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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.

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This answer is inspired by the discussion at this question. 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.

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  • $\begingroup$ Sorry for obvious question. Why is $\Gamma(i_! \mathcal{O}_U) \oplus \Gamma(i_! \mathcal{O}_V) = 0$? $\endgroup$
    – user113988
    Commented Dec 1, 2016 at 13:49
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    $\begingroup$ The lower shriek is this case is just extension by zero, so $\Gamma(W, i_! \mathcal{O}_U) = \Gamma(W,\mathcal{O}_U)$ if $W\subset U$, and $0$ otherwise. So in this case $\Gamma(i_! \mathcal{O}_U) = \Gamma(i_! \mathcal{O}_V) = 0$, since $U$ and $V$ are strict subsets of $X$. $\endgroup$
    – Daniel Miller
    Commented Dec 1, 2016 at 16:48

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