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](https://en.wikipedia.org/wiki/Uniform_module#Uniform_dimension_of_a_module) is well complemented.



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> **Question.** Is the regular right module $R\_R$ of a von Neumann regular ring $R$ well complemented?
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As a recall, a ring $R$ is *von Neumann regular* if, for every $x \in R$, there exists $y \in R$ such that $x = xyx$.



Every non-zero submodule of $R_R$ has an indecomposable direct summand: True when $R$ is von Neumann regular?
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).


