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

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 $$x = xyx$$.

• I think if you take an atomless boolean algebra the corresponding boolean ring is regular but has no primitive idempotents so no indecomposable summands. Jul 4, 2021 at 19:59

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).
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.