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The Stacks project

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


Comments (2)

Comment #1192 by Mohamed Hashi on

In the statement, should be a projective -module.

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