# Is this a true weakening of the quasi-coherence property?

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?

• Isn't (#) satisfied for all submodules of quasicoherent $O_X$-modules? Commented May 21, 2022 at 5:58
• @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. Commented May 22, 2022 at 0:41

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.
• 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. Commented May 22, 2022 at 23:01