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. $\square$

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BZK. Beware of the difference between the letter 'O' and the digit '0'.