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I've been studying projective modules in Rotman, as well as the topic of localization. Now on the Wikipedia article about them, there's an example of a locally free module that is not projective. The module in question is $R/I$ where $R$ is a direct product of countably infinite copies of $\mathbb{F}_2$ and $I$ is a direct sum of countably infinite copies of $\mathbb{F}_2$.

I understand why it is locally free, but in order to explain why it is not projective they mention the following theorem: If $I$ is an ideal of a commutative ring $R$ such that $R/I$ is a projective $R$-module, then $I$ is a principal ideal.

I'm not sure how to prove this (more general) theorem. I feel like I'm missing an obvious map to show that $I$ not principal implies $R/I$ not projective.

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Let $\pi:R\to R/I$ be the natural projection map. This is an $R$-module homomorphism (as well as a ring homomorphism). If $R/I$ is projective, then this map splits. Call such a splitting $\varphi:R/I\to R$. So we have $R= \varphi(R/I)\oplus \ker(\pi)$ (as internal direct sums of $R$-modules). But $\ker(\pi)=I$, and this shows that $I$ is cyclic. [In particular, $I$ will be generated by the idempotent of $R$ which corresponds to the direct sum decomposition.]

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    $\begingroup$ Ok I think I got it thanks. To get used to the different notations, basically $R/I$ projective means $(1)=R=R/I \oplus I=(1+I)\oplus I$ so $I$ cyclic as well? $\endgroup$
    – R. Morty
    Commented Dec 1, 2016 at 14:44
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    $\begingroup$ Basically, yes, except that $R/I$ is only isomorphic (not equal) to a submodule of $R$. $\endgroup$
    – Pace Nielsen
    Commented Dec 1, 2016 at 14:50
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    $\begingroup$ Ah yes sloppy notation on my part. Thanks again! $\endgroup$
    – R. Morty
    Commented Dec 1, 2016 at 15:19

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