Lemma 17.10.5. Let $(X, \mathcal{O}_ X)$ be ringed space. Let $\alpha : R \to \Gamma (X, \mathcal{O}_ X)$ be a ring homomorphism from a ring $R$ into the ring of global sections on $X$. Let $M$ be an $R$-module. The following three constructions give canonically isomorphic sheaves of $\mathcal{O}_ X$-modules:

Let $\pi : (X, \mathcal{O}_ X) \longrightarrow (\{ *\} , R)$ be the morphism of ringed spaces with $\pi : X \to \{ *\} $ the unique map and with $\pi $-map $\pi ^\sharp $ the given map $\alpha : R \to \Gamma (X, \mathcal{O}_ X)$. Set $\mathcal{F}_1 = \pi ^*M$.

Choose a presentation $\bigoplus _{j \in J} R \to \bigoplus _{i \in I} R \to M \to 0$. Set

\[ \mathcal{F}_2 = \mathop{\mathrm{Coker}}\left( \bigoplus \nolimits _{j \in J} \mathcal{O}_ X \to \bigoplus \nolimits _{i \in I} \mathcal{O}_ X \right). \]Here the map on the component $\mathcal{O}_ X$ corresponding to $j \in J$ given by the section $\sum _ i \alpha (r_{ij})$ where the $r_{ij}$ are the matrix coefficients of the map in the presentation of $M$.

Set $\mathcal{F}_3$ equal to the sheaf associated to the presheaf $U \mapsto \mathcal{O}_ X(U) \otimes _ R M$, where the map $R \to \mathcal{O}_ X(U)$ is the composition of $\alpha $ and the restriction map $\mathcal{O}_ X(X) \to \mathcal{O}_ X(U)$.

This construction has the following properties:

The resulting sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}_ M = \mathcal{F}_1 = \mathcal{F}_2 = \mathcal{F}_3$ is quasi-coherent.

The construction gives a functor from the category of $R$-modules to the category of quasi-coherent sheaves on $X$ which commutes with arbitrary colimits.

For any $x \in X$ we have $\mathcal{F}_{M, x} = \mathcal{O}_{X, x} \otimes _ R M$ functorial in $M$.

Given any $\mathcal{O}_ X$-module $\mathcal{G}$ we have

\[ \mathop{Mor}\nolimits _{\mathcal{O}_ X}(\mathcal{F}_ M, \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _ R(M, \Gamma (X, \mathcal{G})) \]where the $R$-module structure on $\Gamma (X, \mathcal{G})$ comes from the $\Gamma (X, \mathcal{O}_ X)$-module structure via $\alpha $.

## Comments (2)

Comment #24 by Pieter Belmans on

Comment #25 by Johan on