TITLE: Does every compact Hausdorff ring admit a decomposition into primitive idempotents?
QUESTION [5 upvotes]: 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 $1$.

REPLY [2 votes]: I have just found an answer - at least for the commutative case - and it is YES.
Proof of 2.): Since $\mathbf{R}$ is a compact Hausdorff ring, it is profinite due to a surprising theorem in ["Profinite Groups" by Ribes and Zalesskii]. Another theorem in this book states that every profinite commutative ring is isomorphic to a product of profinite local rings. Thus, we can assume that $\mathbf{R} = \prod_{i \in I} \mathbf{R_i}$ for a family of profinite local rings $\mathbf{R_i}$ $(i \in I)$. For each $i \in I$, let $e_{i}$ denote the element of $\mathbf{R}$ given by $e_{i}(j) = \delta_{ij}$ for $j \in J$. As each of the $\mathbf{R_i}$'s is local, $E := \{ e_{i} \mid i \in I \}$ is a set of primitive idempotents in $\mathbf{R}$. Evidently, any two distinct elements of $E$ are orthogonal, and we have $1 = \sum_{i \in I} e_{i}$. This completes the proof.