**Proof.**
Note that the equivalent conditions of both Lemma 15.89.1 and Lemma 15.89.2 are satisfied. We will use these without further mention. We first prove (1). Let $M$ be any $R$-module. Set $M' = M/M[I^\infty ]$ and consider the exact sequence

\[ 0 \to M[I^\infty ] \to M \to M' \to 0 \]

As $M[I^\infty ] = M[I^\infty ] \otimes _ R S$ we see that it suffices to show that $(M' \otimes _ R S)[(IS)^\infty ] = 0$. Write $I = (f_1, \ldots , f_ t)$. By Lemma 15.88.4 we see that $M'[I^\infty ] = 0$. Hence for every $n > 0$ the map

\[ M' \longrightarrow \bigoplus \nolimits _{i = 1, \ldots t} M', \quad x \longmapsto (f_1^ n x, \ldots , f_ t^ n x) \]

is injective. As $S$ is flat over $R$ also the corresponding map $M' \otimes _ R S \to \bigoplus _{i = 1, \ldots t} M' \otimes _ R S$ is injective. This means that $(M' \otimes _ R S)[I^ n] = 0$ as desired.

Next we prove (2). If $N$ is $I$-power torsion, then $N \otimes _ R S = N$ and the displayed map of (2) is an isomorphism by Algebra, Lemma 10.14.3. If $M$ is $I$-power torsion, then the image of any map $M \to N$ factors through $N[I^\infty ]$ and the image of any map $M \otimes _ R S \to N \otimes _ R S$ factors through $(N \otimes _ R S)[(IS)^\infty ]$. Hence in this case part (1) guarantees that we may replace $N$ by $N[I^\infty ]$ and the result follows from the case where $N$ is $I$-power torsion we just discussed.

Next we prove (3). The functor is fully faithful by (2). For essential surjectivity, we simply note that for any $IS$-power torsion $S$-module $N$, the natural map $N \otimes _ R S \to N$ is an isomorphism.
$\square$

## Comments (2)

Comment #8830 by Wataru on

Comment #9255 by Stacks project on

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