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 36.8.4. 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 10.22.2. 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 15.92.2. 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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