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A (unital) $R$-module $M$ can also be viewed as given by a (unital) ring morphism $\rho:R\to \text{End}(M)$. This point of view can be extended to many other examples of "compatible actions".

The definition of an $\mathcal{O}_X$-module $F$ is a sheaf of abelian group $F$ equipped with actions $\mathcal{O}_X(U)\times F(U)\to F(U)$ on fibers that are compatible with restriction.

Mimicking the above point of view, I tried to give it an equivalent definition:

An $\mathcal{O}_X$-module $F$ is a sheaf of abelian groups with a morphism of sheafs of rings $\rho:\mathcal{O}_X\to \mathcal{Hom}(F,F)$ from $\mathcal{O}_X$ to the sheaf Hom.

Is it reasonable?

cngzz1
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mez
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    Yes exactly, and this also sheds light on your previous question http://math.stackexchange.com/questions/1098613/sheafs-of-abelian-groups-are-the-same-as-underline-mathbbz-modules: The constant presheaf on ${\mathbb Z}$ is the initial ring object in the category of presheaves, so you get a canonical morphism ${\mathbb Z}^{\text{pre}}\to\text{Hom}(F,F)$, which then canonically extends to a sheaf morphism ${\mathbb Z}\to\text{Hom}(F,F)$ from the constant sheaf on ${\mathbb Z}$. Of course the compatibility with composition still has to be checked. – Hanno Jan 14 '15 at 09:13

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Yes it is reasonable, except that you maybe mean $\mathscr{O}_X$ instead of $\mathscr{O}_x$ ? ($\mathscr{O}_x$ is the stalk of $X$ at the point $x$, so it is not per se a sheaf on $X$.)

Olórin
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