The Stacks project

Proof. By definition and Dualizing Complexes, Lemma 47.15.6 for every $x \in X$ the complex $\omega _{X, x}^\bullet $ is a dualizing complex over $\mathcal{O}_{X, x}$. By Dualizing Complexes, Lemma 47.20.2 we see that (2) holds.

To see (3) assume that $\mathcal{O}_{X, x}$ is Cohen-Macaulay. Let $n_ x$ be the unique integer such that $H^{n_{x}}(\omega _{X, x}^\bullet )$ is nonzero. For an affine neighbourhood $V \subset X$ of $x$ we have $\omega _ X^\bullet |_ V$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ V)$ hence there are finitely many nonzero coherent modules $H^ i(\omega _ X^\bullet )|_ V$. Thus after shrinking $V$ we may assume only $H^{n_ x}$ is nonzero, see Modules, Lemma 17.9.5. In this way we see that $\mathcal{O}_{X, v}$ is Cohen-Macaulay for every $v \in V$. This proves that $U$ is open as well as a Cohen-Macaulay scheme.

Proof of (1). The implication $\Leftarrow $ follows from (2). The implication $\Rightarrow $ follows from the discussion in the previous paragraph, where we showed that if $\mathcal{O}_{X, x}$ is Cohen-Macaulay, then in a neighbourhood of $x$ the complex $\omega _ X^\bullet $ has only one nonzero cohomology sheaf.

Assume $X$ is connected and Cohen-Macaulay. The above shows that the map $x \mapsto n_ x$ is locally constant. Since $X$ is connected it is constant, say equal to $n$. Setting $\omega _ X = H^ n(\omega _ X^\bullet )$ we see that the lemma holds because $\omega _ X$ is Cohen-Macaulay by Dualizing Complexes, Lemma 47.20.2 (and Cohomology of Schemes, Definition 30.11.4). $\square$