The Stacks project

Proof. Consider the base change $f' : X' \to \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ of $f$ by the morphism $g : \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \to S$ and denote $g' : X' \to X$ the other projection. Then

The first equality because $g$ induces an isomorphism on local rings at $s$ and the second by flat base change (Cohomology of Schemes, Lemma 30.5.2). Of course $s \in \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ is not in the image of $f'$. Thus we may assume $S$ is the spectrum of a local ring $(A, \mathfrak m)$ and $s$ corresponds to $\mathfrak m$. By Schemes, Lemma 26.24.1 the sheaf $f_*\mathcal{F}$ is quasi-coherent, say corresponding to the $A$-module $M$. As $s$ is not in the image of $f$ we see that $X = \bigcup _{a \in \mathfrak m} f^{-1}D(a)$ is an open covering. Since $X$ is quasi-compact we can find $a_1, \ldots , a_ n \in \mathfrak m$ such that $X = f^{-1}D(a_1) \cup \ldots \cup f^{-1}D(a_ n)$. It follows that

is injective. Hence for any nonzero element $m$ of the stalk $M_\mathfrak p$ there exists an $i$ such that $a_ i^ n m$ is nonzero for all $n \geq 0$. Thus $\mathfrak m$ is not weakly associated to $M$. $\square$