Lemma 109.33.1. Let $R$ be a ring. Let $I \subset R$ be an ideal generated by a countable collection of idempotents. Then $I$ is projective as an $R$-module.

Proof. Say $I = (e_1, e_2, e_3, \ldots )$ with $e_ n$ an idempotent of $R$. After inductively replacing $e_{n + 1}$ by $e_ n + (1 - e_ n)e_{n + 1}$ we may assume that $(e_1) \subset (e_2) \subset (e_3) \subset \ldots$ and hence $I = \bigcup _{n \geq 1} (e_ n) = \mathop{\mathrm{colim}}\nolimits _ n e_ nR$. In this case

$\mathop{\mathrm{Hom}}\nolimits _ R(I, M) = \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{colim}}\nolimits _ n e_ nR, M) = \mathop{\mathrm{lim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits _ R(e_ nR, M) = \mathop{\mathrm{lim}}\nolimits _ n e_ nM$

Note that the transition maps $e_{n + 1}M \to e_ nM$ are given by multiplication by $e_ n$ and are surjective. Hence by Algebra, Lemma 10.86.4 the functor $\mathop{\mathrm{Hom}}\nolimits _ R(I, M)$ is exact, i.e., $I$ is a projective $R$-module. $\square$

Comment #1451 by H on

"given by multiplication by en are are surjective"

duplicate "are".

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