Hi.

In his Algebraic Geometry, Hartshorne proves that for any ringed spaces $(X,\mathcal O_X)$, category $Mod(X)$ of sheaves of $\mathcal O_X$-modules has enough injectives. If we tako $\mathcal O_X$ to be constant sheaf of rings $\mathbb Z$ (i.e. sheaf associated to a constant presheaf $\mathbb Z$), Hartshorne claims that $Mod(X)=Ab(X):=$ category of sheaves of abelian groups on X.

Why does this last claim hold? Of course, if one takes open subset $U \subseteq X$ that is connected, then section $\mathcal O_X(U)$ actually is ring isomorphic to $\mathbb Z$. But, for more 'complicated' open subsets $U \subseteq X$, sections $\mathcal O_X(U)$ become more complicated rings than $\mathbb Z$. So generally, many sections of a sheaf of $\mathcal O_X$-modules have more complicated structure than merely abelian group (i.e. $\mathbb Z$-module).

So, why is it obvious, according to Hartshorne, that $Mod(X)=Ab(X)$? I can only see that $Mod(X)$ is subcategory of $Ab(X)$.

In his "Algebraic Geometry", Hartshorne proves that for any ringed spaces $(X,\mathcal O_X)$, category $Mod(X)$ of sheaves of $\mathcal O_X$-modules has enough injectives. If we take $\mathcal O_X$ to be constant sheaf of rings $\underline{\mathbb Z}_X^{\natural}$ (i.e. sheaf associated to a constant presheaf $\underline{\mathbb Z}_X$), Hartshorne claims that $Mod(X)=Ab(X):=$ category of sheaves of abelian groups on X.

Why does this last claim hold? Of course, if one takes open subset $U \subseteq X$ that is connected, then section $\mathcal O_X(U)$ actually is ring isomorphic to $\mathbb Z$. But, for more 'complicated' open subsets $U \subseteq X$, sections $\mathcal O_X(U)$ become more complicated rings than $\mathbb Z$. So generally, many sections of a sheaf of $\mathcal O_X$-modules have more complicated structure than merely abelian group (i.e. $\mathbb Z$-module).

So, why is it obvious, according to Hartshorne, that $Mod(X)=Ab(X)$? I can only see that $Mod(X)$ is subcategory of $Ab(X)$.