The Stacks project

Lemma 47.15.3. 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 $D_{\textit{Coh}}(A)$. From Lemma 47.15.2 we see $R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet )$ is an object of $D_{\textit{Coh}}(A)$. By More on Algebra, Lemma 15.98.2 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 (5)

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.

Comment #8374 by Haohao Liu on

I am sorry if this question is naïve. Is it trivial that sends to ? The proof ignores this part.

Comment #8376 by Nicolás on

Maybe this follows from a spectral sequence argument? Something like Here has finite injective dimension, hence the is zero for outside of a fixed interval.

Comment #8980 by on

Directly from the definition, if has finite injective dimension and if is bounded below (resp above), then is bounded above (resp below). Namely, you can represent by a bounded below (resp above) complex and by a bounded complex of injectives, and then the is computed by the hom complex which has the requisite boundedness.

Maybe this should be added to Section 15.69?

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  • 2 comment(s) on Section 47.15: Dualizing complexes

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