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Let $R$ be a commutative Noetherian ring, and $X=$Spec$(R)$ the associated affine scheme. Let $F$ be a sheaf of $O_X$-modules. Consider the following condition

  • (#) For all containments $V \subseteq U$ of affine open subschemes of $X$, the natural map $O(V) \otimes_{O(U)} F(U) \rightarrow F(V)$ of $O(V)$-modules is injective.

One can reduce to the case where $V = D(f)$ where $f\in \Gamma(U,O_X)$.

One of the equivalent conditions for quasi-coherence is that the maps in (#) are isomorphisms. Curiously, though, the examples I know of sheaves that are not quasi-coherent also fail the condition (#).

My question is: Are there any (natural) examples of $O_X$-module sheaves that satisfy (#) but fail to be quasi-coherent? And if this is impossible, would the answer be different if the requirement that $X$ be affine were relaxed?

Also, does anyone know a name for this condition?

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    $\begingroup$ Isn't (#) satisfied for all submodules of quasicoherent $O_X$-modules? $\endgroup$
    – Laurent Moret-Bailly
    May 21, 2022 at 5:58
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    $\begingroup$ @LaurentMoret-Bailly Oh that's really good. Thank you! And now that I work through the commutative diagram, I guess a subsheaf of an $O_X$-module that satisfies (#) also satisfies (#). If you post this as an answer (with a brief explanation and an example), I'd be happy to accept it. $\endgroup$
    – Neil Epstein
    May 22, 2022 at 0:41

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Any submodule of a quasicoherent $O_X$-module satisfies (#): this is clear via reduction to principal open sets, and the fact that localization is exact. More generally, as Neil observes, if $F$ satisfies (#) then so does every submodule of $F$.

For instance, if $F$ is quasicoherent on $X$ and $j:U\hookrightarrow X$ is open, then the extension by zero $j_!(F_{\mid U})$ satisfies (#) but is not quasicoherent in general.

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  • $\begingroup$ And for the uninitiated (as I was until a couple days ago), a good example is the following. Let $X = $Spec$(R)$, $R$ a DVR, $U=$the singleton set consisting of the generic point. Let $F=O_X$ and $G=j_!(F|_U)$. Then $G(X)=0$ and $G(U)$ is the fraction field of $R$. Not quasi-coherent, but it is a sheaf of ideals. $\endgroup$
    – Neil Epstein
    May 22, 2022 at 23:01

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