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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