Lemma 31.9.3. Let $S$ be a scheme. Let $\mathcal{F}$ be a finite type, quasi-coherent $\mathcal{O}_ S$-module. Let $Z_0 \subset S$ be the closed subscheme cut out by $\text{Fit}_0(\mathcal{F})$. Let $Z \subset S$ be the scheme theoretic support of $\mathcal{F}$. Then

1. $Z \subset Z_0 \subset S$ as closed subschemes,

2. $Z = Z_0 = \text{Supp}(\mathcal{F})$ as closed subsets,

3. there exists a finite type, quasi-coherent $\mathcal{O}_{Z_0}$-module $\mathcal{G}_0$ with

$(Z_0 \to X)_*\mathcal{G}_0 = \mathcal{F}.$

Proof. Recall that $Z$ is locally cut out by the annihilator of $\mathcal{F}$, see Morphisms, Definition 29.5.5 (which uses Morphisms, Lemma 29.5.4 to define $Z$). Hence we see that $Z \subset Z_0$ scheme theoretically by More on Algebra, Lemma 15.8.4 part (6). On the other hand we have $Z = \text{Supp}(\mathcal{F})$ set theoretically by Morphisms, Lemma 29.5.4 and we have $Z_0 = Z$ set theoretically by More on Algebra, Lemma 15.8.4 part (7). Finally, to get $\mathcal{G}_0$ as in part (3) we can either use that we have $\mathcal{G}$ on $Z$ as in Morphisms, Lemma 29.5.4 and set $\mathcal{G}_0 = (Z \to Z_0)_*\mathcal{G}$ or we can use Morphisms, Lemma 29.4.1 and the fact that $\text{Fit}_0(\mathcal{F})$ annihilates $\mathcal{F}$ by More on Algebra, Lemma 15.8.4 part (6). $\square$

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