
Lemma 10.76.5. Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Let $\overline{P}$ be a finite projective $R/I$-module. Then there exists a finite projective $R$-module $P$ such that $P/IP \cong \overline{P}$.

Proof. Recall that $\overline{P}$ is a direct summand of a free $R/I$-module $\bigoplus _{\alpha \in A} R/I$ by Lemma 10.76.2. As $\overline{P}$ is finite, it follows that $\overline{P}$ is contained in $\bigoplus _{\alpha \in A'} R/I$ for some $A' \subset A$ finite. Hence we may assume we have a direct sum decomposition $(R/I)^{\oplus n} = \overline{P} \oplus \overline{K}$ for some $n$ and some $R/I$-module $\overline{K}$. Choose a lift $p \in \text{Mat}(n \times n, R)$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$ of $(R/I)^{\oplus n}$. Note that $p^2 - p \in \text{Mat}(n \times n, R)$ is nilpotent: as $I$ is locally nilpotent and the matrix entries $c_{ij}$ of $p^2 - p$ are in $I$ we have $c_{ij}^ t = 0$ for some $t > 0$ and then $(p^2 - p)^{tn^2} = 0$ (by looking at the matrix coefficients). Hence by Lemma 10.31.7 we can modify our choice of $p$ and assume that $p$ is a projector. Set $P = \mathop{\mathrm{Im}}(p)$. $\square$

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