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is the following true: if I have a coherent sheaf $F$ on a noetherian scheme $X$ with a point $x$ and the stalk $F_x$ is zero, then there is a neighborhood $U$ of $x$, such that the restriction of $F$ to $U$ is zero?

Thank you

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up vote 3 down vote accepted

Yes. For the locality of the problem, you can assume that $X$ is affine: $X = \mathrm{Spec} A$ and $F = \tilde{M}$, where $A$ is a noetherian ring and $M$ is a finite $A$-module. Let $P$ a prime such that $M_P = 0$. Let $\{ x_1, \dots, x_n \}$ a set of generators of $M$ as an $A$-module. Then exist $s_i \in A \setminus P$ such that $s_i x_i = 0$. Pick $s = s_1 \cdots s_n$, then $F \vert_{D(s)} = 0$.

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Thanks a lot, Andrea! – Descartes Aug 13 '11 at 15:07
You're welcome! With a similar proof, you may try to solve Exercise II.5.8(a) of Hartshorne. – Andrea Aug 13 '11 at 16:44
Do we really need Noetherianity here? It seems that all is used is that the module is finitely generated. – AreaMan Apr 6 at 18:04

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