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 10.166.1 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 15.78.7. $\square$

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