Lemma 47.15.6. Let $A$ be a Noetherian ring. Let $B = S^{-1}A$ be a localization. If $\omega _ A^\bullet$ is a dualizing complex, then $\omega _ A^\bullet \otimes _ A B$ is a dualizing complex for $B$.

Proof. Let $\omega _ A^\bullet \to I^\bullet$ be a quasi-isomorphism with $I^\bullet$ a bounded complex of injectives. Then $S^{-1}I^\bullet$ is a bounded complex of injective $B = S^{-1}A$-modules (Lemma 47.3.8) representing $\omega _ A^\bullet \otimes _ A B$. Thus $\omega _ A^\bullet \otimes _ A B$ has finite injective dimension. Since $H^ i(\omega _ A^\bullet \otimes _ A B) = H^ i(\omega _ A^\bullet ) \otimes _ A B$ by flatness of $A \to B$ we see that $\omega _ A^\bullet \otimes _ A B$ has finite cohomology modules. Finally, the map

$B \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet \otimes _ A B, \omega _ A^\bullet \otimes _ A B)$

is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma 15.92.2. $\square$

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