Does every strongly $\pi$-regular ring have artinian prime factors? A ring $R$ is called strongly $\pi$-regular if for every element $r \in R$ there exists an element $x \in R$ such that $r^{n+1}x = r^n$ for some positive integer $n$.  Meanwhile, a ring $R$ is said to have artinian prime factors if $R/P$ is artinian for every prime ideal $P$ of $R$.  It is known that every ring with artinian prime factors is strongly $\pi$-regular.  Does every strongly $\pi$-regular ring have artinian prime factors?
 A: No.
Note that a full matrix ring $M_n(k)$ over a field is simple, 
so its only prime ideal is $\{0\}$, and it is artinian.
Hence it has artinian prime factors; hence is strongly $\pi$-regular.
The latter property is preserved under passing to direct
limits, hence the union $R$ of a chain of such matrix rings,
say $M_2(k)\subset M_4(k)\subset\dots\subset M_{2^n}(k)\subset\dots$
(where $M_{2^n}(k)$ is embedded in $M_{2^{n+1}}(k)$ by sending each
matrix to the diagonal sum of two copies of itself) is strongly
$\pi$-regular.
This ring $R$ is again simple, but no longer artinian, so it does not
have artinian prime factors.
(A variant example is the $k$-algebra
$R$ of $\mathbb{N}\times\mathbb{N}$
matrices over $k$ spanned by the identity matrix and the matrices
$e_{ij}$.  This has two prime ideals, $\{0\}$ and the span of the
$e_{ij}$.  It can be shown to be strongly $\pi$-regular by
writing every element $r$ as the sum of an $n\times n$ matrix
for some $n$, and a scalar multiple of "the identity
$(\mathbb{N}-n)\times(\mathbb{N}-n)$ matrix", and using the fact
that $M_n(k)$ is artinian.  But again, for the prime ideal $P=\{0\}$,
the ring $R/P=R$ is non-artinian.)
