Lemma 10.77.6. 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 10.77.2. 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 10.32.7 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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