Let $S$ be a smooth curve and $X$ a noetherian scheme (over $\mathbb{C}$). Let $\Sigma\subset S$ be a nonempty set of closed points. Let $\mathcal{E}$ be a coherent sheaf on $X\times S$. We have then a restriction morphism $$\Phi_\Sigma:\mathcal{E}\longrightarrow\prod_{s\in\Sigma}\mathcal{E}_s \ .$$ My question: if $\Sigma$ is infinite, is $\Phi_\Sigma$ injective ? Or must we add supplementary hypotheses, such that "$\mathcal{E}$ is flat on $S$) ?
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This is false. Take your sheaf to be $\mathcal{O}_{X\times S}/m_s^2$ where $m_s$ is the ideal sheaf of a closed point $s\in S$. Then, the map you described vanishes identically on $m_s\mathcal{O}_{X\times S}/m_s^2 \mathcal{O}_{X\times S}$, because at the $s$ component of the product it is reduction modulo $m_s$ while at every other fiber its identically $0$. This shows that even if you take all closed points it may not be injective.
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$\begingroup$ You are right. So another hypothesis such that flatness over $S$ is needed. $\endgroup$ Commented Jan 15 at 9:59
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1$\begingroup$ Yes, and flatness over $S$ suffices. As if $e$ is a section which vanishes on all the infinite collection of fibers $X_s$, then since $\mathcal{E}$ is coherent, the section $e$ must vanish on the fiber at the generic point as well (the collection of fibers on which $e$ vanishes has to be a constructible set), which contradicts flatness over $S$ since it s then not torsion free as an $\mathcal{O}_S$-module. $\endgroup$ Commented Jan 15 at 13:35
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$\begingroup$ . Thank you very much. But why is the collection of fibers on which $e$ vanishes a constructible set ? $\endgroup$ Commented Jan 16 at 15:41