3
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.]

$\endgroup$
  • $\begingroup$ 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 Dec 1 '16 at 14:44
  • $\begingroup$ Basically, yes, except that $R/I$ is only isomorphic (not equal) to a submodule of $R$. $\endgroup$ – Pace Nielsen Dec 1 '16 at 14:50
  • $\begingroup$ Ah yes sloppy notation on my part. Thanks again! $\endgroup$ – R. Morty Dec 1 '16 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.