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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).