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If $R$ is a left Artin ring with unity, we can conclude that the regular module $_R R$ can be decomposed into a direct sum of indecomposable projective modules. Since $R$ is left Artin, $_R R$ is a finite length module, and any finite length module can be decomposed into a direct sum of indecomposable modules.

Is there a counterexample for a non-Artin ring? What about a Noetherian ring?

Liang Chen
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  • This problem looks like a set theory problem. Assume you are decomposing every submodules of $R$, you continue to decompose the decomposable ones. Maybe you will never stop decomposing a regular module. But once you show that $R$ is a direct sum of indecomposable modules, they must be projective because they are direct summand of free modules. – Functor May 31 '24 at 08:36
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    Eric Wofsey explains a class of counterexamples here: https://math.stackexchange.com/a/1575706/232 We can take $R = F^{\mathbb{N}}$ where $F$ is a field. – Qiaochu Yuan May 31 '24 at 08:56
  • @Qiaochu Yuan Thank you very much for sharing the information. Is there a counterexample involving a Noetherian ring? – Liang Chen May 31 '24 at 09:21
  • @QiaochuYuan, the example in [math.stackexchange.com/a/1575706/232] uses $R$-module $M$ where $M$ is not isomorphic to $R$. But this question is about regular module $R$. – Functor May 31 '24 at 10:54
  • @Functor, the link given by Qiaochu Yuan in the third row of the answer provides an example involving the regular module $_R R = K^T=:M$. – Liang Chen May 31 '24 at 11:21
  • I just found out the answer, and it is always true for Noetherian rings. According to Proposition 10.14 on page 128 of the book Rings and Categories of Modules, any Artinian or Noetherian module can be decomposed into a direct sum of finitely many indecomposable modules. – Liang Chen May 31 '24 at 15:02

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