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Let's say that a (right) module $M$ is well complemented if every non-zero submodule of $M$ has an indecomposable direct summand (by the way, is there a better or more standard name for this property?). For instance, every module of finite uniform dimension is well complemented.

Question. Is the regular right module $R_R$ of a von Neumann regular ring $R$ well complemented?

As a recall, a ring $R$ is von Neumann regular if, for every $x \in R$, there exists $y \in R$ such that $x = xyx$.

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  • $\begingroup$ I think if you take an atomless boolean algebra the corresponding boolean ring is regular but has no primitive idempotents so no indecomposable summands. $\endgroup$ Jul 4 at 19:59
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The answer is no. Take a compact totally disconnected space $X$ with no isolated points, like the Cantor set. Let $K$ be any field and let $R$ be the ring of locally constant functions $f\colon X\to K$ with pointwise operations. This is a commutative von Neumann regular ring. The idempotents of $R$ are precisely the characteristic functions $1_K$ of clopen sets $K$. An orthogonal decomposition of $1_K$ into idempotents corresponds to writing $K$ as a disjoint union of clopen sets. Since $X$ has no isolated points, if $K$ is a nonempty clopen set, there are $x\neq y\in K$. Then we can find a clopen subset $K'$ of $K$ with $x\in K'$ and $y\notin K'$. Thus $1_K = 1_{K'}+1_{K\setminus K'}$ is a decomposition into orthogonal idempotents. If follows that $R$ has no primitive idempotents and hence no indecomposable summands (as an indecomposable summand is of the form $eR$ with $e$ primitive).

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No, a free Boolean algebra $R$ on an infinite cardinal $\kappa$ (e.g., if $\kappa = \aleph_0$, $R$ is the Cantor algebra), is a commutative von Neumann regular ring which is not well complemented as an $R$-module.

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  • $\begingroup$ +1. But I can only accept one answer, and I'm accepting Benjamin Steinberg's answer since it came first and has a few more details. $\endgroup$ Jul 4 at 20:13
  • $\begingroup$ @SalvoTringali No worry, that's totally fair. $\endgroup$
    – Luc Guyot
    Jul 4 at 20:16

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