Lemma 10.77.5. Let $R$ be a ring. Let $I \subset R$ be a nilpotent ideal. Let $\overline{P}$ be a projective $R/I$-module. Then there exists a projective $R$-module $P$ such that $P/IP \cong \overline{P}$.

Proof. By Lemma 10.77.2 we can choose a set $A$ and a direct sum decomposition $\bigoplus _{\alpha \in A} R/I = \overline{P} \oplus \overline{K}$ for some $R/I$-module $\overline{K}$. Write $F = \bigoplus _{\alpha \in A} R$ for the free $R$-module on $A$. Choose a lift $p : F \to F$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$ of $\bigoplus _{\alpha \in A} R/I$. Note that $p^2 - p \in \text{End}_ R(F)$ is a nilpotent endomorphism of $F$ (as $I$ is nilpotent and the matrix entries of $p^2 - p$ are in $I$; more precisely, if $I^ n = 0$, then $(p^2 - p)^ n = 0$). Hence by Lemma 10.32.7 we can modify our choice of $p$ and assume that $p$ is a projector. Set $P = \mathop{\mathrm{Im}}(p)$. $\square$

Comment #1192 by Mohamed Hashi on

In the statement, $\overline{P}$ should be a projective $R/I$-module.

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