Proposition 15.89.19. Let $R$ be a Noetherian ring. Let $f \in R$ be an element. Let $R^\wedge$ be the $f$-adic completion of $R$. Then the functor $M \mapsto (M^\wedge , M_ f, \text{can})$ defines an equivalence

$\text{Mod}^{fg}_ R \longrightarrow \text{Mod}^{fg}_{R^\wedge } \times _{\text{Mod}^{fg}_{(R^\wedge )_ f}} \text{Mod}^{fg}_{R_ f}$

Proof. The ring map $R \to R^\wedge$ is flat by Algebra, Lemma 10.97.2. It is clear that $R/fR = R^\wedge /fR^\wedge$. By Algebra, Lemma 10.97.1 the completion of a finite $R$-module $M$ is equal to $M \otimes _ R R^\wedge$. Hence the displayed functor of the proposition is equal to the functor occurring in Theorem 15.89.18. In particular it is fully faithful. Let $(M_1, M_2, \psi )$ be an object of the right hand side. By Theorem 15.89.18 there exists an $R$-module $M$ such that $M_1 = M \otimes _ R R^\wedge$ and $M_2 = M_ f$. As $R \to R^\wedge \times R_ f$ is faithfully flat we conclude from Algebra, Lemma 10.23.2 that $M$ is finitely generated, i.e., $M \in \text{Mod}^{fg}_ R$. This proves the proposition. $\square$

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