Let $X$ be an affine scheme and $\mathcal{E}$ a finitely generated locally free sheaf on $X$. It is obvious that $\mathcal{E}$ is a projective object in the category Qcoh$(X)$ since we can pass to rings and modules. 


My question is: is $\mathcal{E}$ still projective when we consider it in the larger category $\mathcal{O}\_X$-mod which consists of all $\mathcal{O}\_X$-modules? If it is true, how to prove it in a "cohomological way"?


I'm not sure whether this question is suitable for mathoverflow. Please feel free to move it if necessary.



Is a locally free sheaf projective in the category of $\mathcal{O}_X$-modules when $X$ is an affine scheme?
This answer is inspired by the discussion [at this question](https://mathoverflow.net/questions/5378/when-are-there-enough-projective-sheaves-on-a-space-x). Let $X$ be an integral affine scheme admitting an open cover $X=U\cup V$ with $U$, $V$ and $X$ all distinct. I claim that $\mathscr O\_X$ is not projective in $\operatorname{Mod}(\mathscr O\_X)$. Let $i\colon U\hookrightarrow X$ and $j\colon V\hookrightarrow X$ be the inclusion morphisms. There is an obvious surjection $p\colon i\_! \mathscr O\_U\oplus j\_! \mathscr O\_V\twoheadrightarrow \mathscr O\_X$ (check surjectivity on stalks). If $\mathscr O\_X$ were projective, $p$ would split, so we would have $\Gamma(\mathscr O\_X)\hookrightarrow \Gamma(i\_!\mathscr O\_U)\oplus \Gamma(j\_! \mathscr O\_V)=0$, a contradiction. 


