Lemma 15.89.2. Assume $(\varphi : R \to S, I)$ satisfies the equivalent conditions of Lemma 15.89.1. The following are equivalent

1. for any $I$-power torsion module $M$, the natural map $M \to M \otimes _ R S$ is an isomorphism, and

2. $R/I \to S/IS$ is an isomorphism.

Proof. The implication (1) $\Rightarrow$ (2) is immediate. Assume (2). First assume that $M$ is annihilated by $I$. In this case, $M$ is an $R/I$-module. Hence, we have an isomorphism

$M \otimes _ R S = M \otimes _{R/I} S/IS = M \otimes _{R/I} R/I = M$

proving the claim. Next we prove by induction that $M \to M \otimes _ R S$ is an isomorphism for any module $M$ is annihilated by $I^ n$. Assume the induction hypothesis holds for $n$ and assume $M$ is annihilated by $I^{n + 1}$. Then we have a short exact sequence

$0 \to I^ nM \to M \to M/I^ nM \to 0$

and as $R \to S$ is flat this gives rise to a short exact sequence

$0 \to I^ nM \otimes _ R S \to M \otimes _ R S \to M/I^ nM \otimes _ R S \to 0$

Using that the canonical map is an isomorphism for $M' = I^ nM$ and $M'' = M/I^ nM$ (by induction hypothesis) we conclude the same thing is true for $M$. Finally, suppose that $M$ is a general $I$-power torsion module. Then $M = \bigcup M_ n$ where $M_ n$ is annihilated by $I^ n$ and we conclude using that tensor products commute with colimits. $\square$

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