The Stacks project

Lemma 47.15.2. Let $A$ be a Noetherian ring. If $\omega _ A^\bullet $ is a dualizing complex, then the functor

\[ D : K \longmapsto R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ) \]

is an anti-equivalence $D_{\textit{Coh}}(A) \to D_{\textit{Coh}}(A)$ which exchanges $D^+_{\textit{Coh}}(A)$ and $D^-_{\textit{Coh}}(A)$ and induces an anti-equivalence $D^ b_{\textit{Coh}}(A) \to D^ b_{\textit{Coh}}(A)$. Moreover $D \circ D$ is isomorphic to the identity functor.

Proof. Let $K$ be an object of $D_{\textit{Coh}}(A)$. Pick an integer $n$ and consider the distinguished triangle

\[ \tau _{\leq n}K \to K \to \tau _{\geq n + 1}K \to \tau _{\leq n}K[1] \]

see Derived Categories, Remark 13.12.4. Since $\omega _ A^\bullet $ has finite injective dimension we see that $R\mathop{\mathrm{Hom}}\nolimits _ A(\tau _{\geq n + 1}K, \omega _ A^\bullet )$ has vanishing cohomology in degrees $\geq n - c$ for some constant $c$. On the other hand, we obtain a spectral sequence

\[ \mathop{\mathrm{Ext}}\nolimits _ A^ p(H^{-q}(\tau _{\leq n}K), \omega _ A^\bullet ) \Rightarrow \mathop{\mathrm{Ext}}\nolimits _ A^{p + q}(\tau _{\leq n}K, \omega _ A^\bullet ) = H^{p + q}(R\mathop{\mathrm{Hom}}\nolimits _ A(\tau _{\leq n}K, \omega _ A^\bullet )) \]

which shows that these cohomology modules are finite. Since for $n > p + q + c$ this is equal to $H^{p + q}(R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ))$ we see that $R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet )$ is indeed an object of $D_{\textit{Coh}}(A)$. By More on Algebra, Lemma 15.91.2 and the assumptions on the dualizing complex we obtain a canonical isomorphism

\[ K = R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet ) \otimes _ A^\mathbf {L} K \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ), \omega _ A^\bullet ) \]

Thus our functor has a quasi-inverse and the proof is complete. $\square$


Comments (2)

Comment #3623 by Janos Kollar on

A small remark, but if you say "anti-equivalence" in line 3 then probably should do the same in line 4.

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