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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 $1$.

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I suspect Alfred Foster came up with an algebraic analysis of this about 70 or so years ago. It may have been a precursor to his work on the Chinese Remainder Theorem for large classes of algebras. I suspect compactness gives you the result you want, and that some chain condition is the algebraic condition needed. Gerhard "Ask Me About System Design" Paseman, 2012.02.06 – Gerhard Paseman Feb 6 '12 at 16:13
up vote 3 down vote accepted

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.

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Thank you very much. This is exactly what I wanted (and it is certainly a great result). – Niemi Feb 7 '12 at 10:28
You are welcome. – Benjamin Steinberg Feb 7 '12 at 21:10

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.

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However, I would be interested in a more direct proof. Does anybody have an idea? Moreover, this only works for commutative rings. Can someone give a counterexample for the noncommutative case? – Niemi Feb 6 '12 at 14:32
I think even for Boolean algebras a really direct proof not appealing to Peter-Weyl theorem is not known. Some simplifications based on work of Dikranian, Prodanov and Stoyanov are provided in "On the Proof that Compact Hausdorff Boolean Algebras are Powersets" by Bezhanishvili and Harding – მამუკა ჯიბლაძე Feb 7 at 17:58

Let $S$ be the set of all families $(e_j)$ of pairwise orthogonal idempotents such that $1=\sum_je_j$. Define a partial order on $S$ defined by $(e_j)\ge (f_i)$ iff for every $j$ there exists a $i$ such that $e_jf_i=e_j$. We want to apply Zorn's Lemma to $S$. Let $L\subset S$ be a linearly ordered subset. Consider a net $f_l$ of idempotents, indexed by $L$, with the condition that $f_l$ occurs in $l$ and that $f_{l'}f_l=f_{l'}$ if $l'\ge l$. As $R$ is compact, this net has a convergent subnet. The family $(e_j)$ consisting of all limits of all subnets of all nets of this form, constitutes an upper bound for $L$. Thus Zorn's Lemma gives you a maximal family $(e_j)$ of idempotents which must consist of primitives.

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I had exactly the same idea, but here is the problem that I had and that I do not see answered by your post: You are certainly correct that your constructed family is an upper bound w.r.t the partial order and that the elements are pairwise orthogonal. HOWEVER, the sum conidtion ∑…=1 must also be satisfied, and I do not see that issue addressed in your answer. In fact, that was precisely the point where I decided to post the question here on MO. If you could show that the sum condition holds you would have answered my question...and you would have helped me a lot. – Niemi Feb 5 '12 at 14:55
In fact, I actually doubt that the sum condition (which is the crux of the whole thing) is satisfied in the given construction. – Niemi Feb 5 '12 at 14:59
The sum condition is satisfied, because, suppose the sum $s$ is not one, then there must exist some $l\in L$ and some $f$ occurring in $l$ such that $sf\ne f$. Let $g=f-sf$. Next consider a net as above with $f_lg=f_l$. Its limit will not occur in $s$, a contradiction. – Anton Deitmar Feb 5 '12 at 18:17
I am not (yet) enterily convinced. You say "Next consider a net as above with flg=fl. Its limit will not occur in s, a contradiction." However, I do not see a reason why this limit has to be nonzero (if it is zero, you clearly do not obtain a contradiction). Could you shed some light on this? – Niemi Feb 6 '12 at 9:13
The more I think about this, the more sure I become that it simply does not work that way (by which I mean that one cannot ensure a nonzero limit). But, of course, I will would be happy to stand corrected. – Niemi Feb 6 '12 at 13:06

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