Lemma 47.27.5. Let R \to A be a flat ring map of finite presentation. Let K be a relative dualizing complex. Then A \to R\mathop{\mathrm{Hom}}\nolimits _ A(K, K) is an isomorphism.
Proof. By Algebra, Lemma 10.168.1 there exists a finite type \mathbf{Z} subalgebra R_0 \subset R and a flat finite type ring map R_0 \to A_0 such that A = A_0 \otimes _{R_0} R. By Lemmas 47.27.2, 47.27.3, and 47.27.4 there exists a relative dualizing complex K_0 \in D(A_0) and its derived base change is K. This reduces us to the situation discussed in the next paragraph.
Assume R Noetherian and let K be a relative dualizing complex for R \to A. Given a ring map R \to R' set A' = A \otimes _ R R' and K' = K \otimes _ A^\mathbf {L} A'. To finish the proof we show R\mathop{\mathrm{Hom}}\nolimits _{A'}(K', K') = A'. By Lemma 47.25.2 we know this is true whenever R' is Noetherian. Since a general R' is a filtered colimit of Noetherian R-algebras, we find the result holds by More on Algebra, Lemma 15.83.7. \square
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