Lemma 47.24.2. Let $\varphi : R \to A$ be a finite type homomorphism of Noetherian rings.

1. $\varphi ^!$ maps $D^+(R)$ into $D^+(A)$ and $D^+_{\textit{Coh}}(R)$ into $D^+_{\textit{Coh}}(A)$.

2. if $\varphi$ is perfect, then $\varphi ^!$ maps $D^-(R)$ into $D^-(A)$, $D^-_{\textit{Coh}}(R)$ into $D^-_{\textit{Coh}}(A)$, and $D^ b_{\textit{Coh}}(R)$ into $D^ b_{\textit{Coh}}(A)$.

Proof. Choose a factorization $R \to P \to A$ as in the definition of $\varphi ^!$. The functor $- \otimes _ R^\mathbf {L} : D(R) \to D(P)$ preserves the subcategories $D^+, D^+_{\textit{Coh}}, D^-, D^-_{\textit{Coh}}, D^ b_{\textit{Coh}}$. The functor $R\mathop{\mathrm{Hom}}\nolimits (A, -) : D(P) \to D(A)$ preserves $D^+$ and $D^+_{\textit{Coh}}$ by Lemma 47.13.4. If $R \to A$ is perfect, then $A$ is perfect as a $P$-module, see More on Algebra, Lemma 15.82.2. Recall that the restriction of $R\mathop{\mathrm{Hom}}\nolimits (A, K)$ to $D(P)$ is $R\mathop{\mathrm{Hom}}\nolimits _ P(A, K)$. By More on Algebra, Lemma 15.74.15 we have $R\mathop{\mathrm{Hom}}\nolimits _ P(A, K) = E \otimes _ P^\mathbf {L} K$ for some perfect $E \in D(P)$. Since we can represent $E$ by a finite complex of finite projective $P$-modules it is clear that $R\mathop{\mathrm{Hom}}\nolimits _ P(A, K)$ is in $D^-(P), D^-_{\textit{Coh}}(P), D^ b_{\textit{Coh}}(P)$ as soon as $K$ is. Since the restriction functor $D(A) \to D(P)$ reflects these subcategories, the proof is complete. $\square$

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).