Does there exist a ring such that all modules are decomposable?

If we consider an infinite dimensional $k$-linear space $V$, then $V\cong V \oplus V $, then we can show that $R \cong R \oplus R$ as $R$-modules, where $R: = End_k(V)$ (this is an example from Rotman <An introduction to homological algebra> Example 2.36). We know $R$ is a von Neumann regular ring which does not have IBN. Maybe $R$ satisfy the condition but I can't prove it.


1 Answer 1


Not for rings with identity, because all rings with identity have simple modules which are indecomposable. (These are found by applying Zorn's Lemma to the proper right ideals of $R$ and finding a maximal right ideal $T$, after which $R/T$ is a simple right $R$ module.)

This is certainly the case for the ring $R$ which you have chosen, since endomorphism rings always have the identity endomorphism.

  • 2
    $\begingroup$ @JyrkiLahtonen I added elaboration about why there are simple modules. The OP's ring is an endomorphism ring, hence it has an identity, hence it has simple right and simple left modules... The title asks "is there a ring that all modules are decomposable" which I take to mean "has no indecomposable modules." $\endgroup$
    – rschwieb
    Mar 26 at 13:19
  • $\begingroup$ According to this answer, for a non-unital ring $R$ there exists an unital ring $S$ such that their categories of modules are isomorphic, so this answer should still be true even for rings without identity. $\endgroup$ Mar 26 at 18:31
  • $\begingroup$ Thank you very much, I complicated this question before. $\endgroup$
    – Well
    Mar 27 at 2:55

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