Suppose $\left( M,\mathcal{O}_M \right)$ is a complex analytic space and let $\mathcal{F}$ be a coherent sheaf on $M$. Suppose that $s$ is a global section of $\mathcal{F}$ such that $s\left(x\right)=0$ in the fiber $\mathcal{F}_x\otimes k\left(x\right)$ for every $x \in M$.
Under what assumptions on $\mathcal{F}$ and $M$ can one show that $s=0$?
I have an argument for $M$ reduced and $\mathcal{F}$ torsion-free. Are there more general assumptions?
For $M$ reduced the assumptions imply there exists an open dense subset $V$ such that $s=0$ and the sheaf $\mathcal{F}$ is locally free on $V$. The section $s$ defines a morphism $\phi_s\colon\mathcal{O}_M\to \mathcal{F}$. Supposing there exists $p\in M\setminus V$ such that $s_p\neq 0$ then $\phi_{s,p}$ is injective because $s$ is not torsion. This would imply that $\phi_s$ is injective in an open neighborhood of $p$. But $V$ is dense and thus that is a contradiction.
Is there a reference that discusses this situation?
This question is related to this question.
Doesn't this follow from Nakayama's Lemma?
Let $M$ be a finitely generated $A$-module. For all $\mathfrak{p} \in \mathrm{Spec}(A)$ assume that $$ M_{\mathfrak{p}} \otimes_{A_{\mathfrak{p}} }A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \cong M \otimes_A A/\mathfrak{p}=0$$
then $$(\mathfrak{p}A_{\mathfrak{p}}) M_{\mathfrak{p}}=M_{\mathfrak{p}}$$
and by Nakayama's Lemma we get $M_{\mathfrak{p}}$ and by assumption it follows that $M=0$.