Non-commutative abelian rings.

Question

Let $R$ be a ring with unity.

An element $e\in R$ is called idempotent if $e^2=e$. Clearly, $0,1$ are idempotents.

An element $e\in R$ is called central if $er=re$ for all $r\in R$.

Recall that a ring $R$ is called abelian if every idempotent is central.

Indeed, every commutative ring is abelian.

An example of noncommutative abelian ring is $k[x,y]$, where $k$ is any field and $xy\neq yx$. But the idempotents of $k[x,y]$ are the trivial idempotents $0$ and $1$.

My question is: Does there exist a noncommutative abelian ring with nontrivial idempotents?.

Thanks in advance.

Answer 1

Well, you could just use $\mathbb H\times \mathbb H$ where $\mathbb H$ is Hamilton's quaternions.

Answer 2

Let $R = F\langle x, y \rangle$ be the free algebra with two variables over any field. Then $R$ has no nontrivial ideompotent. The ring $R\times R$ has four idempotents $\{0, 1\}\times\{0, 1\}$, and all of them are central.