Lemma 10.101.4. Let $\varphi : R \to R'$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

$M/IM$ is flat over $R/I$, and

$R' \otimes _ R M$ is flat over $R'$.

Set $I_2 = \varphi ^{-1}(\varphi (I^2)R')$. Then $M/I_2M$ is flat over $R/I_2$.

**Proof.**
We may replace $R$, $M$, and $R'$ by $R/I_2$, $M/I_2M$, and $R'/\varphi (I)^2R'$. Then $I^2 = 0$ and $\varphi $ is injective. By Lemma 10.99.8 and the fact that $I^2 = 0$ it suffices to prove that $\text{Tor}^ R_1(R/I, M) = K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$ is zero. Set $M' = M \otimes _ R R'$ and $I' = IR'$. By assumption the map $I' \otimes _{R'} M' \to M'$ is injective. Hence $K$ maps to zero in

\[ I' \otimes _{R'} M' = I' \otimes _ R M = I' \otimes _{R/I} M/IM. \]

Then $I \to I'$ is an injective map of $R/I$-modules. Since $M/IM$ is flat over $R/I$ the map

\[ I \otimes _{R/I} M/IM \longrightarrow I' \otimes _{R/I} M/IM \]

is injective. This implies that $K$ is zero in $I \otimes _ R M = I \otimes _{R/I} M/IM$ as desired.
$\square$

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