Lemma 10.101.5. Let $\varphi : R \to R'$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

$I$ is nilpotent,

$R \to R'$ is injective,

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

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

Then $M$ is flat over $R$.

**Proof.**
Define inductively $I_1 = I$ and $I_{n + 1} = \varphi ^{-1}(\varphi (I_ n)^2R')$ for $n \geq 1$. Note that by Lemma 10.101.4 we find that $M/I_ nM$ is flat over $R/I_ n$ for each $n \geq 1$. It is clear that $\varphi (I_ n) \subset \varphi (I)^{2^ n}R'$. Since $I$ is nilpotent we see that $\varphi (I_ n) = 0$ for some $n$. As $\varphi $ is injective we conclude that $I_ n = 0$ for some $n$ and we win.
$\square$

## Comments (1)

Comment #9807 by Branislav Sobot on

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