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We say a category $\mathcal{N}$ is exact if it is additive and is endowed with an exact structure. In brief, it is an additive category with a predetermined class of short exact sequences in its ambient abelian category $\mathcal{A}$. It is idempotent complete if for every idempotent endomorphism $p$ ($p^2 = p$) in $\mathcal{N}$; $ker(p) \in \mathcal{N}$ (up to isomorphism).

As an example, we can take $P(R)$ the category of finitely generated projective $R$-modules ($R$ is a commutative ring with unity). Since for any idempotent endomorphism $p : P \rightarrow P$, the exact sequence $0\rightarrow ker(p) \rightarrow P \rightarrow Im(p) \rightarrow 0$ splits.

My question is if I take the category $P(X)$; which we define as the category of locally free $\mathcal{O}_X$-modules of finite rank, where $X$ is a quasi-compact scheme. Then will $P(X)$ be idempotent complete? I understand that if I take $X = Spec(R)$ the category $P(X)$ coincides with $P(R)$, but is that sufficient to conclude that $P(X)$ is idempotent complete?

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    $\begingroup$ $P(X)$ is a limit of categories of the form $P(R)$, and limits preserve idempotent completeness. $\endgroup$ Commented Sep 12, 2023 at 7:49
  • $\begingroup$ If you could please elaborate a little it would mean a lot to me. $\endgroup$
    – user443060
    Commented Sep 12, 2023 at 10:04
  • $\begingroup$ $X\mapsto P(X)$ is a sheaf of categories, so $P(X)$ is the limit of $P(U)$ as $U$ ranges over all affine opens of $X$. On the other hand, since idempotents and their images are preserved by arbitrary functors, a limit of idempotent complete categories is again idempotent complete. Thus $P(X)$ is idempotent complete. $\endgroup$ Commented Sep 13, 2023 at 14:42

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