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 $M=0$?
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1$\begingroup$ Please don't change your question in such a way that it invalidates an existing answer. Instead, please post a new question with the correct hypotheses. See here on Meta for our policies about this. $\endgroup$– KReiserCommented 16 hours ago
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$\begingroup$ Ah my bad. Thanks for letting me know. $\endgroup$– Calculus101Commented 15 hours ago
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No, we can’t.
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 $\mathbb{Q}_p$.
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1$\begingroup$ 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? $\endgroup$ Commented yesterday
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$\begingroup$ @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 $2$… $\endgroup$– AphelliCommented yesterday