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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

3 Answers 3

up vote 2 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

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
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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. –  doug 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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