If $e$ is idempotent and both $eRe$ and $(1-e)R(1-e)$ are orthogonally finite, is $R$ orthogonally finite? Let $R$ be a ring with $1$ possessing a non-zero idempotent $e$. It usually happens that if $eRe$ and $(1-e)R(1-e)$ have a property $P$ then so does $R$.
A ring is said to be orthogonally finite if it contains no infinite set of nonzero orthogonal idempotents.  I want to know whether $R$ is orthogonally finite if  $eRe$ and $(1-e)R(1-e)$ are both orthogonally finite.
Thanks for any cooperation!
 A: A counterexample occurring in nature (for some rather generous definition of "nature") comes from the fact that there are abelian groups $A$ that have direct sum decompositions both as a direct sum $B\oplus C$ of two indecomposable groups, and as the direct sum of infinitely many non-zero groups. If $R$ is the endomorphism ring of such a group $A$, and $e$ is projection onto $B$, then $R$ is a counterexample.
The existence of such groups was proved in the (one-and-a-half page) paper: A. L. S. Corner (1969). A note on rank and direct decompositions of torsion-free Abelian groups. II. Mathematical Proceedings of the Cambridge Philosophical Society, 66, pp 239-240. and can also be found in Fuchs' book(s) on Abelian Groups.)
A more contrived but "generic" example is the ring
$$R=\mathbb{Z}\langle e,e_1,e_2,\dots\vert e^2=e, e_i^2=e_i, e_ie_j=0 (i\neq j)\rangle.$$
It's not hard to show that $R$ has a $\mathbb{Z}$-basis consisting of the words in $e$ and the $e_i$ that alternate $e$s and $e_i$s (such as $e$, $e_1$, $ee_2e$, $e_1ee_2e$), including the empty word as the multiplicative identity, and that $eRe$ is the free ring generated by $\{ee_ie\vert i=1,2,\dots\}$, with multiplicative identity $e$, which has no idempotents other than $0$ and $e$. Similarly $(1-e)R(1-e)$ has no idempotents other than $0$ and $1-e$, but of course $R$ has the infinite set $\{e_1,e_2,\dots\}$ of orthogonal idempotents.
