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 $\mathcal{F}_ M \cong \mathcal{F}|_ U$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (4)

Comment #1781 by Keenan Kidwell on

Comment #1820 by Johan on

Comment #7707 by Ryo Suzuki on

Comment #7967 by Stacks Project on