Is a locally free sheaf of modules free on some affine open basis? Let $(X,O_X)$ be a scheme, let $F$ be a locally free sheaf of $O_X$-modules on $X$.
According to this MSE answer, it is not necessarily true that for any affine open $U \subset X$ we have $F|_U$ is (isomorphic to) a direct sum of $O_U$'s.
However, is it possible to find an affine open base $\{U_i\}_{i\in I}$ of $X$,
such that $F|_{U_i}$ is isomorphic to $\bigoplus_{j\in J_i}(O_{U_i})$ for some index set $J_i$, for each $i \in I$?

My thinking so far is, I would like to show if $F$ is "free over" some open set $U$, then it is also "free over" any open subset of $U$ (or at least, for each $x \in U$ there is an open $V$ with $x \in V \subset U$ such that $F|_U$ is free over $V$).
Since a localization of a free module is free, I see why the above is true when $U$ is affine, but I'm not sure what to do for the non-affine case. Could anyone explain how to approach this? Or, is my thinking incorrect here?
 A: a few remarks that are maybe helpful:

*

*a quasi-coherent sheaf $\mathcal{F}$ on $X$ of $\mathcal{O}_X$-modules is locally free if there exists for every $x \in X$ an open neighbourhood $x\in U$ and an index set $I$ such that $\oplus_{i\in I} \mathcal{O}_X|_U \cong \mathcal{F}|_U$. So as you already noticed, $I$ depends on $x$.

*this open $U$ can be chosen to be affine, as affine opens form a basis for the Zariski topology on a scheme $X$.

So without loss of generality you can choose your neighbourhood $U$ around $x$ to be affine, but you don't need that:

*

*If you have any opens $j: V\subset U$ and given $\oplus_{i\in I} \mathcal{O}_X|_U \cong \mathcal{F}|_U$ then restricting to $V$ is just applying $j^*$ so$\oplus_{i\in I} \mathcal{O}_X|_V = j^*(\oplus_{i\in I}\mathcal{O}_X|_U )\cong j^*\mathcal{F}|_U = \mathcal{F}|_V$ because the functor $j^*$ preserves isos (any functor does) and commutes with direct sums.

A: Assuming you mean locally of finite dimension, the answer is yes.
We can write $X = \bigcup \{V \subseteq X \mid V$ open and $F|_V$ is isomorphic as a sheaf of vector spaces to $\mathcal{O}|_V^n$ for some $n\}$. For every such $V$, we can write $V = \bigcup \{U\subseteq V \mid U$ open and affine$\}$. Note that any $U \subseteq V$ automatically satisfies $F|_U \cong \mathcal{O}|_U^n$, so we can rewrite this as $V = \bigcup \{U\subseteq V \mid U$ open and affine and $F|_U \cong \mathcal{O}|_U^n\}$.
So we have $X = \bigcup \{ \bigcup \{U\subseteq V \mid U$ open and affine and $F|_U \cong \mathcal{O}|_U^n\} \mid V \subseteq X$ open, $n \in \mathbb{N}$, and $F|_V \cong \mathcal{O}|_V^n\} = \bigcup \{U\subseteq V \mid U$ open and affine and $F|_U \cong \mathcal{O}|_U^n$ and $V \subseteq X$ open, $n \in \mathbb{N}$, and $F|_V \cong \mathcal{O}|_V^n\} \subseteq \bigcup \{U \subseteq X \mid U$ open and affine, and there is some $n \in \mathbb{N}$ such that $F|_U \cong \mathcal{O}|_U^n\} \subseteq X$.
This shows that such affine sets do indeed cover $X$.
In slightly more generality, the set of affine opens forms a basis for the topology, while the set of $V$ where $F|_V$ is of finite dimension forms a downward closed basis. The intersection of a basis and a downward closed basis is always a basis.
