The Stacks project

Proof. Say the morphism $f : X \to S$ is given by the ring map $A \to B$ and that $\mathcal{F}$ is the quasi-coherent sheaf associated to the $B$-module $M$. By Morphisms, Lemma 29.15.2 we know that $A \to B$ is a finite type ring map, and by Properties, Lemma 28.16.1 we know that $M$ is a finite $B$-module. In particular the support of $\mathcal{F}$ is the closed subscheme of $\mathop{\mathrm{Spec}}(B)$ cut out by the annihilator $I = \{ x \in B \mid xm = 0\ \forall m \in M\} $ of $M$, see Algebra, Lemma 10.40.5. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $x$ and let $\mathfrak p \subset A$ be the prime ideal corresponding to $s$. Note that $X_ s = \mathop{\mathrm{Spec}}(B \otimes _ A \kappa (\mathfrak p))$ and that $\mathcal{F}_ s$ is the quasi-coherent sheaf associated to the $B \otimes _ A \kappa (\mathfrak p)$ module $M \otimes _ A \kappa (\mathfrak p)$. By Morphisms, Lemma 29.5.3 the support of $\mathcal{F}_ s$ is equal to $V(I(B \otimes _ A \kappa (\mathfrak p)))$. Since $B \otimes _ A \kappa (\mathfrak p)$ is of finite type over $\kappa (\mathfrak p)$ there exist finitely many elements $f_1, \ldots , f_ m \in I$ such that

Denote $i : Z \to X$ the closed subscheme cut out by $(f_1, \ldots , f_ m)$, in a formula $Z = \mathop{\mathrm{Spec}}(B/(f_1, \ldots , f_ m))$. Since $M$ is annihilated by $I$ we can think of $M$ as an $B/(f_1, \ldots , f_ m)$-module. In other words, $\mathcal{F}$ is the pushforward of a finite type module on $Z$. As $Z_ s = \text{Supp}(\mathcal{F}_ s)$ by construction, this proves the lemma. $\square$