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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

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2 Answers 2

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It is just not true that given a locally free sheaf $\mathcal G$ of rank $r$ and an arbitrary affine open subset $U\subset X$ the restriction $\mathcal G\vert U$ is isomorphic to the free $\mathcal O_U$-Module $\mathcal O_U^{\oplus r}$.
Here is a counterexample:

Consider a smooth projective curve $\overline X$ of positive genus (over $\mathbb C$, say) and a point $P\in \overline X$.
The curve $X=\overline X\setminus \{P\}$ is then affine.
Now let $Q\in X$ be an arbitrary point and consider the line bundle $\mathcal G=\mathcal O(Q)$, which is a locally free rank of rank one.
Although $U=X$ is affine, that line bundle is not trivial, i.e . is not isomorphic to $\mathcal O_X$:
Indeed, if it were there would exist a rational function $f\in Rat(X)$ with divisor $div f=1.Q$ and that rational function would extend to a rational function $\overline f\in Rat(\overline X)$ with divisor necessarily of the form $div (\overline f)=-1.P+1.Q$ (recall that the divisor of a rational function on $\overline X $ must have degree zero).
But this is a contradiction: on a smooth projective curve of positive genus two distinct points cannot be linearly equivalent.

Algebraic remark
In the dictionary mentioned in the question translating quasi-coherent sheaves on $X$ into $A$-modules the result above says that there exist a finitely generated projective module $\Gamma(X,\mathcal O_X(Q))$ of rank $1$ over $A=\Gamma(X,\mathcal O_X)$ which isn't isomorphic to $A$ as an $A$-module.

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  • $\begingroup$ This reminds me a theorem saying any holomorphic line bundle on a noncompact Riemann surface is homomorphically trivial. But there is no contradiction because noncompact R.S. is not projective. This gives us an example that algebraic Picard group not equal to analytic Picard group. $\endgroup$
    – AG learner
    Jan 5, 2019 at 22:37
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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.

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