Lemma 49.2.12. Let $A \to B$ be a quasi-finite homomorphism of Noetherian rings. Let $\omega _{B/A}^\bullet \in D(B)$ be the algebraic relative dualizing complex discussed in Dualizing Complexes, Section 47.25. Then there is a (nonunique) isomorphism $\omega _{B/A} = H^0(\omega _{B/A}^\bullet )$.
Proof. Choose a factorization $A \to B' \to B$ where $A \to B'$ is finite and $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(B)$ is an open immersion. Then $\omega _{B/A}^\bullet = \omega _{B'/A}^\bullet \otimes _ B^\mathbf {L} B'$ by Dualizing Complexes, Lemmas 47.24.7 and 47.24.9 and the definition of $\omega _{B/A}^\bullet $. Hence it suffices to show there is an isomorphism when $A \to B$ is finite. In this case we can use Dualizing Complexes, Lemma 47.24.8 to see that $\omega _{B/A}^\bullet = R\mathop{\mathrm{Hom}}\nolimits (B, A)$ and hence $H^0(\omega ^\bullet _{B/A}) = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ as desired. $\square$
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