Lemma 47.24.11. Let \varphi : A \to B be a finite type homomorphism of Noetherian rings. Let \omega _ A^\bullet be a dualizing complex for A. Set \omega _ B^\bullet = \varphi ^!(\omega _ A^\bullet ). Denote D_ A(K) = R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ) for K \in D_{\textit{Coh}}(A) and D_ B(L) = R\mathop{\mathrm{Hom}}\nolimits _ B(L, \omega _ B^\bullet ) for L \in D_{\textit{Coh}}(B). Then there is a functorial isomorphism
\varphi ^!(K) = D_ B(D_ A(K) \otimes _ A^\mathbf {L} B)
for K \in D_{\textit{Coh}}(A).
Proof.
Observe that \omega _ B^\bullet is a dualizing complex for B by Lemma 47.24.3. Let A \to B \to C be finite type homomorphisms of Noetherian rings. If the lemma holds for A \to B and B \to C, then the lemma holds for A \to C. This follows from Lemma 47.24.7 and the fact that D_ B \circ D_ B \cong \text{id} by Lemma 47.15.3. Thus it suffices to prove the lemma in case A \to B is a surjection and in the case where B is a polynomial ring over A.
Assume B = A[x_1, \ldots , x_ n]. Since D_ A \circ D_ A \cong \text{id}, it suffices to prove D_ B(K \otimes _ A B) \cong D_ A(K) \otimes _ A B[n] for K in D_{\textit{Coh}}(A). Choose a bounded complex I^\bullet of injectives representing \omega _ A^\bullet . Choose a quasi-isomorphism I^\bullet \otimes _ A B \to J^\bullet where J^\bullet is a bounded complex of B-modules. Given a complex K^\bullet of A-modules, consider the obvious map of complexes
\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , I^\bullet ) \otimes _ A B[n] \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet \otimes _ A B, J^\bullet [n])
The left hand side represents D_ A(K) \otimes _ A B[n] and the right hand side represents D_ B(K \otimes _ A B). Thus it suffices to prove this map is a quasi-isomorphism if the cohomology modules of K^\bullet are finite A-modules. Observe that the cohomology of the complex in degree r (on either side) only depends on finitely many of the K^ i. Thus we may replace K^\bullet by a truncation, i.e., we may assume K^\bullet represents an object of D^-_{\textit{Coh}}(A). Then K^\bullet is quasi-isomorphic to a bounded above complex of finite free A-modules. Therefore we may assume K^\bullet is a bounded above complex of finite free A-modules. In this case it is easy to that the displayed map is an isomorphism of complexes which finishes the proof in this case.
Assume that A \to B is surjective. Denote i_* : D(B) \to D(A) the restriction functor and recall that \varphi ^!(-) = R\mathop{\mathrm{Hom}}\nolimits (A, -) is a right adjoint to i_* (Lemma 47.13.1). For F \in D(B) we have
\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ B(F, D_ B(D_ A(K) \otimes _ A^\mathbf {L} B)) & = \mathop{\mathrm{Hom}}\nolimits _ B((D_ A(K) \otimes _ A^\mathbf {L} B) \otimes _ B^\mathbf {L} F, \omega _ B^\bullet ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ A(D_ A(K) \otimes _ A^\mathbf {L} i_*F, \omega _ A^\bullet ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ A(i_*F, D_ A(D_ A(K))) \\ & = \mathop{\mathrm{Hom}}\nolimits _ A(i_*F, K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ B(F, \varphi ^!(K)) \end{align*}
The first equality follows from More on Algebra, Lemma 15.73.1 and the definition of D_ B. The second equality by the adjointness mentioned above and the equality i_*((D_ A(K) \otimes _ A^\mathbf {L} B) \otimes _ B^\mathbf {L} F) = D_ A(K) \otimes _ A^\mathbf {L} i_*F (More on Algebra, Lemma 15.60.1). The third equality follows from More on Algebra, Lemma 15.73.1. The fourth because D_ A \circ D_ A = \text{id}. The final equality by adjointness again. Thus the result holds by the Yoneda lemma.
\square
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