**Proof.**
Assume $R \to A$ is perfect. Choose $R \to P \to A$ as in the definition of $\varphi ^!$. Then $A$ is perfect as a $P$-modue (More on Algebra, Lemma 15.82.2). This shows that $\omega _{A/R}^\bullet $ is in $D^ b_{\textit{Coh}}(A)$ by Lemma 47.24.2. This proves (1)(a). To show $\omega _{A/R}^\bullet $ has finite tor dimension as a complex of $R$-modules, observe that $\omega _{A/R}^\bullet = \varphi ^!(R) = R\mathop{\mathrm{Hom}}\nolimits (A, P)[n]$ maps to $R\mathop{\mathrm{Hom}}\nolimits _ P(A, P)[n]$ in $D(P)$, which is perfect in $D(P)$ (More on Algebra, Lemma 15.74.15), hence has finite tor dimension in $D(R)$ as $R \to P$ is flat. This proves (1)(b). The object $R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _{A/R}^\bullet , \omega _{A/R}^\bullet )$ of $D(A)$ maps in $D(P)$ to

\begin{align*} R\mathop{\mathrm{Hom}}\nolimits _ P(\omega _{A/R}^\bullet , R\mathop{\mathrm{Hom}}\nolimits (A, P)[n]) & = R\mathop{\mathrm{Hom}}\nolimits _ P(R\mathop{\mathrm{Hom}}\nolimits _ P(A, P)[n], P)[n] \\ & = R\mathop{\mathrm{Hom}}\nolimits _ P(R\mathop{\mathrm{Hom}}\nolimits _ P(A, P), P) \end{align*}

This is equal to $A$ by the already used More on Algebra, Lemma 15.74.15. This proves (1)(c).

Assume $\varphi $ is flat. Then $R \to A$ is a perfect ring map (More on Algebra, Lemma 15.82.4) and we see that (1)(a), (1)(b), and (1)(c) hold. Of course, then $\omega _{A/R}^\bullet $ is $R$-perfect by (1)(a) and (1)(b) and the definitions. Let $R \to k$ be as in (2)(b). By Lemma 47.25.1 there is an isomorphism

\[ \omega _{A/R}^\bullet \otimes _ A^\mathbf {L} (A \otimes _ R k) \cong \omega ^\bullet _{A \otimes _ R k/k} \]

and the right hand side is a dualizing complex by Lemma 47.24.3. This finishes the proof.
$\square$

## Comments (2)

Comment #8352 by Bogdan on

Comment #8959 by Stacks project on