**Proof.**
For invertible objects of $D(A)$, see More on Algebra, Lemma 15.115.4 and the discussion in Section 47.15.

By Lemma 47.15.6 for every $\mathfrak p$ the complex $(\omega _ A^\bullet )_\mathfrak p$ is a dualizing complex over $A_\mathfrak p$. By definition and uniqueness of dualizing complexes (Lemma 47.15.5) we see that (1)(b) holds.

To see (1)(c) assume that $A_\mathfrak p$ is Gorenstein. Let $n_ x$ be the unique integer such that $H^{n_{x}}((\omega _ A^\bullet )_\mathfrak p)$ is nonzero and isomorphic to $A_\mathfrak p$. Since $\omega _ A^\bullet $ is in $D^ b_{\textit{Coh}}(A)$ there are finitely many nonzero finite $A$-modules $H^ i(\omega _ A^\bullet )$. Thus there exists some $f \in A$, $f \not\in \mathfrak p$ such that only $H^{n_ x}((\omega _ A^\bullet )_ f)$ is nonzero and generated by $1$ element over $A_ f$. Since dualizing complexes are faithful (by definition) we conclude that $A_ f \cong H^{n_ x}((\omega _ A^\bullet )_ f)$. In this way we see that $A_\mathfrak q$ is Gorenstein for every $\mathfrak q \in D(f)$. This proves that the set in (1)(c) is open.

Proof of (1)(a). The implication $\Leftarrow $ follows from (1)(b). The implication $\Rightarrow $ follows from the discussion in the previous paragraph, where we showed that if $A_\mathfrak p$ is Gorenstein, then for some $f \in A$, $f \not\in \mathfrak p$ the complex $(\omega _ A^\bullet )_ f$ has only one nonzero cohomology module which is invertible.

If $A[0]$ is a dualizing complex then $A$ is Gorenstein by part (1). Conversely, we see that part (1) shows that $\omega _ A^\bullet $ is locally isomorphic to a shift of $A$. Since being a dualizing complex is local (Lemma 47.15.7) the result is clear.
$\square$

## Comments (2)

Comment #4360 by comment_bot on

Comment #4502 by Johan on