Lemma 47.15.8 (0AX0)—The Stacks project

Lemma 47.15.8. Let $A \to B$ be a finite ring map of Noetherian rings. Let $\omega _ A^\bullet $ be a dualizing complex. Then $R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$ is a dualizing complex for $B$.

Proof. Let $\omega _ A^\bullet \to I^\bullet $ be a quasi-isomorphism with $I^\bullet $ a bounded complex of injectives. Then $\mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )$ is a bounded complex of injective $B$-modules (Lemma 47.3.4) representing $R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$. Thus $R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$ has finite injective dimension. By Lemma 47.13.4 it is an object of $D_{\textit{Coh}}(B)$. Finally, we compute

\[ \mathop{\mathrm{Hom}}\nolimits _{D(B)}(R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet ), R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet ), \omega _ A^\bullet ) = B \]

and for $n \not= 0$ we compute

\[ \mathop{\mathrm{Hom}}\nolimits _{D(B)}(R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet ), R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )[n]) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet ), \omega _ A^\bullet [n]) = 0 \]

which proves the last property of a dualizing complex. In the displayed equations, the first equality holds by Lemma 47.13.1 and the second equality holds by Lemma 47.15.3. $\square$

Comments (2)

Comment #3581 by Kestutis Cesnavicius on September 23, 2018 at 08:24

The dualizing complex is by definition a literal complex, whereas $R\mathrm{Hom}(B, \omega_A^\bullet)$ is only an object of some derived category, so it seems not very good to write in the statement that the latter is a dualizing complex. There is a similar issue in the statement of the next lemma.

Comment #3705 by Johan on October 22, 2018 at 17:38

Hmm... yes. I'm going to let this stand for now and change it if more people complain. For me an object of $D(B)$ is a complex of $B$ -modules. Moreover, if we have two complexes of $B$ -modules $K^\bullet$ and $L^\bullet$ which are quasi-isomorphic, then one is a dualizing complex if and only if the other one is. So actually, for me the terminology makes sense.

But I personally also experience cognitive dissonance when I try to think of $R\Hom(B, \omega_A^\bullet)$ as an object of $D(B)$ in the sense above. Namely, I prefer not to think of it as an actual complex of $B$ -modules, but rather as a certain ind-object of the category of complexes of $B$ -modules. (Since this is what you get when you read back to the definition of right derived functors between derived categories.)

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