Products of idempotents in separative regular rings. (Q2878806)

Let \(R\) be an associative ring with identity. The ring \(R\) is separative if the condition \(A\oplus A\cong A\oplus B\cong B\oplus B\) implies \(A\cong B\) for all finitely generated projective \(R\)-modules \(A\) and \(B\). The ring \(R\) is regular if for each element \(a\in R\) there exists an element \(b\in R\) such that \(a=aba\). If \(b\) can be chosen to be a unit in \(R\), then \(a\) is unit-regular. For \(a\in R\) let \(r(a)\) and \(l(a)\) be the right and the left annihilators in \(R\), respectively.NEWLINENEWLINE It is known [\textit{P. Ara, K. R. Goodearl, K. C. O'Meara} and \textit{E. Pardo}, Isr. J. Math. 105, 105-137 (1998; Zbl 0908.16002)] that a regular ring \(R\) is separative if and only if each \(a\in R\), satisfying the condition (*) \(Rr(a)=l(a)R=R(1-a)R\), is unit-regular in \(R\). \textit{J. Hannah} and the author [J. Algebra 123, No. 1, 223-239 (1989; Zbl 0673.16007)] showed that the condition (*) characterizes when an element \(a\) of a regular ring \(R\) is a product of idempotents in several important classes of regular rings, including unit-regular rings and right self-injective regular rings.NEWLINENEWLINE The present paper closes that gap by showing that the (*) characterization for products of idempotents holds exactly for separative regular rings. Consequently, there are no regular rings in which the characterization is known to fail.