Lemma 29.5.3. Let $\mathcal{F}$ be a finite type quasi-coherent module on a scheme $X$. Then

1. The support of $\mathcal{F}$ is closed.

2. For $x \in X$ we have

$x \in \text{Supp}(\mathcal{F}) \Leftrightarrow \mathcal{F}_ x \not= 0 \Leftrightarrow \mathcal{F}_ x \otimes _{\mathcal{O}_{X, x}} \kappa (x) \not= 0.$
3. For any morphism of schemes $f : Y \to X$ the pullback $f^*\mathcal{F}$ is of finite type as well and we have $\text{Supp}(f^*\mathcal{F}) = f^{-1}(\text{Supp}(\mathcal{F}))$.

Proof. Part (1) is a reformulation of Modules, Lemma 17.9.6. You can also combine Lemma 29.5.1, Properties, Lemma 28.16.1, and Algebra, Lemma 10.40.5 to see this. The first equivalence in (2) is the definition of support, and the second equivalence follows from Nakayama's lemma, see Algebra, Lemma 10.20.1. Let $f : Y \to X$ be a morphism of schemes. Note that $f^*\mathcal{F}$ is of finite type by Modules, Lemma 17.9.2. For the final assertion, let $y \in Y$ with image $x \in X$. Recall that

$(f^*\mathcal{F})_ y = \mathcal{F}_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{Y, y},$

see Sheaves, Lemma 6.26.4. Hence $(f^*\mathcal{F})_ y \otimes \kappa (y)$ is nonzero if and only if $\mathcal{F}_ x \otimes \kappa (x)$ is nonzero. By (2) this implies $x \in \text{Supp}(\mathcal{F})$ if and only if $y \in \text{Supp}(f^*\mathcal{F})$, which is the content of assertion (3). $\square$

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