“Almost Boolean” rings are commutative

Posted: March 12, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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 $\mathbb{Z}/2\mathbb{Z}.$