“Almost Boolean” rings are commutative

Posted: March 12, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

It is easy to prove that if every element of a ring is idempotent, then the ring is commutative. This fact can be generalized as follows.

Problem. 1) Let R be a ring with identity and suppose that every element of R is a product of idempotent  elements. Prove that R is commutative.
2)  Give an example of a noncommutative ring with identity R such that every element of R is a product of some elements of the set \{r \in R: \ r^n=r, \ \text{for some} \ n \geq 2 \}.

Solution. 1) Obviously we only need to prove that every idempotent is central. Suppose first that ab = 1, for some a,b \in R. We claim that a = b = 1. So suppose the claim is false. Then a = e_1e_2 \cdots e_k, where e_j are idempotents and e_1 \neq 1. Let e = e_2 \cdots e_kb. Then e_1e = 1 and hence 1 - e_1 = (1 - e_1)e_1e = 0. Thus e_1 = 1. Contradiction! Now suppose that x^2 = 0, for some x \in R. Then (1 - x)(1 + x) = 1 and therefore x = 0, by what we just proved. Finally, since (ey-eye)^2=(ye-eye)^2=0 for any idempotent e \in R and any y \in R, we have ey = ye and so e is central.
2) One example is the ring of 2 \times 2 upper triangular matrices with entries from \mathbb{Z}/2\mathbb{Z}.

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