Lemma 47.15.5 (0A7F)—The Stacks project

Lemma 47.15.5. Let $A$ be a Noetherian ring. If $\omega _ A^\bullet $ and $(\omega '_ A)^\bullet $ are dualizing complexes, then $(\omega '_ A)^\bullet $ is quasi-isomorphic to $\omega _ A^\bullet \otimes _ A^\mathbf {L} L$ for some invertible object $L$ of $D(A)$.

Proof. By Lemmas 47.15.3 and 47.15.4 the functor $K \mapsto R\mathop{\mathrm{Hom}}\nolimits _ A(R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ), (\omega _ A')^\bullet )$ maps $A$ to an invertible object $L$. In other words, there is an isomorphism

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

Since $L$ has finite tor dimension, this means that we can apply More on Algebra, Lemma 15.98.2 to see that

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

is an isomorphism for $K$ in $D^ b_{\textit{Coh}}(A)$. In particular, setting $K = \omega _ A^\bullet $ finishes the proof. $\square$

Comments (2)

Comment #3584 by Kestutis Cesnavicius on September 25, 2018 at 02:53

Continuing the somewhat pedantic theme of my comment on https://stacks.math.columbia.edu/tag/0AX0, I think in this lemma it would be better to say that $(\omega_A')^\bullet$ and $\omega_A^\bullet \otimes_A^L L$ are isomorphic as objects of $D(A)$ rather than quasi-isomorphic (the latter somehow entails a lift to a map of actual complexes that induces an isomorphism on cohomology).

Comment #3708 by Johan on October 22, 2018 at 17:54

OK, yes, this is another one of these types of things. Please see Derived Categories, Section 13.11 where we introduce the following sloppy terminology: If $K^\bullet$ and $L^\bullet$ are complexes of $\mathcal{A}$ then we sometimes say $K^\bullet$ is quasi-isomorphic to $L^\bullet$ to indicate that $K^\bullet$ and $L^\bullet$ are isomorphic objects of $D(\mathcal{A})$ .

Today I went to a talk by Bhargav Bhatt where he used this word in exactly this way, so I feel I am in good company! But still, you should continue to complain and if you get multiple mathematical friends to be on your side, then I will consider changing this.

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