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In a ring $R$ with unity, every element can be written as product of finitely many idempotents. Can one show that the ring is commutative?

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closed as off-topic by user26857, Claude Leibovici, user91500, Daniel W. Farlow, C. Falcon Jul 11 '16 at 17:32

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Lemma 1: The only unit is the identity.

If $e$ is the leftmost idempotent in a factorization of a unit $u$, then $(1-e)u=0$ implies $1-e=0$ and $e=1$. After finitely many steps we have proven $u=1$.

Lemma 2: The ring is reduced, that is, it has no nonzero nilpotent elements. The only nilpotent element is $0$.

If $x$ is nilpotent, then $1-x$ is a unit, and that must be $1$ by the first lemma. Thus $x=0$.

Lemma 3: Idempotents in a reduced ring commute with all elements.

For any $r\in R$, both $er-ere$ and $re-ere$ square to zero, so they are both zero, and $er=ere=re$.

Proof of the original post:

everything is a product of idempotents, all of which commute with each other, so the ring is commutative. Moreover, every product of commuting idempotents is idempotent, so we can even conclude the ring is a Boolean ring.

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