Let $\mathbf{R} = (R,\mathcal{T},+,\cdot,0,1)$ be a compact Hausdorff topological unitary ring, and consider the set $I(\mathbf{R}) := \{ e \in R \mid e \cdot e = e \}$ (of idempotents in $\mathbf{R}$). In the sequel, two idempotents $e,f \in I(\mathbf{R})$ are said to be *orthogonal* if $e \cdot f = f \cdot e = 0$, and an idempotent $e \in I(\mathbf{R})$ is called *primitive* if it is nonzero and cannot be written as a sum of two orthogonal nonzero idempotents. Now, my questions are the following:


1.) Does there have to exist a primitive idempotent in $\mathbf{R}$?


2.) Does $\mathbf{R}$ necessarily admit a set $E \subseteq I(\mathbf{R})$ of pairwise orthogonal, primitive idempotents such that $1 = \sum\_{e \in E} e$ (that is, the family $(e)\_{e \in E}$ is summable in $\mathbf{R}$ and the corresponding limit is $1$)?


If so, can you outline a proof or at least give a suitable reference? Otherwise, do you know any counterexamples? Are the statements true for the ring compactification of $(\mathbb{Z},\mathfrak{P}(\mathbb{Z}),+,\cdot,0,1)$?


Since all the rings I am dealing with are commutative, I am particularly interested in the commutative case. Personally, I expect a negative answer even in the commutative case, but that is just a feeling. Hence, I would be most enthuasiastic if somebody could provide me with a counterexample for the commutative case.


Remark: As $(e)\_{e \in E}$ is a family of pairwise orthogonal idempotents, it is summable, anyway. [see "Topological Rings Satisfying Compactness Conditions", page 139, by Mihail Ursul] Hence, in the second question, we only need to require that the limit value is $1$.



Does every compact Hausdorff ring admit a decomposition into primitive idempotents?
This is proved more generally for pseudo-compact rings by Gabriel in Gabriel, Pierre
Des catégories abéliennes, Bull. Soc. Math. France 90 1962 323–448; see Page 393 Corollaries 1 and 2. 


