Suppose $X$ is a scheme. I have been studying (finite rank) locally free $\mathcal{O}_{X}$-modules, and more generally, quasi-coherent sheaves on $X$ mainly from Ravi Vakil's excellent notes as well as Hartshorne. Let $\mathcal{F}$ be any quasi-coherent $\mathcal{O}_{X}$-module and let $\mathcal{G}$ be a locally free $\mathcal{O}_{X}$-module. I understand that for any affine $U = \text{spec}A \subset X$, we can fine an $A$-module $M$ such that $\mathcal{F} \vert_{U} \simeq \widetilde{M}.$ Hence this property also holds for locally free $\mathcal{O}_{X}$-modules. However, how do I know that for any such affine subset that there holds $\mathcal{G} \vert_{U} \simeq \mathcal{O}_{X}^{\oplus^{n}}$? I know that there exists a cover (not necessarily affine) $\left\lbrace U_{i} \right\rbrace_{i \in I}$ such that this holds locally at each $U_{i}$, but I am not sure how to show that the "freeness" property must then hold for any affine subset.
I strongly suspect an argument can be made from the transition functions, but I haven't been able to make any progress.
Any help is appreciated.
Thanks
I don't think this is true. There exist (finitely generated) $A$-modules $M$ which are locally free but not free themselves, which means that $\widetilde M$ is a locally free $\mathcal O_{Spec(A)}$-module on $Spec(A)$, but not a free $\mathcal O_{Spec(A)}$-module. That is, there is a cover of $Spec(A)$ by affine opens (the distinguished open sets) on which the restriction of $\widetilde M$ is free, but $\widetilde M$ is not free on all affine opens because it's not free on $Spec(A)$ itself.