Lemma 30.11.5. Let $X$ be a regular scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. The following are equivalent
$\mathcal{F}$ is Cohen-Macaulay and $\text{Supp}(\mathcal{F}) = X$,
$\mathcal{F}$ is finite locally free of rank $> 0$.
Proof. Let $x \in X$. If (2) holds, then $\mathcal{F}_ x$ is a free $\mathcal{O}_{X, x}$-module of rank $> 0$. Hence $\text{depth}(\mathcal{F}_ x) = \dim (\mathcal{O}_{X, x})$ because a regular local ring is Cohen-Macaulay (Algebra, Lemma 10.106.3). Conversely, if (1) holds, then $\mathcal{F}_ x$ is a maximal Cohen-Macaulay module over $\mathcal{O}_{X, x}$ (Algebra, Definition 10.103.8). Hence $\mathcal{F}_ x$ is free by Algebra, Lemma 10.106.6. $\square$
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