Lemma 10.77.7. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume
$I$ is nilpotent,
$M/IM$ is a projective $R/I$-module,
$M$ is a flat $R$-module.
Then $M$ is a projective $R$-module.
Proof. By Lemma 10.77.5 we can find a projective $R$-module $P$ and an isomorphism $P/IP \to M/IM$. We are going to show that $M$ is isomorphic to $P$ which will finish the proof. Because $P$ is projective we can lift the map $P \to P/IP \to M/IM$ to an $R$-module map $P \to M$ which is an isomorphism modulo $I$. Since $I^ n = 0$ for some $n$, we can use the filtrations
\begin{align*} 0 = I^ nM \subset I^{n - 1}M \subset \ldots \subset IM \subset M \\ 0 = I^ nP \subset I^{n - 1}P \subset \ldots \subset IP \subset P \end{align*}
to see that it suffices to show that the induced maps $I^ aP/I^{a + 1}P \to I^ aM/I^{a + 1}M$ are bijective. Since both $P$ and $M$ are flat $R$-modules we can identify this with the map
\[ I^ a/I^{a + 1} \otimes _{R/I} P/IP \longrightarrow I^ a/I^{a + 1} \otimes _{R/I} M/IM \]
induced by $P \to M$. Since we chose $P \to M$ such that the induced map $P/IP \to M/IM$ is an isomorphism, we win. $\square$
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