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X is a non singular projective variety over an infinite field k. How to prove there are no projectives with a surjective map to the structure sheaf O_X in Qch(X) and Coh(X). Coh(X) is the category of coherent sheaves on X.

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  • $\begingroup$ @SashaP Nope. Choices are few since you work in Qch and Coh. $\endgroup$ Commented Dec 31, 2017 at 9:55
  • $\begingroup$ Oh, I see, my bad $\endgroup$
    – SashaP
    Commented Dec 31, 2017 at 10:53

1 Answer 1

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Let $X$ be a projective variety over a field. (A variety is reduced and irreducible.) I'm going to use some elementary properties of the category of quasi-coherent $\mathcal{O}_X$-modules without mention in the discussion below.

If $\dim(X) = 0$, then $X$ is affine and we see that the category of quasi-coherent $\mathcal{O}_X$-modules does have enough projectives. Similarly for the category of coherent $\mathcal{O}_X$-modules.

If $\dim(X) > 0$, then I claim the only projective object of the category of quasi-coherent $\mathcal{O}_X$-modules is the zero object. Namely, let $\mathcal{F}$ be a nonzero quasi-coherent $\mathcal{O}_X$-module. Let $\mathcal{O}_X(1)$ be an ample invertible $\mathcal{O}_X$-module. For some $a \in \mathbf{Z}$ there exists a nonzero map $\mathcal{O}_X(a) \to \mathcal{F}$. Since $X$ is a variety and $\dim(X) > 0$, for every $b < a$ there is no nonzero map $\mathcal{O}_X(a) \to \mathcal{O}_X(b)$. On the other hand, we can find a set $I$, a function $I \to \{b \in \mathbf{Z} \mid b < a\}, i \mapsto b_i$, and a surjection $$ \bigoplus\nolimits_{i \in I} \mathcal{O}_X(b_i) \longrightarrow \mathcal{F} $$ By the remark above this map cannot have a splitting. Thus $\mathcal{F}$ cannot be projective in the category of quasi-coherent $\mathcal{O}_X$-modules. (A fortiori we find that $\mathcal{F}$ is not projective in the category of all $\mathcal{O}_X$-modules.)

For coherent $\mathcal{O}_X$-modules the argument is exactly the same.

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    $\begingroup$ +1 for the username. $\endgroup$ Commented Dec 31, 2017 at 15:33
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    $\begingroup$ Suggestion: accept this answer. $\endgroup$ Commented Jan 1, 2018 at 15:06

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