is a complex of A-modules such that (\omega _ A^\bullet )_{f_ i} is a dualizing complex for A_{f_ i} for all i, then \omega _ A^\bullet is a dualizing complex for A.
Proof. Consider the double complex
The associated total complex is quasi-isomorphic to \omega _ A^\bullet for example by Descent, Remark 35.3.10 or by Derived Categories of Schemes, Lemma 36.9.4. By assumption the complexes (\omega _ A^\bullet )_{f_ i} have finite injective dimension as complexes of A_{f_ i}-modules. This implies that each of the complexes (\omega _ A^\bullet )_{f_{i_0} \ldots f_{i_ p}}, p > 0 has finite injective dimension over A_{f_{i_0} \ldots f_{i_ p}}, see Lemma 47.3.8. This in turn implies that each of the complexes (\omega _ A^\bullet )_{f_{i_0} \ldots f_{i_ p}}, p > 0 has finite injective dimension over A, see Lemma 47.3.2. Hence \omega _ A^\bullet has finite injective dimension as a complex of A-modules (as it can be represented by a complex endowed with a finite filtration whose graded parts have finite injective dimension). Since H^ n(\omega _ A^\bullet )_{f_ i} is a finite A_{f_ i} module for each i we see that H^ i(\omega _ A^\bullet ) is a finite A-module, see Algebra, Lemma 10.23.2. Finally, the (derived) base change of the map A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet ) to A_{f_ i} is the map A_{f_ i} \to R\mathop{\mathrm{Hom}}\nolimits _ A((\omega _ A^\bullet )_{f_ i}, (\omega _ A^\bullet )_{f_ i}) by More on Algebra, Lemma 15.99.2. Hence we deduce that A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet ) is an isomorphism and the proof is complete. \square
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