Lemma 17.10.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $x \in X$ be a point. Assume that $x$ has a fundamental system of quasi-compact neighbourhoods. Consider any quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$. Then there exists an open neighbourhood $U$ of $x$ such that $\mathcal{F}|_ U$ is isomorphic to the sheaf of modules $\mathcal{F}_ M$ on $(U, \mathcal{O}_ U)$ associated to some $\Gamma (U, \mathcal{O}_ U)$-module $M$.

**Proof.**
First we may replace $X$ by an open neighbourhood of $x$ and assume that $\mathcal{F}$ is isomorphic to the cokernel of a map

The problem is that this map may not be given by a “matrix”, because the module of global sections of a direct sum is in general different from the direct sum of the modules of global sections.

Let $x \in E \subset X$ be a quasi-compact neighbourhood of $x$ (note: $E$ may not be open). Let $x \in U \subset E$ be an open neighbourhood of $x$ contained in $E$. Next, we proceed as in the proof of Lemma 17.3.5. For each $j \in J$ denote $s_ j \in \Gamma (X, \bigoplus \nolimits _{i \in I} \mathcal{O}_ X)$ the image of the section $1$ in the summand $\mathcal{O}_ X$ corresponding to $j$. There exists a finite collection of opens $U_{jk}$, $k \in K_ j$ such that $E \subset \bigcup _{k \in K_ j} U_{jk}$ and such that each restriction $s_ j|_{U_{jk}}$ is a finite sum $\sum _{i \in I_{jk}} f_{jki}$ with $I_{jk} \subset I$, and $f_{jki}$ in the summand $\mathcal{O}_ X$ corresponding to $i \in I$. Set $I_ j = \bigcup _{k \in K_ j} I_{jk}$. This is a finite set. Since $U \subset E \subset \bigcup _{k \in K_ j} U_{jk}$ the section $s_ j|_ U$ is a section of the finite direct sum $\bigoplus _{i \in I_ j} \mathcal{O}_ X$. By Lemma 17.3.2 we see that actually $s_ j|_ U$ is a sum $\sum _{i \in I_ j} f_{ij}$ and $f_{ij} \in \mathcal{O}_ X(U) = \Gamma (U, \mathcal{O}_ U)$.

At this point we can define a module $M$ as the cokernel of the map

with matrix given by the $(f_{ij})$. By construction (2) of Lemma 17.10.5 we see that $\mathcal{F}_ M$ has the same presentation as $\mathcal{F}|_ U$ and therefore $\mathcal{F}_ M \cong \mathcal{F}|_ U$. $\square$

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## Comments (2)

Comment #1781 by Keenan Kidwell on

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