Lemma 26.5.4. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\widetilde M$ be the sheaf of $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$-modules associated to $M$.

1. We have $\Gamma (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)}) = R$.

2. We have $\Gamma (\mathop{\mathrm{Spec}}(R), \widetilde M) = M$ as an $R$-module.

3. For every $f \in R$ we have $\Gamma (D(f), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)}) = R_ f$.

4. For every $f\in R$ we have $\Gamma (D(f), \widetilde M) = M_ f$ as an $R_ f$-module.

5. Whenever $D(g) \subset D(f)$ the restriction mappings on $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$ and $\widetilde M$ are the maps $R_ f \to R_ g$ and $M_ f \to M_ g$ from Lemma 26.5.1.

6. Let $\mathfrak p$ be a prime of $R$, and let $x \in \mathop{\mathrm{Spec}}(R)$ be the corresponding point. We have $\mathcal{O}_{\mathop{\mathrm{Spec}}(R), x} = R_{\mathfrak p}$.

7. Let $\mathfrak p$ be a prime of $R$, and let $x \in \mathop{\mathrm{Spec}}(R)$ be the corresponding point. We have $\widetilde M_ x = M_{\mathfrak p}$ as an $R_{\mathfrak p}$-module.

Moreover, all these identifications are functorial in the $R$ module $M$. In particular, the functor $M \mapsto \widetilde M$ is an exact functor from the category of $R$-modules to the category of $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$-modules.

Proof. Assertions (1) - (7) are clear from the discussion above. The exactness of the functor $M \mapsto \widetilde M$ follows from the fact that the functor $M \mapsto M_{\mathfrak p}$ is exact and the fact that exactness of short exact sequences may be checked on stalks, see Modules, Lemma 17.3.1. $\square$

Comment #3810 by Carles Sáez on

In part (7) it should be "We have $\widetilde M_x = M_\mathfrak{p}$."

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