# If a quasi-coherent sheaf is zero when restricted to each fibre then is it zero? [duplicate]

Let $$A$$ be a ring and $$M$$ an $$A$$-module. Suppose for every field $$k$$ with a ring map $$A \rightarrow k$$ the module $$M \otimes_A k = 0$$. Can we conclude $$M=0$$?

• Ah my bad. Thanks for letting me know. Commented 15 hours ago

Consider the module $$M=\mathbb{Q}_p/\mathbb{Z}_p$$ as a $$A=\mathbb{Z}_p$$-module. Then $$M=pM$$, so $$M \otimes \mathbb{F}_p=0$$. But $$M$$ is torsion, so $$M \otimes \mathbb{Q}_p=0$$.
Note that any morphism $$\mathbb{Z}_p \rightarrow k$$ for a field $$k$$ factors through $$\mathbb{F}_p$$ or $$\mathbb{Q}_p$$.
• Thank you, that is a wonderful counterexample. However, if I require that the $\mathrm{\textbf{derived}}$ restriction to each fibre be zero (i.e. $M \otimes^L_A k = 0$ for every $A \rightarrow k$), does that change the outcome? Commented yesterday
• @Calculus101: I do not know. This certainly works when $R$ is a PID (because the derived tensor product controls $Tor_1$, which describes torsion), and therefore when $R$ is Dedekind. We can probably assume that $R$ is local, and I suppose the next step is to check what happens for $R$ local regular of dimension $2$… Commented yesterday